Mixing
Rare Integral $int_0^1 frac{cosh left( alpha cos ^{-1}x right)cos left( alpha sinh ^{-1}x...
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I need to prove if this statement is true, some ideas?
$$int_0^1 frac{coshleft(alpha cos ^{-1}xright)cos left( alpha sinh^{-1} x right)}{sqrt{1-x^2}} , dx = frac pi 4 + frac 1 {2alpha }cdot sinh frac{alpha pi } 2$$
calculus integration definite-integrals
edited Sep 3 '17 at 13:12
pisco
11.9k21743
asked Aug 24 '17 at 16:46
whitexlotuswhitexlotus
37916
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add a comment |
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I need to prove if this statement is true, some ideas?
$$int_0^1 frac{coshleft(alpha cos ^{-1}xright)cos left( alpha sinh^{-1} x right)}{sqrt{1-x^2}} , dx = frac pi 4 + frac 1 {2alpha }cdot sinh frac{alpha pi } 2$$
calculus integration definite-integrals
edited Sep 3 '17 at 13:12
pisco
11.9k21743
asked Aug 24 '17 at 16:46
whitexlotuswhitexlotus
37916
$endgroup$
5
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Do you have some reason to think it's true? Did you derive it yourself or find it in some book or what?
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– Michael Hardy
Aug 24 '17 at 17:05
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By substituting $x=costheta$ we are left with $$ int_{0}^{pi/2}cosh(atheta)cosleft(alog(costheta+sqrt{1+cos^2theta})right),dtheta $$ with by parity equals $$ frac{1}{2}int_{-pi/2}^{pi/2}cosh(atheta),underbrace{cosleft(alog(costheta+sqrt{1+cos^2theta})right)}_{text{has the same value at }theta,,theta+pi,,theta+2pitext{ and so on}.},dtheta $$ Maybe this is relevant.
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– Jack D'Aurizio
Aug 24 '17 at 19:27
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@JackD'Aurizio I was too hasty and there was a serious flaw in my now-deleted-post. Apology for posting without checking. I'm pursuing another tact.
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– Mark Viola
Aug 24 '17 at 19:31
3
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@MarkViola: no need for apologies. For the OP: a bit of extra context would not be bad. Where does this integral come from?
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– Jack D'Aurizio
Aug 24 '17 at 20:53
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Hey, that's an integral I first posted on a site that was then called "integrals and series" in 2016: tapatalk.com/groups/integralsandseries/… It's even written in exactly the same format as I wrote it there.
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– nospoon
Jun 5 '18 at 19:49
add a comment |
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I need to prove if this statement is true, some ideas?
$$int_0^1 frac{coshleft(alpha cos ^{-1}xright)cos left( alpha sinh^{-1} x right)}{sqrt{1-x^2}} , dx = frac pi 4 + frac 1 {2alpha }cdot sinh frac{alpha pi } 2$$
calculus integration definite-integrals
edited Sep 3 '17 at 13:12
pisco
11.9k21743
asked Aug 24 '17 at 16:46
whitexlotuswhitexlotus
37916
$endgroup$
I need to prove if this statement is true, some ideas?
$$int_0^1 frac{coshleft(alpha cos ^{-1}xright)cos left( alpha sinh^{-1} x right)}{sqrt{1-x^2}} , dx = frac pi 4 + frac 1 {2alpha }cdot sinh frac{alpha pi } 2$$
calculus integration definite-integrals
calculus integration definite-integrals
edited Sep 3 '17 at 13:12
pisco
11.9k21743
asked Aug 24 '17 at 16:46
whitexlotuswhitexlotus
37916
edited Sep 3 '17 at 13:12
pisco
11.9k21743
asked Aug 24 '17 at 16:46
whitexlotuswhitexlotus
37916
edited Sep 3 '17 at 13:12
pisco
11.9k21743
edited Sep 3 '17 at 13:12
pisco
11.9k21743
edited Sep 3 '17 at 13:12
pisco
11.9k21743
11.9k21743
asked Aug 24 '17 at 16:46
whitexlotuswhitexlotus
37916
asked Aug 24 '17 at 16:46
whitexlotuswhitexlotus
37916
asked Aug 24 '17 at 16:46
whitexlotuswhitexlotus
37916
37916
5
$begingroup$
Do you have some reason to think it's true? Did you derive it yourself or find it in some book or what?
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– Michael Hardy
Aug 24 '17 at 17:05
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By substituting $x=costheta$ we are left with $$ int_{0}^{pi/2}cosh(atheta)cosleft(alog(costheta+sqrt{1+cos^2theta})right),dtheta $$ with by parity equals $$ frac{1}{2}int_{-pi/2}^{pi/2}cosh(atheta),underbrace{cosleft(alog(costheta+sqrt{1+cos^2theta})right)}_{text{has the same value at }theta,,theta+pi,,theta+2pitext{ and so on}.},dtheta $$ Maybe this is relevant.
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– Jack D'Aurizio
Aug 24 '17 at 19:27
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@JackD'Aurizio I was too hasty and there was a serious flaw in my now-deleted-post. Apology for posting without checking. I'm pursuing another tact.
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– Mark Viola
Aug 24 '17 at 19:31
3
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@MarkViola: no need for apologies. For the OP: a bit of extra context would not be bad. Where does this integral come from?
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– Jack D'Aurizio
Aug 24 '17 at 20:53
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Hey, that's an integral I first posted on a site that was then called "integrals and series" in 2016: tapatalk.com/groups/integralsandseries/… It's even written in exactly the same format as I wrote it there.
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– nospoon
Jun 5 '18 at 19:49
add a comment |
5
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Do you have some reason to think it's true? Did you derive it yourself or find it in some book or what?
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– Michael Hardy
Aug 24 '17 at 17:05
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By substituting $x=costheta$ we are left with $$ int_{0}^{pi/2}cosh(atheta)cosleft(alog(costheta+sqrt{1+cos^2theta})right),dtheta $$ with by parity equals $$ frac{1}{2}int_{-pi/2}^{pi/2}cosh(atheta),underbrace{cosleft(alog(costheta+sqrt{1+cos^2theta})right)}_{text{has the same value at }theta,,theta+pi,,theta+2pitext{ and so on}.},dtheta $$ Maybe this is relevant.
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– Jack D'Aurizio
Aug 24 '17 at 19:27
$begingroup$
@JackD'Aurizio I was too hasty and there was a serious flaw in my now-deleted-post. Apology for posting without checking. I'm pursuing another tact.
$endgroup$
– Mark Viola
Aug 24 '17 at 19:31
3
$begingroup$
@MarkViola: no need for apologies. For the OP: a bit of extra context would not be bad. Where does this integral come from?
$endgroup$
– Jack D'Aurizio
Aug 24 '17 at 20:53
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Hey, that's an integral I first posted on a site that was then called "integrals and series" in 2016: tapatalk.com/groups/integralsandseries/… It's even written in exactly the same format as I wrote it there.
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– nospoon
Jun 5 '18 at 19:49
5
5
$begingroup$
Do you have some reason to think it's true? Did you derive it yourself or find it in some book or what?
$endgroup$
– Michael Hardy
Aug 24 '17 at 17:05
$begingroup$
Do you have some reason to think it's true? Did you derive it yourself or find it in some book or what?
$endgroup$
– Michael Hardy
Aug 24 '17 at 17:05
$begingroup$
By substituting $x=costheta$ we are left with $$ int_{0}^{pi/2}cosh(atheta)cosleft(alog(costheta+sqrt{1+cos^2theta})right),dtheta $$ with by parity equals $$ frac{1}{2}int_{-pi/2}^{pi/2}cosh(atheta),underbrace{cosleft(alog(costheta+sqrt{1+cos^2theta})right)}_{text{has the same value at }theta,,theta+pi,,theta+2pitext{ and so on}.},dtheta $$ Maybe this is relevant.
$endgroup$
– Jack D'Aurizio
Aug 24 '17 at 19:27
$begingroup$
By substituting $x=costheta$ we are left with $$ int_{0}^{pi/2}cosh(atheta)cosleft(alog(costheta+sqrt{1+cos^2theta})right),dtheta $$ with by parity equals $$ frac{1}{2}int_{-pi/2}^{pi/2}cosh(atheta),underbrace{cosleft(alog(costheta+sqrt{1+cos^2theta})right)}_{text{has the same value at }theta,,theta+pi,,theta+2pitext{ and so on}.},dtheta $$ Maybe this is relevant.
$endgroup$
– Jack D'Aurizio
Aug 24 '17 at 19:27
$begingroup$
@JackD'Aurizio I was too hasty and there was a serious flaw in my now-deleted-post. Apology for posting without checking. I'm pursuing another tact.
$endgroup$
– Mark Viola
Aug 24 '17 at 19:31
$begingroup$
@JackD'Aurizio I was too hasty and there was a serious flaw in my now-deleted-post. Apology for posting without checking. I'm pursuing another tact.
$endgroup$
– Mark Viola
Aug 24 '17 at 19:31
3
3
$begingroup$
@MarkViola: no need for apologies. For the OP: a bit of extra context would not be bad. Where does this integral come from?
$endgroup$
– Jack D'Aurizio
Aug 24 '17 at 20:53
$begingroup$
@MarkViola: no need for apologies. For the OP: a bit of extra context would not be bad. Where does this integral come from?
$endgroup$
– Jack D'Aurizio
Aug 24 '17 at 20:53
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Hey, that's an integral I first posted on a site that was then called "integrals and series" in 2016: tapatalk.com/groups/integralsandseries/… It's even written in exactly the same format as I wrote it there.
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– nospoon
Jun 5 '18 at 19:49
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Hey, that's an integral I first posted on a site that was then called "integrals and series" in 2016: tapatalk.com/groups/integralsandseries/… It's even written in exactly the same format as I wrote it there.
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– nospoon
Jun 5 '18 at 19:49
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3 Answers
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This is not a trivial integral. It has a beautiful, yet subtle, symmetry which is the key for its elegant closed form. Let
$$
F(alpha) = int_0^1 frac{coshleft(alpha cos ^{-1}xright)cos left( alpha sinh^{-1} x right)}{sqrt{1-x^2}} dx$$
Note that $F(0) = pi/2$, and $F(alpha)$ is an even entire function, thus it suffices to prove that
$$F^{(2n)}(0) = frac{1}{{2(2n + 1)}}{(frac{pi }{2})^{2n + 1}} $$
for all $ngeq 1$.
The observation
$$F(alpha ) = frac{1}{2}int_{ - frac{pi }{2}}^{frac{pi }{2}} {cosh left[ {alpha left(x - iln (cos x + sqrt {1 + {{cos }^2}x} ) right)} right]dx} $$
implies that
$$F^{(2n)}(0) = frac{1}{2}int_{ - frac{pi }{2}}^{frac{pi }{2}} {{{left[ {x - iln (cos x + sqrt {1 + {{cos }^2}x} )} right]}^{2n}}dx} $$
Denote $$f(z) = frac{1}{z}{ln ^{2n}}left( {frac{{z + 1 + sqrt {{z^2} + 6z + 1} }}{2}} right)$$ where $ln$ is the principal branch of logarithm, and square root is the one with positive real part. Then $f(z)$ is holormophic in the unit disk except the slit from $-1$ to $-rho$, with $rho = 3-2sqrt{2} approx 0.17$, because $-rho$ is a root of the equation $z^2+6z+1=0$.
We integrate $f(z)$ with keyhole contour, with $-rho$ and $-1$ outside the contour. Around the unit circle $C$, we have $$begin{aligned}int_C {f(z)dz} &= 2iint_{ - pi/2}^{pi/2} {{{ln }^{2n}}left( {frac{{{e^{2ix}} + 1 + sqrt {{e^{4ix}} + 6{e^{2ix}} + 1} }}{2}} right)dx} \
& = 2iint_{ - pi/2}^{pi/2} {{{left[ {ix + ln (cos x + sqrt {1 + {{cos }^2}x} )} right]}^{2n}}dx} \
&= 4i{( - 1)^n}{F^{(2n)}}(0) end{aligned}$$
some algebraic manipulations are required at the 2nd line.
The integral above and below the branch cut is:
$$int_{ - 1}^{ - rho } {frac{1}{x}left[ {{{ln }^{2n}}left( {frac{{x + 1 + isqrt { - ({x^2} + 6x + 1)} }}{2}} right) - {{ln }^{2n}}left( {frac{{x + 1 - isqrt { - ({x^2} + 6x + 1)} }}{2}} right)} right]dx} $$
which says
$$-4i{( - 1)^n}{F^{(2n)}}(0) = underbrace{int_rho ^1 {frac{1}{x}{{ln }^{2n}}left( {frac{{1 - x - isqrt {6x - {x^2} - 1} }}{2}} right)dx}}_{I_1} - underbrace{ int_rho ^1 {frac{1}{x}{{ln }^{2n}}left( {frac{{1 - x + isqrt {6x - {x^2} - 1} }}{2}} right)dx}}_{I_2} $$
Now comes the miracle, when $x$ increases from $rho$ to $1$, the curve $ (1 - x pm isqrt {6x - {x^2} - 1})/2$ traces out a curve $Gamma$ in the right half plane. It passes through $(sqrt{2}-1,0)$ and the two ends are $(0,1)$ and $(0,-1)$. Note that
$$ u = frac{{1 - x pm isqrt {6x - {x^2} - 1} }}{2} implies x = frac{u(1-u)}{1+u} $$
thus $dx = (1-2u-u^2)/(1+u)^2$. Therefore
$$I_1 = int_{Gamma_1} {frac{{1 - 2u - {u^2}}}{{u(1 - u)(1 + u)}}{{ln }^{2n}}udu} $$
$$I_2 = color{red}{-}int_{Gamma_2} {frac{{1 - 2u - {u^2}}}{{u(1 - u)(1 + u)}}{{ln }^{2n}}udu} $$
where $Gamma_2$ is the portion of $Gamma$ lying in first quadrant, $Gamma_1$ is the portion lying in fourth quadrant, both clockwise.
Now we close $Gamma$ by adding a vertical segment from $i$ to $-i$, and a small indention to the right at the origin, denote these collectively as $Gamma_3$, with direction from $-i$ to $i$. Note that $$frac{{1 - 2u - {u^2}}}{{u(1 - u)(1 + u)}} = frac{1}{u} - frac{2}{(1+u)(1-u)}$$
with $$begin{aligned}
int_{Gamma_3} frac{ln^{2n} u}{u} du &= int_{C'} frac{ln^{2n} u}{u} du \
&= i int_{-pi/2}^{pi/2} ln^{2n} (e^{itheta}) dtheta \
&= i int_{-pi/2}^{pi/2} (itheta)^{2n} dtheta = frac{{2i{{( - 1)}^n}}}{{2n + 1}}{(frac{pi }{2})^{2n + 1}} end{aligned}$$
where $C'$ is the right portion of the unit circle.
Also $$int_{{Gamma_3}} {frac{{{{ln }^{2n}}u}}{{(1 - u)(1 + u)}}du} = iint_0^1 {frac{1}{{1 + {u^2}}}left[ {{{left( { - frac{pi }{2}i + ln u} right)}^{2n}} + {{left( {frac{pi }{2}i + ln u} right)}^{2n}}} right]du} $$
By integrating $${frac{{1 - 2z - {z^2}}}{{z(1 - z)(1 + z)}}{{ln }^{2z}}z}$$ around the contour formed by $Gamma_1, Gamma_2, Gamma_3$, we have
$$-2iunderbrace{int_0^1 {frac{1}{{1 + {u^2}}}left[ {{{left( { - frac{pi }{2}i + ln u} right)}^{2n}} + {{left( {frac{pi }{2}i + ln u} right)}^{2n}}} right]du}}_{J} + frac{{2i{{( - 1)}^n}}}{{2n + 1}}{(frac{pi }{2})^{2n + 1}} = 4i{( - 1)^n}{F^{(2n)}}(0)$$
The final step is showing $J=0$ when $ngeq 1$. Invoking the famous Euler numbers $E_{2k}$, we have
$$begin{aligned}
J &= 2sumlimits_{k = 0}^n {binom{2n}{2k}{{(frac{pi }{2}i)}^{2n - 2k}}color{blue}{int_0^1 frac{{{{ln }^{2k}}u}}{{1 + {u^2}}}du} } \
&= 2sumlimits_{k = 0}^n {binom{2n}{2k}{{(frac{pi }{2}i)}^{2n - 2k}}color{blue}{frac{1}{2}{{(frac{pi }{2})}^{2k + 1}}left| {{E_{2k}}} right|}} \
&= {(frac{pi }{2})^{2n + 1}}(2n)!sumlimits_{k = 0}^n {frac{{{(-1)^{n - k}}}}{{(2n - 2k)!}}frac{{left| {{E_{2k}}} right|}}{{(2k)!}}} end{aligned}$$
Since $|E_{2k}|/(2k)!$ are coefficient of $sec x$, the above sum is a Cauchy product between $sec x$ and $cos x$, hence it is $0$ when $ngeq 1$. The proof is finally completed.
answered Sep 2 '17 at 17:47
piscopisco
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Oh, I think I found a nasty trick. If we denote the integral in the LHS as $I(alpha)$, it should not be difficult to check that $f(alpha)=I(alpha)$ is an entire function in the complex variable $alpha$, and it should not be difficult to prove that its order is $1$. $f(0)=frac{pi}{2}$ is trivial and $f(z)=frac{pi}{4}$ at any $zin 2imathbb{Z}setminus{0}$ just a bit less. $f'(0)=0$ since $f$ is an even function, and
$$ f''(0)=int_{0}^{1}frac{arccos(x)^2-text{arcsinh}(x)^2}{sqrt{1-x^2}},dx =frac{pi^3}{24}-int_{0}^{pi/2}text{arcsinh}^2(sintheta),dtheta$$
equals:
$$ frac{pi^3}{24}-frac{1}{2}int_{0}^{pi/2}sum_{ngeq 1}frac{4^n (-1)^{n+1}(sintheta)^{2n}}{n^2binom{2n}{n}},dtheta=frac{pi^3}{24}-frac{pi^3}{48}=frac{pi^3}{48} $$
by the Taylor series of the squared arcsine and the well-known $int_{0}^{pi/2}(sintheta)^{2n},dtheta=frac{pi}{2cdot 4^n}binom{2n}{n}$, $sum_{ngeq 1}frac{(-1)^{n+1}}{n^2}=frac{pi^2}{12}$. So, long story short, if we manage to prove that $f(z)-tfrac{pi}{4}$ only vanishes at $2imathbb{Z}setminus{0}$, the claim follows from the Weierstrass product of the sine function and Herglotz' trick. As an alternative, we may try to compute every value of $f^{(2k)}(0)$, but that looks harder at first sight.
As a second alternative, we may just show that, by integration by parts, $f(alpha)-tfrac{pi}{4}$ is the solution of a differential equation of the form $ zcdot g(z) = frac{pi^2}{4}left(frac{d^2}{dz^2} z,g(z)right)$, then prove the statement by invoking the unicity part of the Cauchy-Lipschitz Theorem.
answered Aug 24 '17 at 20:05
Jack D'AurizioJack D'Aurizio
290k33284666
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Have you tried using the product-to-sum trigonometric formulas for complex values of the argument, in light of $cos(it)=cosh t$ ?
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– Lucian
Aug 30 '17 at 0:49
2
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Missing the "wow" upvote button :)
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– nbubis
Sep 3 '17 at 13:38
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Hey Jack, you can check my proof of this integral which I posted more than a year before this question was asked: tapatalk.com/groups/integralsandseries/… for some reason the TeX does not render but a chrome extension can fix that.
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– nospoon
Jun 5 '18 at 19:54
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Here I unscrambled the prove introduced by $textit{nospoon}$, which I really recommend for it only requires elementary resources
let $x=costheta$, notice that the integrated function is an even function
$$begin{aligned}
int_{0}^{1} {frac{cosh(zarccos x)cos(zoperatorname{arsinh} x)}{sqrt{1-x^2}} mathrm{d}x}
& = int_{0}^{pi/2} {cosh(ztheta)cos(zoperatorname{arsinh}(costheta)) mathrm{d}theta}\
& = frac1{2} int_{-pi/2}^{pi/2} {cosh(ztheta)cos(zoperatorname{arsinh}(costheta)) mathrm{d}theta}\
& = frac1{2} int_{-pi/2}^{pi/2} {e^{ztheta} cos(zoperatorname{arsinh}(costheta)) mathrm{d}theta}\
& quad (text{let } varphi=pi/2-theta)\
& = frac{e^{pi z/2}}{2} int_{0}^{pi} {e^{-zvarphi} cos(zoperatorname{arsinh}(sinvarphi)) mathrm{d}varphi}\
end{aligned}$$
we start with
$$f(x) = cos(zoperatorname{arsinh}x) = sum_{n=0}^{infty} {a_{n} x^{2n}}$$
easily get $a_{0}=1$ for $x=0$, from the derivative of $f(x)$ we find this equation
$$(x^{2}+1)f''(x)+xf'(x)+z^{2}f(x)=0$$
which indicates
$$(2n+2)(2n+1) a_{n+1} + (z^{2}+(2n)^{2}) a_{n} = 0$$
thus
$$a_{n} = frac{(-1)^{n}}{(2n)!} prod_{k=0}^{n-1} {(z^{2}+(2k)^{2})}$$
considering a well known integral $I_{n}(z) = int_{0}^{infty} {e^{-zvarphi} sin^{2n}!varphi mathrm{d}varphi}$
$$begin{aligned}
& int_{0}^{infty} {e^{-zvarphi} sin^{2n}!varphi mathrm{d}varphi}\
= & -frac{e^{-zvarphi}}{z} sin^{2n}!varphi bigr|_{varphi=0}^{infty} + frac{2n}{z} int_{0}^{infty} {e^{-zvarphi} sin^{2n-1}!varphi cosvarphi mathrm{d}varphi}\
= & -frac{2n}{z^{2}} sin^{2n-1}!varphi cosvarphi bigr|_{varphi=0}^{infty} + frac{2n(2n-1)}{z^{2}} int_{0}^{infty} {e^{-zvarphi} sin^{2n-2}!varphi cos^{2}!varphi mathrm{d}varphi} - frac{2n}{z^{2}} int_{0}^{infty} {e^{-zvarphi} sin^{2n}!varphi mathrm{d}varphi}\
= & frac{2n(2n-1)}{z^{2}} int_{0}^{infty} {e^{-zvarphi} sin^{2n-2}!varphi mathrm{d}varphi} - frac{(2n)^2}{z^{2}} int_{0}^{infty} {e^{-zvarphi} sin^{2n}!varphi mathrm{d}varphi}\
end{aligned}$$
which deduces the recurrence relation
$$I_{n}(z) = frac{2n(2n-1)}{z^{2}+(2n)^{2}}I_{n-1}(z)$$
with $I_{0}(z)=1/z$
$$I_{n}(z) = frac{(2n)!}{prod_{k=1}^{n} {(z^{2}+(2n)^{2})}} I_{0}(z) = frac{(2n)!}{zprod_{k=1}^{n} {(z^{2}+(2k)^{2})}}$$
thus we find this integral
$$begin{aligned}
int_{0}^{infty} {e^{-zvarphi} cos(zoperatorname{arsinh}(sinvarphi)) mathrm{d}varphi}
& = int_{0}^{infty} {e^{-zvarphi} sum_{n=0}^{infty}{a_{n}sin^{2n}!varphi} mathrm{d}varphi}\
& = sum_{n=0}^{infty} {a_{n} int_{0}^{infty} {e^{-zvarphi} sin^{2n}!varphi mathrm{d}varphi}}\
& = sum_{n=0}^{infty} {a_{n} I_{n}(z)}\
& = sum_{n=0}^{infty} {(-1)^{n}frac{z}{z^{2}+(2n)^{2}}}\
& = frac1{z} + frac1{4} sum_{n=1}^{infty} {(-1)^{n}frac{z}{(z/2)^{2}+(n)^{2}}}
end{aligned}$$
since (actually, this identity can be proven with the Herglotz trick, also suggested in other answers)
$$frac{pi}{sinpi z} = frac1{z} + sum_{n=1}^{infty} {(-1)^{n}frac{2z}{z^{2}-n^{2}}}$$
let $zto iz/2$ with $sinh(z)=isin(iz)$ we have
$$int_{0}^{infty} {e^{-zvarphi} cos(zoperatorname{arsinh}(sinvarphi)) mathrm{d}varphi} = frac1{2z} + frac1{4}frac{pi}{sinh(pi z/2)}$$
the last step is to write
$$begin{aligned}
int_{0}^{infty} {e^{-zvarphi} cos(zoperatorname{arsinh}(sinvarphi)) mathrm{d}varphi}
& = sum_{k=0}^{infty} {int_{kpi}^{(k+1)pi} {e^{-zvarphi} cos(zoperatorname{arsinh}(sinvarphi)) mathrm{d}varphi}}\
& = sum_{k=0}^{infty} {e^{-kpi z} int_{0}^{pi} {e^{-zvarphi} cos(zoperatorname{arsinh}(sinvarphi)) mathrm{d}varphi}}\
& = frac1{1-e^{-pi z}} int_{0}^{pi} {e^{-zvarphi} cos(zoperatorname{arsinh}(sinvarphi)) mathrm{d}varphi}\
& = frac{e^{pi z/2}}{2sinh(pi z/2)} int_{0}^{pi} {e^{-zvarphi} cos(zoperatorname{arsinh}(sinvarphi)) mathrm{d}varphi}
end{aligned}$$
where the original integral can be proven by times $sinh(pi z/2)$ in both side of the identity above
answered Jan 20 at 20:49
NanayajitzukiNanayajitzuki
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$begingroup$
This is not a trivial integral. It has a beautiful, yet subtle, symmetry which is the key for its elegant closed form. Let
$$
F(alpha) = int_0^1 frac{coshleft(alpha cos ^{-1}xright)cos left( alpha sinh^{-1} x right)}{sqrt{1-x^2}} dx$$
Note that $F(0) = pi/2$, and $F(alpha)$ is an even entire function, thus it suffices to prove that
$$F^{(2n)}(0) = frac{1}{{2(2n + 1)}}{(frac{pi }{2})^{2n + 1}} $$
for all $ngeq 1$.
The observation
$$F(alpha ) = frac{1}{2}int_{ - frac{pi }{2}}^{frac{pi }{2}} {cosh left[ {alpha left(x - iln (cos x + sqrt {1 + {{cos }^2}x} ) right)} right]dx} $$
implies that
$$F^{(2n)}(0) = frac{1}{2}int_{ - frac{pi }{2}}^{frac{pi }{2}} {{{left[ {x - iln (cos x + sqrt {1 + {{cos }^2}x} )} right]}^{2n}}dx} $$
Denote $$f(z) = frac{1}{z}{ln ^{2n}}left( {frac{{z + 1 + sqrt {{z^2} + 6z + 1} }}{2}} right)$$ where $ln$ is the principal branch of logarithm, and square root is the one with positive real part. Then $f(z)$ is holormophic in the unit disk except the slit from $-1$ to $-rho$, with $rho = 3-2sqrt{2} approx 0.17$, because $-rho$ is a root of the equation $z^2+6z+1=0$.
We integrate $f(z)$ with keyhole contour, with $-rho$ and $-1$ outside the contour. Around the unit circle $C$, we have $$begin{aligned}int_C {f(z)dz} &= 2iint_{ - pi/2}^{pi/2} {{{ln }^{2n}}left( {frac{{{e^{2ix}} + 1 + sqrt {{e^{4ix}} + 6{e^{2ix}} + 1} }}{2}} right)dx} \
& = 2iint_{ - pi/2}^{pi/2} {{{left[ {ix + ln (cos x + sqrt {1 + {{cos }^2}x} )} right]}^{2n}}dx} \
&= 4i{( - 1)^n}{F^{(2n)}}(0) end{aligned}$$
some algebraic manipulations are required at the 2nd line.
The integral above and below the branch cut is:
$$int_{ - 1}^{ - rho } {frac{1}{x}left[ {{{ln }^{2n}}left( {frac{{x + 1 + isqrt { - ({x^2} + 6x + 1)} }}{2}} right) - {{ln }^{2n}}left( {frac{{x + 1 - isqrt { - ({x^2} + 6x + 1)} }}{2}} right)} right]dx} $$
which says
$$-4i{( - 1)^n}{F^{(2n)}}(0) = underbrace{int_rho ^1 {frac{1}{x}{{ln }^{2n}}left( {frac{{1 - x - isqrt {6x - {x^2} - 1} }}{2}} right)dx}}_{I_1} - underbrace{ int_rho ^1 {frac{1}{x}{{ln }^{2n}}left( {frac{{1 - x + isqrt {6x - {x^2} - 1} }}{2}} right)dx}}_{I_2} $$
Now comes the miracle, when $x$ increases from $rho$ to $1$, the curve $ (1 - x pm isqrt {6x - {x^2} - 1})/2$ traces out a curve $Gamma$ in the right half plane. It passes through $(sqrt{2}-1,0)$ and the two ends are $(0,1)$ and $(0,-1)$. Note that
$$ u = frac{{1 - x pm isqrt {6x - {x^2} - 1} }}{2} implies x = frac{u(1-u)}{1+u} $$
thus $dx = (1-2u-u^2)/(1+u)^2$. Therefore
$$I_1 = int_{Gamma_1} {frac{{1 - 2u - {u^2}}}{{u(1 - u)(1 + u)}}{{ln }^{2n}}udu} $$
$$I_2 = color{red}{-}int_{Gamma_2} {frac{{1 - 2u - {u^2}}}{{u(1 - u)(1 + u)}}{{ln }^{2n}}udu} $$
where $Gamma_2$ is the portion of $Gamma$ lying in first quadrant, $Gamma_1$ is the portion lying in fourth quadrant, both clockwise.
Now we close $Gamma$ by adding a vertical segment from $i$ to $-i$, and a small indention to the right at the origin, denote these collectively as $Gamma_3$, with direction from $-i$ to $i$. Note that $$frac{{1 - 2u - {u^2}}}{{u(1 - u)(1 + u)}} = frac{1}{u} - frac{2}{(1+u)(1-u)}$$
with $$begin{aligned}
int_{Gamma_3} frac{ln^{2n} u}{u} du &= int_{C'} frac{ln^{2n} u}{u} du \
&= i int_{-pi/2}^{pi/2} ln^{2n} (e^{itheta}) dtheta \
&= i int_{-pi/2}^{pi/2} (itheta)^{2n} dtheta = frac{{2i{{( - 1)}^n}}}{{2n + 1}}{(frac{pi }{2})^{2n + 1}} end{aligned}$$
where $C'$ is the right portion of the unit circle.
Also $$int_{{Gamma_3}} {frac{{{{ln }^{2n}}u}}{{(1 - u)(1 + u)}}du} = iint_0^1 {frac{1}{{1 + {u^2}}}left[ {{{left( { - frac{pi }{2}i + ln u} right)}^{2n}} + {{left( {frac{pi }{2}i + ln u} right)}^{2n}}} right]du} $$
By integrating $${frac{{1 - 2z - {z^2}}}{{z(1 - z)(1 + z)}}{{ln }^{2z}}z}$$ around the contour formed by $Gamma_1, Gamma_2, Gamma_3$, we have
$$-2iunderbrace{int_0^1 {frac{1}{{1 + {u^2}}}left[ {{{left( { - frac{pi }{2}i + ln u} right)}^{2n}} + {{left( {frac{pi }{2}i + ln u} right)}^{2n}}} right]du}}_{J} + frac{{2i{{( - 1)}^n}}}{{2n + 1}}{(frac{pi }{2})^{2n + 1}} = 4i{( - 1)^n}{F^{(2n)}}(0)$$
The final step is showing $J=0$ when $ngeq 1$. Invoking the famous Euler numbers $E_{2k}$, we have
$$begin{aligned}
J &= 2sumlimits_{k = 0}^n {binom{2n}{2k}{{(frac{pi }{2}i)}^{2n - 2k}}color{blue}{int_0^1 frac{{{{ln }^{2k}}u}}{{1 + {u^2}}}du} } \
&= 2sumlimits_{k = 0}^n {binom{2n}{2k}{{(frac{pi }{2}i)}^{2n - 2k}}color{blue}{frac{1}{2}{{(frac{pi }{2})}^{2k + 1}}left| {{E_{2k}}} right|}} \
&= {(frac{pi }{2})^{2n + 1}}(2n)!sumlimits_{k = 0}^n {frac{{{(-1)^{n - k}}}}{{(2n - 2k)!}}frac{{left| {{E_{2k}}} right|}}{{(2k)!}}} end{aligned}$$
Since $|E_{2k}|/(2k)!$ are coefficient of $sec x$, the above sum is a Cauchy product between $sec x$ and $cos x$, hence it is $0$ when $ngeq 1$. The proof is finally completed.
answered Sep 2 '17 at 17:47
piscopisco
11.9k21743
$endgroup$
add a comment |
$begingroup$
This is not a trivial integral. It has a beautiful, yet subtle, symmetry which is the key for its elegant closed form. Let
$$
F(alpha) = int_0^1 frac{coshleft(alpha cos ^{-1}xright)cos left( alpha sinh^{-1} x right)}{sqrt{1-x^2}} dx$$
Note that $F(0) = pi/2$, and $F(alpha)$ is an even entire function, thus it suffices to prove that
$$F^{(2n)}(0) = frac{1}{{2(2n + 1)}}{(frac{pi }{2})^{2n + 1}} $$
for all $ngeq 1$.
The observation
$$F(alpha ) = frac{1}{2}int_{ - frac{pi }{2}}^{frac{pi }{2}} {cosh left[ {alpha left(x - iln (cos x + sqrt {1 + {{cos }^2}x} ) right)} right]dx} $$
implies that
$$F^{(2n)}(0) = frac{1}{2}int_{ - frac{pi }{2}}^{frac{pi }{2}} {{{left[ {x - iln (cos x + sqrt {1 + {{cos }^2}x} )} right]}^{2n}}dx} $$
Denote $$f(z) = frac{1}{z}{ln ^{2n}}left( {frac{{z + 1 + sqrt {{z^2} + 6z + 1} }}{2}} right)$$ where $ln$ is the principal branch of logarithm, and square root is the one with positive real part. Then $f(z)$ is holormophic in the unit disk except the slit from $-1$ to $-rho$, with $rho = 3-2sqrt{2} approx 0.17$, because $-rho$ is a root of the equation $z^2+6z+1=0$.
We integrate $f(z)$ with keyhole contour, with $-rho$ and $-1$ outside the contour. Around the unit circle $C$, we have $$begin{aligned}int_C {f(z)dz} &= 2iint_{ - pi/2}^{pi/2} {{{ln }^{2n}}left( {frac{{{e^{2ix}} + 1 + sqrt {{e^{4ix}} + 6{e^{2ix}} + 1} }}{2}} right)dx} \
& = 2iint_{ - pi/2}^{pi/2} {{{left[ {ix + ln (cos x + sqrt {1 + {{cos }^2}x} )} right]}^{2n}}dx} \
&= 4i{( - 1)^n}{F^{(2n)}}(0) end{aligned}$$
some algebraic manipulations are required at the 2nd line.
The integral above and below the branch cut is:
$$int_{ - 1}^{ - rho } {frac{1}{x}left[ {{{ln }^{2n}}left( {frac{{x + 1 + isqrt { - ({x^2} + 6x + 1)} }}{2}} right) - {{ln }^{2n}}left( {frac{{x + 1 - isqrt { - ({x^2} + 6x + 1)} }}{2}} right)} right]dx} $$
which says
$$-4i{( - 1)^n}{F^{(2n)}}(0) = underbrace{int_rho ^1 {frac{1}{x}{{ln }^{2n}}left( {frac{{1 - x - isqrt {6x - {x^2} - 1} }}{2}} right)dx}}_{I_1} - underbrace{ int_rho ^1 {frac{1}{x}{{ln }^{2n}}left( {frac{{1 - x + isqrt {6x - {x^2} - 1} }}{2}} right)dx}}_{I_2} $$
Now comes the miracle, when $x$ increases from $rho$ to $1$, the curve $ (1 - x pm isqrt {6x - {x^2} - 1})/2$ traces out a curve $Gamma$ in the right half plane. It passes through $(sqrt{2}-1,0)$ and the two ends are $(0,1)$ and $(0,-1)$. Note that
$$ u = frac{{1 - x pm isqrt {6x - {x^2} - 1} }}{2} implies x = frac{u(1-u)}{1+u} $$
thus $dx = (1-2u-u^2)/(1+u)^2$. Therefore
$$I_1 = int_{Gamma_1} {frac{{1 - 2u - {u^2}}}{{u(1 - u)(1 + u)}}{{ln }^{2n}}udu} $$
$$I_2 = color{red}{-}int_{Gamma_2} {frac{{1 - 2u - {u^2}}}{{u(1 - u)(1 + u)}}{{ln }^{2n}}udu} $$
where $Gamma_2$ is the portion of $Gamma$ lying in first quadrant, $Gamma_1$ is the portion lying in fourth quadrant, both clockwise.
Now we close $Gamma$ by adding a vertical segment from $i$ to $-i$, and a small indention to the right at the origin, denote these collectively as $Gamma_3$, with direction from $-i$ to $i$. Note that $$frac{{1 - 2u - {u^2}}}{{u(1 - u)(1 + u)}} = frac{1}{u} - frac{2}{(1+u)(1-u)}$$
with $$begin{aligned}
int_{Gamma_3} frac{ln^{2n} u}{u} du &= int_{C'} frac{ln^{2n} u}{u} du \
&= i int_{-pi/2}^{pi/2} ln^{2n} (e^{itheta}) dtheta \
&= i int_{-pi/2}^{pi/2} (itheta)^{2n} dtheta = frac{{2i{{( - 1)}^n}}}{{2n + 1}}{(frac{pi }{2})^{2n + 1}} end{aligned}$$
where $C'$ is the right portion of the unit circle.
Also $$int_{{Gamma_3}} {frac{{{{ln }^{2n}}u}}{{(1 - u)(1 + u)}}du} = iint_0^1 {frac{1}{{1 + {u^2}}}left[ {{{left( { - frac{pi }{2}i + ln u} right)}^{2n}} + {{left( {frac{pi }{2}i + ln u} right)}^{2n}}} right]du} $$
By integrating $${frac{{1 - 2z - {z^2}}}{{z(1 - z)(1 + z)}}{{ln }^{2z}}z}$$ around the contour formed by $Gamma_1, Gamma_2, Gamma_3$, we have
$$-2iunderbrace{int_0^1 {frac{1}{{1 + {u^2}}}left[ {{{left( { - frac{pi }{2}i + ln u} right)}^{2n}} + {{left( {frac{pi }{2}i + ln u} right)}^{2n}}} right]du}}_{J} + frac{{2i{{( - 1)}^n}}}{{2n + 1}}{(frac{pi }{2})^{2n + 1}} = 4i{( - 1)^n}{F^{(2n)}}(0)$$
The final step is showing $J=0$ when $ngeq 1$. Invoking the famous Euler numbers $E_{2k}$, we have
$$begin{aligned}
J &= 2sumlimits_{k = 0}^n {binom{2n}{2k}{{(frac{pi }{2}i)}^{2n - 2k}}color{blue}{int_0^1 frac{{{{ln }^{2k}}u}}{{1 + {u^2}}}du} } \
&= 2sumlimits_{k = 0}^n {binom{2n}{2k}{{(frac{pi }{2}i)}^{2n - 2k}}color{blue}{frac{1}{2}{{(frac{pi }{2})}^{2k + 1}}left| {{E_{2k}}} right|}} \
&= {(frac{pi }{2})^{2n + 1}}(2n)!sumlimits_{k = 0}^n {frac{{{(-1)^{n - k}}}}{{(2n - 2k)!}}frac{{left| {{E_{2k}}} right|}}{{(2k)!}}} end{aligned}$$
Since $|E_{2k}|/(2k)!$ are coefficient of $sec x$, the above sum is a Cauchy product between $sec x$ and $cos x$, hence it is $0$ when $ngeq 1$. The proof is finally completed.
answered Sep 2 '17 at 17:47
piscopisco
11.9k21743
$endgroup$
add a comment |
$begingroup$
This is not a trivial integral. It has a beautiful, yet subtle, symmetry which is the key for its elegant closed form. Let
$$
F(alpha) = int_0^1 frac{coshleft(alpha cos ^{-1}xright)cos left( alpha sinh^{-1} x right)}{sqrt{1-x^2}} dx$$
Note that $F(0) = pi/2$, and $F(alpha)$ is an even entire function, thus it suffices to prove that
$$F^{(2n)}(0) = frac{1}{{2(2n + 1)}}{(frac{pi }{2})^{2n + 1}} $$
for all $ngeq 1$.
The observation
$$F(alpha ) = frac{1}{2}int_{ - frac{pi }{2}}^{frac{pi }{2}} {cosh left[ {alpha left(x - iln (cos x + sqrt {1 + {{cos }^2}x} ) right)} right]dx} $$
implies that
$$F^{(2n)}(0) = frac{1}{2}int_{ - frac{pi }{2}}^{frac{pi }{2}} {{{left[ {x - iln (cos x + sqrt {1 + {{cos }^2}x} )} right]}^{2n}}dx} $$
Denote $$f(z) = frac{1}{z}{ln ^{2n}}left( {frac{{z + 1 + sqrt {{z^2} + 6z + 1} }}{2}} right)$$ where $ln$ is the principal branch of logarithm, and square root is the one with positive real part. Then $f(z)$ is holormophic in the unit disk except the slit from $-1$ to $-rho$, with $rho = 3-2sqrt{2} approx 0.17$, because $-rho$ is a root of the equation $z^2+6z+1=0$.
We integrate $f(z)$ with keyhole contour, with $-rho$ and $-1$ outside the contour. Around the unit circle $C$, we have $$begin{aligned}int_C {f(z)dz} &= 2iint_{ - pi/2}^{pi/2} {{{ln }^{2n}}left( {frac{{{e^{2ix}} + 1 + sqrt {{e^{4ix}} + 6{e^{2ix}} + 1} }}{2}} right)dx} \
& = 2iint_{ - pi/2}^{pi/2} {{{left[ {ix + ln (cos x + sqrt {1 + {{cos }^2}x} )} right]}^{2n}}dx} \
&= 4i{( - 1)^n}{F^{(2n)}}(0) end{aligned}$$
some algebraic manipulations are required at the 2nd line.
The integral above and below the branch cut is:
$$int_{ - 1}^{ - rho } {frac{1}{x}left[ {{{ln }^{2n}}left( {frac{{x + 1 + isqrt { - ({x^2} + 6x + 1)} }}{2}} right) - {{ln }^{2n}}left( {frac{{x + 1 - isqrt { - ({x^2} + 6x + 1)} }}{2}} right)} right]dx} $$
which says
$$-4i{( - 1)^n}{F^{(2n)}}(0) = underbrace{int_rho ^1 {frac{1}{x}{{ln }^{2n}}left( {frac{{1 - x - isqrt {6x - {x^2} - 1} }}{2}} right)dx}}_{I_1} - underbrace{ int_rho ^1 {frac{1}{x}{{ln }^{2n}}left( {frac{{1 - x + isqrt {6x - {x^2} - 1} }}{2}} right)dx}}_{I_2} $$
Now comes the miracle, when $x$ increases from $rho$ to $1$, the curve $ (1 - x pm isqrt {6x - {x^2} - 1})/2$ traces out a curve $Gamma$ in the right half plane. It passes through $(sqrt{2}-1,0)$ and the two ends are $(0,1)$ and $(0,-1)$. Note that
$$ u = frac{{1 - x pm isqrt {6x - {x^2} - 1} }}{2} implies x = frac{u(1-u)}{1+u} $$
thus $dx = (1-2u-u^2)/(1+u)^2$. Therefore
$$I_1 = int_{Gamma_1} {frac{{1 - 2u - {u^2}}}{{u(1 - u)(1 + u)}}{{ln }^{2n}}udu} $$
$$I_2 = color{red}{-}int_{Gamma_2} {frac{{1 - 2u - {u^2}}}{{u(1 - u)(1 + u)}}{{ln }^{2n}}udu} $$
where $Gamma_2$ is the portion of $Gamma$ lying in first quadrant, $Gamma_1$ is the portion lying in fourth quadrant, both clockwise.
Now we close $Gamma$ by adding a vertical segment from $i$ to $-i$, and a small indention to the right at the origin, denote these collectively as $Gamma_3$, with direction from $-i$ to $i$. Note that $$frac{{1 - 2u - {u^2}}}{{u(1 - u)(1 + u)}} = frac{1}{u} - frac{2}{(1+u)(1-u)}$$
with $$begin{aligned}
int_{Gamma_3} frac{ln^{2n} u}{u} du &= int_{C'} frac{ln^{2n} u}{u} du \
&= i int_{-pi/2}^{pi/2} ln^{2n} (e^{itheta}) dtheta \
&= i int_{-pi/2}^{pi/2} (itheta)^{2n} dtheta = frac{{2i{{( - 1)}^n}}}{{2n + 1}}{(frac{pi }{2})^{2n + 1}} end{aligned}$$
where $C'$ is the right portion of the unit circle.
Also $$int_{{Gamma_3}} {frac{{{{ln }^{2n}}u}}{{(1 - u)(1 + u)}}du} = iint_0^1 {frac{1}{{1 + {u^2}}}left[ {{{left( { - frac{pi }{2}i + ln u} right)}^{2n}} + {{left( {frac{pi }{2}i + ln u} right)}^{2n}}} right]du} $$
By integrating $${frac{{1 - 2z - {z^2}}}{{z(1 - z)(1 + z)}}{{ln }^{2z}}z}$$ around the contour formed by $Gamma_1, Gamma_2, Gamma_3$, we have
$$-2iunderbrace{int_0^1 {frac{1}{{1 + {u^2}}}left[ {{{left( { - frac{pi }{2}i + ln u} right)}^{2n}} + {{left( {frac{pi }{2}i + ln u} right)}^{2n}}} right]du}}_{J} + frac{{2i{{( - 1)}^n}}}{{2n + 1}}{(frac{pi }{2})^{2n + 1}} = 4i{( - 1)^n}{F^{(2n)}}(0)$$
The final step is showing $J=0$ when $ngeq 1$. Invoking the famous Euler numbers $E_{2k}$, we have
$$begin{aligned}
J &= 2sumlimits_{k = 0}^n {binom{2n}{2k}{{(frac{pi }{2}i)}^{2n - 2k}}color{blue}{int_0^1 frac{{{{ln }^{2k}}u}}{{1 + {u^2}}}du} } \
&= 2sumlimits_{k = 0}^n {binom{2n}{2k}{{(frac{pi }{2}i)}^{2n - 2k}}color{blue}{frac{1}{2}{{(frac{pi }{2})}^{2k + 1}}left| {{E_{2k}}} right|}} \
&= {(frac{pi }{2})^{2n + 1}}(2n)!sumlimits_{k = 0}^n {frac{{{(-1)^{n - k}}}}{{(2n - 2k)!}}frac{{left| {{E_{2k}}} right|}}{{(2k)!}}} end{aligned}$$
Since $|E_{2k}|/(2k)!$ are coefficient of $sec x$, the above sum is a Cauchy product between $sec x$ and $cos x$, hence it is $0$ when $ngeq 1$. The proof is finally completed.
answered Sep 2 '17 at 17:47
piscopisco
11.9k21743
$endgroup$
This is not a trivial integral. It has a beautiful, yet subtle, symmetry which is the key for its elegant closed form. Let
$$
F(alpha) = int_0^1 frac{coshleft(alpha cos ^{-1}xright)cos left( alpha sinh^{-1} x right)}{sqrt{1-x^2}} dx$$
Note that $F(0) = pi/2$, and $F(alpha)$ is an even entire function, thus it suffices to prove that
$$F^{(2n)}(0) = frac{1}{{2(2n + 1)}}{(frac{pi }{2})^{2n + 1}} $$
for all $ngeq 1$.
The observation
$$F(alpha ) = frac{1}{2}int_{ - frac{pi }{2}}^{frac{pi }{2}} {cosh left[ {alpha left(x - iln (cos x + sqrt {1 + {{cos }^2}x} ) right)} right]dx} $$
implies that
$$F^{(2n)}(0) = frac{1}{2}int_{ - frac{pi }{2}}^{frac{pi }{2}} {{{left[ {x - iln (cos x + sqrt {1 + {{cos }^2}x} )} right]}^{2n}}dx} $$
Denote $$f(z) = frac{1}{z}{ln ^{2n}}left( {frac{{z + 1 + sqrt {{z^2} + 6z + 1} }}{2}} right)$$ where $ln$ is the principal branch of logarithm, and square root is the one with positive real part. Then $f(z)$ is holormophic in the unit disk except the slit from $-1$ to $-rho$, with $rho = 3-2sqrt{2} approx 0.17$, because $-rho$ is a root of the equation $z^2+6z+1=0$.
We integrate $f(z)$ with keyhole contour, with $-rho$ and $-1$ outside the contour. Around the unit circle $C$, we have $$begin{aligned}int_C {f(z)dz} &= 2iint_{ - pi/2}^{pi/2} {{{ln }^{2n}}left( {frac{{{e^{2ix}} + 1 + sqrt {{e^{4ix}} + 6{e^{2ix}} + 1} }}{2}} right)dx} \
& = 2iint_{ - pi/2}^{pi/2} {{{left[ {ix + ln (cos x + sqrt {1 + {{cos }^2}x} )} right]}^{2n}}dx} \
&= 4i{( - 1)^n}{F^{(2n)}}(0) end{aligned}$$
some algebraic manipulations are required at the 2nd line.
The integral above and below the branch cut is:
$$int_{ - 1}^{ - rho } {frac{1}{x}left[ {{{ln }^{2n}}left( {frac{{x + 1 + isqrt { - ({x^2} + 6x + 1)} }}{2}} right) - {{ln }^{2n}}left( {frac{{x + 1 - isqrt { - ({x^2} + 6x + 1)} }}{2}} right)} right]dx} $$
which says
$$-4i{( - 1)^n}{F^{(2n)}}(0) = underbrace{int_rho ^1 {frac{1}{x}{{ln }^{2n}}left( {frac{{1 - x - isqrt {6x - {x^2} - 1} }}{2}} right)dx}}_{I_1} - underbrace{ int_rho ^1 {frac{1}{x}{{ln }^{2n}}left( {frac{{1 - x + isqrt {6x - {x^2} - 1} }}{2}} right)dx}}_{I_2} $$
Now comes the miracle, when $x$ increases from $rho$ to $1$, the curve $ (1 - x pm isqrt {6x - {x^2} - 1})/2$ traces out a curve $Gamma$ in the right half plane. It passes through $(sqrt{2}-1,0)$ and the two ends are $(0,1)$ and $(0,-1)$. Note that
$$ u = frac{{1 - x pm isqrt {6x - {x^2} - 1} }}{2} implies x = frac{u(1-u)}{1+u} $$
thus $dx = (1-2u-u^2)/(1+u)^2$. Therefore
$$I_1 = int_{Gamma_1} {frac{{1 - 2u - {u^2}}}{{u(1 - u)(1 + u)}}{{ln }^{2n}}udu} $$
$$I_2 = color{red}{-}int_{Gamma_2} {frac{{1 - 2u - {u^2}}}{{u(1 - u)(1 + u)}}{{ln }^{2n}}udu} $$
where $Gamma_2$ is the portion of $Gamma$ lying in first quadrant, $Gamma_1$ is the portion lying in fourth quadrant, both clockwise.
Now we close $Gamma$ by adding a vertical segment from $i$ to $-i$, and a small indention to the right at the origin, denote these collectively as $Gamma_3$, with direction from $-i$ to $i$. Note that $$frac{{1 - 2u - {u^2}}}{{u(1 - u)(1 + u)}} = frac{1}{u} - frac{2}{(1+u)(1-u)}$$
with $$begin{aligned}
int_{Gamma_3} frac{ln^{2n} u}{u} du &= int_{C'} frac{ln^{2n} u}{u} du \
&= i int_{-pi/2}^{pi/2} ln^{2n} (e^{itheta}) dtheta \
&= i int_{-pi/2}^{pi/2} (itheta)^{2n} dtheta = frac{{2i{{( - 1)}^n}}}{{2n + 1}}{(frac{pi }{2})^{2n + 1}} end{aligned}$$
where $C'$ is the right portion of the unit circle.
Also $$int_{{Gamma_3}} {frac{{{{ln }^{2n}}u}}{{(1 - u)(1 + u)}}du} = iint_0^1 {frac{1}{{1 + {u^2}}}left[ {{{left( { - frac{pi }{2}i + ln u} right)}^{2n}} + {{left( {frac{pi }{2}i + ln u} right)}^{2n}}} right]du} $$
By integrating $${frac{{1 - 2z - {z^2}}}{{z(1 - z)(1 + z)}}{{ln }^{2z}}z}$$ around the contour formed by $Gamma_1, Gamma_2, Gamma_3$, we have
$$-2iunderbrace{int_0^1 {frac{1}{{1 + {u^2}}}left[ {{{left( { - frac{pi }{2}i + ln u} right)}^{2n}} + {{left( {frac{pi }{2}i + ln u} right)}^{2n}}} right]du}}_{J} + frac{{2i{{( - 1)}^n}}}{{2n + 1}}{(frac{pi }{2})^{2n + 1}} = 4i{( - 1)^n}{F^{(2n)}}(0)$$
The final step is showing $J=0$ when $ngeq 1$. Invoking the famous Euler numbers $E_{2k}$, we have
$$begin{aligned}
J &= 2sumlimits_{k = 0}^n {binom{2n}{2k}{{(frac{pi }{2}i)}^{2n - 2k}}color{blue}{int_0^1 frac{{{{ln }^{2k}}u}}{{1 + {u^2}}}du} } \
&= 2sumlimits_{k = 0}^n {binom{2n}{2k}{{(frac{pi }{2}i)}^{2n - 2k}}color{blue}{frac{1}{2}{{(frac{pi }{2})}^{2k + 1}}left| {{E_{2k}}} right|}} \
&= {(frac{pi }{2})^{2n + 1}}(2n)!sumlimits_{k = 0}^n {frac{{{(-1)^{n - k}}}}{{(2n - 2k)!}}frac{{left| {{E_{2k}}} right|}}{{(2k)!}}} end{aligned}$$
Since $|E_{2k}|/(2k)!$ are coefficient of $sec x$, the above sum is a Cauchy product between $sec x$ and $cos x$, hence it is $0$ when $ngeq 1$. The proof is finally completed.
answered Sep 2 '17 at 17:47
piscopisco
11.9k21743
edited Jun 5 '18 at 16:12
answered Sep 2 '17 at 17:47
piscopisco
11.9k21743
answered Sep 2 '17 at 17:47
piscopisco
11.9k21743
answered Sep 2 '17 at 17:47
piscopisco
11.9k21743
11.9k21743
add a comment |
add a comment |
$begingroup$
Oh, I think I found a nasty trick. If we denote the integral in the LHS as $I(alpha)$, it should not be difficult to check that $f(alpha)=I(alpha)$ is an entire function in the complex variable $alpha$, and it should not be difficult to prove that its order is $1$. $f(0)=frac{pi}{2}$ is trivial and $f(z)=frac{pi}{4}$ at any $zin 2imathbb{Z}setminus{0}$ just a bit less. $f'(0)=0$ since $f$ is an even function, and
$$ f''(0)=int_{0}^{1}frac{arccos(x)^2-text{arcsinh}(x)^2}{sqrt{1-x^2}},dx =frac{pi^3}{24}-int_{0}^{pi/2}text{arcsinh}^2(sintheta),dtheta$$
equals:
$$ frac{pi^3}{24}-frac{1}{2}int_{0}^{pi/2}sum_{ngeq 1}frac{4^n (-1)^{n+1}(sintheta)^{2n}}{n^2binom{2n}{n}},dtheta=frac{pi^3}{24}-frac{pi^3}{48}=frac{pi^3}{48} $$
by the Taylor series of the squared arcsine and the well-known $int_{0}^{pi/2}(sintheta)^{2n},dtheta=frac{pi}{2cdot 4^n}binom{2n}{n}$, $sum_{ngeq 1}frac{(-1)^{n+1}}{n^2}=frac{pi^2}{12}$. So, long story short, if we manage to prove that $f(z)-tfrac{pi}{4}$ only vanishes at $2imathbb{Z}setminus{0}$, the claim follows from the Weierstrass product of the sine function and Herglotz' trick. As an alternative, we may try to compute every value of $f^{(2k)}(0)$, but that looks harder at first sight.
As a second alternative, we may just show that, by integration by parts, $f(alpha)-tfrac{pi}{4}$ is the solution of a differential equation of the form $ zcdot g(z) = frac{pi^2}{4}left(frac{d^2}{dz^2} z,g(z)right)$, then prove the statement by invoking the unicity part of the Cauchy-Lipschitz Theorem.
answered Aug 24 '17 at 20:05
Jack D'AurizioJack D'Aurizio
290k33284666
$endgroup$
$begingroup$
Have you tried using the product-to-sum trigonometric formulas for complex values of the argument, in light of $cos(it)=cosh t$ ?
$endgroup$
– Lucian
Aug 30 '17 at 0:49
2
$begingroup$
Missing the "wow" upvote button :)
$endgroup$
– nbubis
Sep 3 '17 at 13:38
$begingroup$
Hey Jack, you can check my proof of this integral which I posted more than a year before this question was asked: tapatalk.com/groups/integralsandseries/… for some reason the TeX does not render but a chrome extension can fix that.
$endgroup$
– nospoon
Jun 5 '18 at 19:54
add a comment |
$begingroup$
Oh, I think I found a nasty trick. If we denote the integral in the LHS as $I(alpha)$, it should not be difficult to check that $f(alpha)=I(alpha)$ is an entire function in the complex variable $alpha$, and it should not be difficult to prove that its order is $1$. $f(0)=frac{pi}{2}$ is trivial and $f(z)=frac{pi}{4}$ at any $zin 2imathbb{Z}setminus{0}$ just a bit less. $f'(0)=0$ since $f$ is an even function, and
$$ f''(0)=int_{0}^{1}frac{arccos(x)^2-text{arcsinh}(x)^2}{sqrt{1-x^2}},dx =frac{pi^3}{24}-int_{0}^{pi/2}text{arcsinh}^2(sintheta),dtheta$$
equals:
$$ frac{pi^3}{24}-frac{1}{2}int_{0}^{pi/2}sum_{ngeq 1}frac{4^n (-1)^{n+1}(sintheta)^{2n}}{n^2binom{2n}{n}},dtheta=frac{pi^3}{24}-frac{pi^3}{48}=frac{pi^3}{48} $$
by the Taylor series of the squared arcsine and the well-known $int_{0}^{pi/2}(sintheta)^{2n},dtheta=frac{pi}{2cdot 4^n}binom{2n}{n}$, $sum_{ngeq 1}frac{(-1)^{n+1}}{n^2}=frac{pi^2}{12}$. So, long story short, if we manage to prove that $f(z)-tfrac{pi}{4}$ only vanishes at $2imathbb{Z}setminus{0}$, the claim follows from the Weierstrass product of the sine function and Herglotz' trick. As an alternative, we may try to compute every value of $f^{(2k)}(0)$, but that looks harder at first sight.
As a second alternative, we may just show that, by integration by parts, $f(alpha)-tfrac{pi}{4}$ is the solution of a differential equation of the form $ zcdot g(z) = frac{pi^2}{4}left(frac{d^2}{dz^2} z,g(z)right)$, then prove the statement by invoking the unicity part of the Cauchy-Lipschitz Theorem.
answered Aug 24 '17 at 20:05
Jack D'AurizioJack D'Aurizio
290k33284666
$endgroup$
$begingroup$
Have you tried using the product-to-sum trigonometric formulas for complex values of the argument, in light of $cos(it)=cosh t$ ?
$endgroup$
– Lucian
Aug 30 '17 at 0:49
2
$begingroup$
Missing the "wow" upvote button :)
$endgroup$
– nbubis
Sep 3 '17 at 13:38
$begingroup$
Hey Jack, you can check my proof of this integral which I posted more than a year before this question was asked: tapatalk.com/groups/integralsandseries/… for some reason the TeX does not render but a chrome extension can fix that.
$endgroup$
– nospoon
Jun 5 '18 at 19:54
add a comment |
$begingroup$
Oh, I think I found a nasty trick. If we denote the integral in the LHS as $I(alpha)$, it should not be difficult to check that $f(alpha)=I(alpha)$ is an entire function in the complex variable $alpha$, and it should not be difficult to prove that its order is $1$. $f(0)=frac{pi}{2}$ is trivial and $f(z)=frac{pi}{4}$ at any $zin 2imathbb{Z}setminus{0}$ just a bit less. $f'(0)=0$ since $f$ is an even function, and
$$ f''(0)=int_{0}^{1}frac{arccos(x)^2-text{arcsinh}(x)^2}{sqrt{1-x^2}},dx =frac{pi^3}{24}-int_{0}^{pi/2}text{arcsinh}^2(sintheta),dtheta$$
equals:
$$ frac{pi^3}{24}-frac{1}{2}int_{0}^{pi/2}sum_{ngeq 1}frac{4^n (-1)^{n+1}(sintheta)^{2n}}{n^2binom{2n}{n}},dtheta=frac{pi^3}{24}-frac{pi^3}{48}=frac{pi^3}{48} $$
by the Taylor series of the squared arcsine and the well-known $int_{0}^{pi/2}(sintheta)^{2n},dtheta=frac{pi}{2cdot 4^n}binom{2n}{n}$, $sum_{ngeq 1}frac{(-1)^{n+1}}{n^2}=frac{pi^2}{12}$. So, long story short, if we manage to prove that $f(z)-tfrac{pi}{4}$ only vanishes at $2imathbb{Z}setminus{0}$, the claim follows from the Weierstrass product of the sine function and Herglotz' trick. As an alternative, we may try to compute every value of $f^{(2k)}(0)$, but that looks harder at first sight.
As a second alternative, we may just show that, by integration by parts, $f(alpha)-tfrac{pi}{4}$ is the solution of a differential equation of the form $ zcdot g(z) = frac{pi^2}{4}left(frac{d^2}{dz^2} z,g(z)right)$, then prove the statement by invoking the unicity part of the Cauchy-Lipschitz Theorem.
answered Aug 24 '17 at 20:05
Jack D'AurizioJack D'Aurizio
290k33284666
$endgroup$
Oh, I think I found a nasty trick. If we denote the integral in the LHS as $I(alpha)$, it should not be difficult to check that $f(alpha)=I(alpha)$ is an entire function in the complex variable $alpha$, and it should not be difficult to prove that its order is $1$. $f(0)=frac{pi}{2}$ is trivial and $f(z)=frac{pi}{4}$ at any $zin 2imathbb{Z}setminus{0}$ just a bit less. $f'(0)=0$ since $f$ is an even function, and
$$ f''(0)=int_{0}^{1}frac{arccos(x)^2-text{arcsinh}(x)^2}{sqrt{1-x^2}},dx =frac{pi^3}{24}-int_{0}^{pi/2}text{arcsinh}^2(sintheta),dtheta$$
equals:
$$ frac{pi^3}{24}-frac{1}{2}int_{0}^{pi/2}sum_{ngeq 1}frac{4^n (-1)^{n+1}(sintheta)^{2n}}{n^2binom{2n}{n}},dtheta=frac{pi^3}{24}-frac{pi^3}{48}=frac{pi^3}{48} $$
by the Taylor series of the squared arcsine and the well-known $int_{0}^{pi/2}(sintheta)^{2n},dtheta=frac{pi}{2cdot 4^n}binom{2n}{n}$, $sum_{ngeq 1}frac{(-1)^{n+1}}{n^2}=frac{pi^2}{12}$. So, long story short, if we manage to prove that $f(z)-tfrac{pi}{4}$ only vanishes at $2imathbb{Z}setminus{0}$, the claim follows from the Weierstrass product of the sine function and Herglotz' trick. As an alternative, we may try to compute every value of $f^{(2k)}(0)$, but that looks harder at first sight.
As a second alternative, we may just show that, by integration by parts, $f(alpha)-tfrac{pi}{4}$ is the solution of a differential equation of the form $ zcdot g(z) = frac{pi^2}{4}left(frac{d^2}{dz^2} z,g(z)right)$, then prove the statement by invoking the unicity part of the Cauchy-Lipschitz Theorem.
answered Aug 24 '17 at 20:05
Jack D'AurizioJack D'Aurizio
290k33284666
edited Aug 24 '17 at 20:20
answered Aug 24 '17 at 20:05
Jack D'AurizioJack D'Aurizio
290k33284666
answered Aug 24 '17 at 20:05
Jack D'AurizioJack D'Aurizio
290k33284666
answered Aug 24 '17 at 20:05
Jack D'AurizioJack D'Aurizio
290k33284666
290k33284666
$begingroup$
Have you tried using the product-to-sum trigonometric formulas for complex values of the argument, in light of $cos(it)=cosh t$ ?
$endgroup$
– Lucian
Aug 30 '17 at 0:49
2
$begingroup$
Missing the "wow" upvote button :)
$endgroup$
– nbubis
Sep 3 '17 at 13:38
$begingroup$
Hey Jack, you can check my proof of this integral which I posted more than a year before this question was asked: tapatalk.com/groups/integralsandseries/… for some reason the TeX does not render but a chrome extension can fix that.
$endgroup$
– nospoon
Jun 5 '18 at 19:54
add a comment |
$begingroup$
Have you tried using the product-to-sum trigonometric formulas for complex values of the argument, in light of $cos(it)=cosh t$ ?
$endgroup$
– Lucian
Aug 30 '17 at 0:49
2
$begingroup$
Missing the "wow" upvote button :)
$endgroup$
– nbubis
Sep 3 '17 at 13:38
$begingroup$
Hey Jack, you can check my proof of this integral which I posted more than a year before this question was asked: tapatalk.com/groups/integralsandseries/… for some reason the TeX does not render but a chrome extension can fix that.
$endgroup$
– nospoon
Jun 5 '18 at 19:54
$begingroup$
Have you tried using the product-to-sum trigonometric formulas for complex values of the argument, in light of $cos(it)=cosh t$ ?
$endgroup$
– Lucian
Aug 30 '17 at 0:49
$begingroup$
Have you tried using the product-to-sum trigonometric formulas for complex values of the argument, in light of $cos(it)=cosh t$ ?
$endgroup$
– Lucian
Aug 30 '17 at 0:49
2
2
$begingroup$
Missing the "wow" upvote button :)
$endgroup$
– nbubis
Sep 3 '17 at 13:38
$begingroup$
Missing the "wow" upvote button :)
$endgroup$
– nbubis
Sep 3 '17 at 13:38
$begingroup$
Hey Jack, you can check my proof of this integral which I posted more than a year before this question was asked: tapatalk.com/groups/integralsandseries/… for some reason the TeX does not render but a chrome extension can fix that.
$endgroup$
– nospoon
Jun 5 '18 at 19:54
$begingroup$
Hey Jack, you can check my proof of this integral which I posted more than a year before this question was asked: tapatalk.com/groups/integralsandseries/… for some reason the TeX does not render but a chrome extension can fix that.
$endgroup$
– nospoon
Jun 5 '18 at 19:54
add a comment |
$begingroup$
Here I unscrambled the prove introduced by $textit{nospoon}$, which I really recommend for it only requires elementary resources
let $x=costheta$, notice that the integrated function is an even function
$$begin{aligned}
int_{0}^{1} {frac{cosh(zarccos x)cos(zoperatorname{arsinh} x)}{sqrt{1-x^2}} mathrm{d}x}
& = int_{0}^{pi/2} {cosh(ztheta)cos(zoperatorname{arsinh}(costheta)) mathrm{d}theta}\
& = frac1{2} int_{-pi/2}^{pi/2} {cosh(ztheta)cos(zoperatorname{arsinh}(costheta)) mathrm{d}theta}\
& = frac1{2} int_{-pi/2}^{pi/2} {e^{ztheta} cos(zoperatorname{arsinh}(costheta)) mathrm{d}theta}\
& quad (text{let } varphi=pi/2-theta)\
& = frac{e^{pi z/2}}{2} int_{0}^{pi} {e^{-zvarphi} cos(zoperatorname{arsinh}(sinvarphi)) mathrm{d}varphi}\
end{aligned}$$
we start with
$$f(x) = cos(zoperatorname{arsinh}x) = sum_{n=0}^{infty} {a_{n} x^{2n}}$$
easily get $a_{0}=1$ for $x=0$, from the derivative of $f(x)$ we find this equation
$$(x^{2}+1)f''(x)+xf'(x)+z^{2}f(x)=0$$
which indicates
$$(2n+2)(2n+1) a_{n+1} + (z^{2}+(2n)^{2}) a_{n} = 0$$
thus
$$a_{n} = frac{(-1)^{n}}{(2n)!} prod_{k=0}^{n-1} {(z^{2}+(2k)^{2})}$$
considering a well known integral $I_{n}(z) = int_{0}^{infty} {e^{-zvarphi} sin^{2n}!varphi mathrm{d}varphi}$
$$begin{aligned}
& int_{0}^{infty} {e^{-zvarphi} sin^{2n}!varphi mathrm{d}varphi}\
= & -frac{e^{-zvarphi}}{z} sin^{2n}!varphi bigr|_{varphi=0}^{infty} + frac{2n}{z} int_{0}^{infty} {e^{-zvarphi} sin^{2n-1}!varphi cosvarphi mathrm{d}varphi}\
= & -frac{2n}{z^{2}} sin^{2n-1}!varphi cosvarphi bigr|_{varphi=0}^{infty} + frac{2n(2n-1)}{z^{2}} int_{0}^{infty} {e^{-zvarphi} sin^{2n-2}!varphi cos^{2}!varphi mathrm{d}varphi} - frac{2n}{z^{2}} int_{0}^{infty} {e^{-zvarphi} sin^{2n}!varphi mathrm{d}varphi}\
= & frac{2n(2n-1)}{z^{2}} int_{0}^{infty} {e^{-zvarphi} sin^{2n-2}!varphi mathrm{d}varphi} - frac{(2n)^2}{z^{2}} int_{0}^{infty} {e^{-zvarphi} sin^{2n}!varphi mathrm{d}varphi}\
end{aligned}$$
which deduces the recurrence relation
$$I_{n}(z) = frac{2n(2n-1)}{z^{2}+(2n)^{2}}I_{n-1}(z)$$
with $I_{0}(z)=1/z$
$$I_{n}(z) = frac{(2n)!}{prod_{k=1}^{n} {(z^{2}+(2n)^{2})}} I_{0}(z) = frac{(2n)!}{zprod_{k=1}^{n} {(z^{2}+(2k)^{2})}}$$
thus we find this integral
$$begin{aligned}
int_{0}^{infty} {e^{-zvarphi} cos(zoperatorname{arsinh}(sinvarphi)) mathrm{d}varphi}
& = int_{0}^{infty} {e^{-zvarphi} sum_{n=0}^{infty}{a_{n}sin^{2n}!varphi} mathrm{d}varphi}\
& = sum_{n=0}^{infty} {a_{n} int_{0}^{infty} {e^{-zvarphi} sin^{2n}!varphi mathrm{d}varphi}}\
& = sum_{n=0}^{infty} {a_{n} I_{n}(z)}\
& = sum_{n=0}^{infty} {(-1)^{n}frac{z}{z^{2}+(2n)^{2}}}\
& = frac1{z} + frac1{4} sum_{n=1}^{infty} {(-1)^{n}frac{z}{(z/2)^{2}+(n)^{2}}}
end{aligned}$$
since (actually, this identity can be proven with the Herglotz trick, also suggested in other answers)
$$frac{pi}{sinpi z} = frac1{z} + sum_{n=1}^{infty} {(-1)^{n}frac{2z}{z^{2}-n^{2}}}$$
let $zto iz/2$ with $sinh(z)=isin(iz)$ we have
$$int_{0}^{infty} {e^{-zvarphi} cos(zoperatorname{arsinh}(sinvarphi)) mathrm{d}varphi} = frac1{2z} + frac1{4}frac{pi}{sinh(pi z/2)}$$
the last step is to write
$$begin{aligned}
int_{0}^{infty} {e^{-zvarphi} cos(zoperatorname{arsinh}(sinvarphi)) mathrm{d}varphi}
& = sum_{k=0}^{infty} {int_{kpi}^{(k+1)pi} {e^{-zvarphi} cos(zoperatorname{arsinh}(sinvarphi)) mathrm{d}varphi}}\
& = sum_{k=0}^{infty} {e^{-kpi z} int_{0}^{pi} {e^{-zvarphi} cos(zoperatorname{arsinh}(sinvarphi)) mathrm{d}varphi}}\
& = frac1{1-e^{-pi z}} int_{0}^{pi} {e^{-zvarphi} cos(zoperatorname{arsinh}(sinvarphi)) mathrm{d}varphi}\
& = frac{e^{pi z/2}}{2sinh(pi z/2)} int_{0}^{pi} {e^{-zvarphi} cos(zoperatorname{arsinh}(sinvarphi)) mathrm{d}varphi}
end{aligned}$$
where the original integral can be proven by times $sinh(pi z/2)$ in both side of the identity above
answered Jan 20 at 20:49
NanayajitzukiNanayajitzuki
3185
$endgroup$
add a comment |
$begingroup$
Here I unscrambled the prove introduced by $textit{nospoon}$, which I really recommend for it only requires elementary resources
let $x=costheta$, notice that the integrated function is an even function
$$begin{aligned}
int_{0}^{1} {frac{cosh(zarccos x)cos(zoperatorname{arsinh} x)}{sqrt{1-x^2}} mathrm{d}x}
& = int_{0}^{pi/2} {cosh(ztheta)cos(zoperatorname{arsinh}(costheta)) mathrm{d}theta}\
& = frac1{2} int_{-pi/2}^{pi/2} {cosh(ztheta)cos(zoperatorname{arsinh}(costheta)) mathrm{d}theta}\
& = frac1{2} int_{-pi/2}^{pi/2} {e^{ztheta} cos(zoperatorname{arsinh}(costheta)) mathrm{d}theta}\
& quad (text{let } varphi=pi/2-theta)\
& = frac{e^{pi z/2}}{2} int_{0}^{pi} {e^{-zvarphi} cos(zoperatorname{arsinh}(sinvarphi)) mathrm{d}varphi}\
end{aligned}$$
we start with
$$f(x) = cos(zoperatorname{arsinh}x) = sum_{n=0}^{infty} {a_{n} x^{2n}}$$
easily get $a_{0}=1$ for $x=0$, from the derivative of $f(x)$ we find this equation
$$(x^{2}+1)f''(x)+xf'(x)+z^{2}f(x)=0$$
which indicates
$$(2n+2)(2n+1) a_{n+1} + (z^{2}+(2n)^{2}) a_{n} = 0$$
thus
$$a_{n} = frac{(-1)^{n}}{(2n)!} prod_{k=0}^{n-1} {(z^{2}+(2k)^{2})}$$
considering a well known integral $I_{n}(z) = int_{0}^{infty} {e^{-zvarphi} sin^{2n}!varphi mathrm{d}varphi}$
$$begin{aligned}
& int_{0}^{infty} {e^{-zvarphi} sin^{2n}!varphi mathrm{d}varphi}\
= & -frac{e^{-zvarphi}}{z} sin^{2n}!varphi bigr|_{varphi=0}^{infty} + frac{2n}{z} int_{0}^{infty} {e^{-zvarphi} sin^{2n-1}!varphi cosvarphi mathrm{d}varphi}\
= & -frac{2n}{z^{2}} sin^{2n-1}!varphi cosvarphi bigr|_{varphi=0}^{infty} + frac{2n(2n-1)}{z^{2}} int_{0}^{infty} {e^{-zvarphi} sin^{2n-2}!varphi cos^{2}!varphi mathrm{d}varphi} - frac{2n}{z^{2}} int_{0}^{infty} {e^{-zvarphi} sin^{2n}!varphi mathrm{d}varphi}\
= & frac{2n(2n-1)}{z^{2}} int_{0}^{infty} {e^{-zvarphi} sin^{2n-2}!varphi mathrm{d}varphi} - frac{(2n)^2}{z^{2}} int_{0}^{infty} {e^{-zvarphi} sin^{2n}!varphi mathrm{d}varphi}\
end{aligned}$$
which deduces the recurrence relation
$$I_{n}(z) = frac{2n(2n-1)}{z^{2}+(2n)^{2}}I_{n-1}(z)$$
with $I_{0}(z)=1/z$
$$I_{n}(z) = frac{(2n)!}{prod_{k=1}^{n} {(z^{2}+(2n)^{2})}} I_{0}(z) = frac{(2n)!}{zprod_{k=1}^{n} {(z^{2}+(2k)^{2})}}$$
thus we find this integral
$$begin{aligned}
int_{0}^{infty} {e^{-zvarphi} cos(zoperatorname{arsinh}(sinvarphi)) mathrm{d}varphi}
& = int_{0}^{infty} {e^{-zvarphi} sum_{n=0}^{infty}{a_{n}sin^{2n}!varphi} mathrm{d}varphi}\
& = sum_{n=0}^{infty} {a_{n} int_{0}^{infty} {e^{-zvarphi} sin^{2n}!varphi mathrm{d}varphi}}\
& = sum_{n=0}^{infty} {a_{n} I_{n}(z)}\
& = sum_{n=0}^{infty} {(-1)^{n}frac{z}{z^{2}+(2n)^{2}}}\
& = frac1{z} + frac1{4} sum_{n=1}^{infty} {(-1)^{n}frac{z}{(z/2)^{2}+(n)^{2}}}
end{aligned}$$
since (actually, this identity can be proven with the Herglotz trick, also suggested in other answers)
$$frac{pi}{sinpi z} = frac1{z} + sum_{n=1}^{infty} {(-1)^{n}frac{2z}{z^{2}-n^{2}}}$$
let $zto iz/2$ with $sinh(z)=isin(iz)$ we have
$$int_{0}^{infty} {e^{-zvarphi} cos(zoperatorname{arsinh}(sinvarphi)) mathrm{d}varphi} = frac1{2z} + frac1{4}frac{pi}{sinh(pi z/2)}$$
the last step is to write
$$begin{aligned}
int_{0}^{infty} {e^{-zvarphi} cos(zoperatorname{arsinh}(sinvarphi)) mathrm{d}varphi}
& = sum_{k=0}^{infty} {int_{kpi}^{(k+1)pi} {e^{-zvarphi} cos(zoperatorname{arsinh}(sinvarphi)) mathrm{d}varphi}}\
& = sum_{k=0}^{infty} {e^{-kpi z} int_{0}^{pi} {e^{-zvarphi} cos(zoperatorname{arsinh}(sinvarphi)) mathrm{d}varphi}}\
& = frac1{1-e^{-pi z}} int_{0}^{pi} {e^{-zvarphi} cos(zoperatorname{arsinh}(sinvarphi)) mathrm{d}varphi}\
& = frac{e^{pi z/2}}{2sinh(pi z/2)} int_{0}^{pi} {e^{-zvarphi} cos(zoperatorname{arsinh}(sinvarphi)) mathrm{d}varphi}
end{aligned}$$
where the original integral can be proven by times $sinh(pi z/2)$ in both side of the identity above
answered Jan 20 at 20:49
NanayajitzukiNanayajitzuki
3185
$endgroup$
add a comment |
$begingroup$
Here I unscrambled the prove introduced by $textit{nospoon}$, which I really recommend for it only requires elementary resources
let $x=costheta$, notice that the integrated function is an even function
$$begin{aligned}
int_{0}^{1} {frac{cosh(zarccos x)cos(zoperatorname{arsinh} x)}{sqrt{1-x^2}} mathrm{d}x}
& = int_{0}^{pi/2} {cosh(ztheta)cos(zoperatorname{arsinh}(costheta)) mathrm{d}theta}\
& = frac1{2} int_{-pi/2}^{pi/2} {cosh(ztheta)cos(zoperatorname{arsinh}(costheta)) mathrm{d}theta}\
& = frac1{2} int_{-pi/2}^{pi/2} {e^{ztheta} cos(zoperatorname{arsinh}(costheta)) mathrm{d}theta}\
& quad (text{let } varphi=pi/2-theta)\
& = frac{e^{pi z/2}}{2} int_{0}^{pi} {e^{-zvarphi} cos(zoperatorname{arsinh}(sinvarphi)) mathrm{d}varphi}\
end{aligned}$$
we start with
$$f(x) = cos(zoperatorname{arsinh}x) = sum_{n=0}^{infty} {a_{n} x^{2n}}$$
easily get $a_{0}=1$ for $x=0$, from the derivative of $f(x)$ we find this equation
$$(x^{2}+1)f''(x)+xf'(x)+z^{2}f(x)=0$$
which indicates
$$(2n+2)(2n+1) a_{n+1} + (z^{2}+(2n)^{2}) a_{n} = 0$$
thus
$$a_{n} = frac{(-1)^{n}}{(2n)!} prod_{k=0}^{n-1} {(z^{2}+(2k)^{2})}$$
considering a well known integral $I_{n}(z) = int_{0}^{infty} {e^{-zvarphi} sin^{2n}!varphi mathrm{d}varphi}$
$$begin{aligned}
& int_{0}^{infty} {e^{-zvarphi} sin^{2n}!varphi mathrm{d}varphi}\
= & -frac{e^{-zvarphi}}{z} sin^{2n}!varphi bigr|_{varphi=0}^{infty} + frac{2n}{z} int_{0}^{infty} {e^{-zvarphi} sin^{2n-1}!varphi cosvarphi mathrm{d}varphi}\
= & -frac{2n}{z^{2}} sin^{2n-1}!varphi cosvarphi bigr|_{varphi=0}^{infty} + frac{2n(2n-1)}{z^{2}} int_{0}^{infty} {e^{-zvarphi} sin^{2n-2}!varphi cos^{2}!varphi mathrm{d}varphi} - frac{2n}{z^{2}} int_{0}^{infty} {e^{-zvarphi} sin^{2n}!varphi mathrm{d}varphi}\
= & frac{2n(2n-1)}{z^{2}} int_{0}^{infty} {e^{-zvarphi} sin^{2n-2}!varphi mathrm{d}varphi} - frac{(2n)^2}{z^{2}} int_{0}^{infty} {e^{-zvarphi} sin^{2n}!varphi mathrm{d}varphi}\
end{aligned}$$
which deduces the recurrence relation
$$I_{n}(z) = frac{2n(2n-1)}{z^{2}+(2n)^{2}}I_{n-1}(z)$$
with $I_{0}(z)=1/z$
$$I_{n}(z) = frac{(2n)!}{prod_{k=1}^{n} {(z^{2}+(2n)^{2})}} I_{0}(z) = frac{(2n)!}{zprod_{k=1}^{n} {(z^{2}+(2k)^{2})}}$$
thus we find this integral
$$begin{aligned}
int_{0}^{infty} {e^{-zvarphi} cos(zoperatorname{arsinh}(sinvarphi)) mathrm{d}varphi}
& = int_{0}^{infty} {e^{-zvarphi} sum_{n=0}^{infty}{a_{n}sin^{2n}!varphi} mathrm{d}varphi}\
& = sum_{n=0}^{infty} {a_{n} int_{0}^{infty} {e^{-zvarphi} sin^{2n}!varphi mathrm{d}varphi}}\
& = sum_{n=0}^{infty} {a_{n} I_{n}(z)}\
& = sum_{n=0}^{infty} {(-1)^{n}frac{z}{z^{2}+(2n)^{2}}}\
& = frac1{z} + frac1{4} sum_{n=1}^{infty} {(-1)^{n}frac{z}{(z/2)^{2}+(n)^{2}}}
end{aligned}$$
since (actually, this identity can be proven with the Herglotz trick, also suggested in other answers)
$$frac{pi}{sinpi z} = frac1{z} + sum_{n=1}^{infty} {(-1)^{n}frac{2z}{z^{2}-n^{2}}}$$
let $zto iz/2$ with $sinh(z)=isin(iz)$ we have
$$int_{0}^{infty} {e^{-zvarphi} cos(zoperatorname{arsinh}(sinvarphi)) mathrm{d}varphi} = frac1{2z} + frac1{4}frac{pi}{sinh(pi z/2)}$$
the last step is to write
$$begin{aligned}
int_{0}^{infty} {e^{-zvarphi} cos(zoperatorname{arsinh}(sinvarphi)) mathrm{d}varphi}
& = sum_{k=0}^{infty} {int_{kpi}^{(k+1)pi} {e^{-zvarphi} cos(zoperatorname{arsinh}(sinvarphi)) mathrm{d}varphi}}\
& = sum_{k=0}^{infty} {e^{-kpi z} int_{0}^{pi} {e^{-zvarphi} cos(zoperatorname{arsinh}(sinvarphi)) mathrm{d}varphi}}\
& = frac1{1-e^{-pi z}} int_{0}^{pi} {e^{-zvarphi} cos(zoperatorname{arsinh}(sinvarphi)) mathrm{d}varphi}\
& = frac{e^{pi z/2}}{2sinh(pi z/2)} int_{0}^{pi} {e^{-zvarphi} cos(zoperatorname{arsinh}(sinvarphi)) mathrm{d}varphi}
end{aligned}$$
where the original integral can be proven by times $sinh(pi z/2)$ in both side of the identity above
answered Jan 20 at 20:49
NanayajitzukiNanayajitzuki
3185
$endgroup$
Here I unscrambled the prove introduced by $textit{nospoon}$, which I really recommend for it only requires elementary resources
let $x=costheta$, notice that the integrated function is an even function
$$begin{aligned}
int_{0}^{1} {frac{cosh(zarccos x)cos(zoperatorname{arsinh} x)}{sqrt{1-x^2}} mathrm{d}x}
& = int_{0}^{pi/2} {cosh(ztheta)cos(zoperatorname{arsinh}(costheta)) mathrm{d}theta}\
& = frac1{2} int_{-pi/2}^{pi/2} {cosh(ztheta)cos(zoperatorname{arsinh}(costheta)) mathrm{d}theta}\
& = frac1{2} int_{-pi/2}^{pi/2} {e^{ztheta} cos(zoperatorname{arsinh}(costheta)) mathrm{d}theta}\
& quad (text{let } varphi=pi/2-theta)\
& = frac{e^{pi z/2}}{2} int_{0}^{pi} {e^{-zvarphi} cos(zoperatorname{arsinh}(sinvarphi)) mathrm{d}varphi}\
end{aligned}$$
we start with
$$f(x) = cos(zoperatorname{arsinh}x) = sum_{n=0}^{infty} {a_{n} x^{2n}}$$
easily get $a_{0}=1$ for $x=0$, from the derivative of $f(x)$ we find this equation
$$(x^{2}+1)f''(x)+xf'(x)+z^{2}f(x)=0$$
which indicates
$$(2n+2)(2n+1) a_{n+1} + (z^{2}+(2n)^{2}) a_{n} = 0$$
thus
$$a_{n} = frac{(-1)^{n}}{(2n)!} prod_{k=0}^{n-1} {(z^{2}+(2k)^{2})}$$
considering a well known integral $I_{n}(z) = int_{0}^{infty} {e^{-zvarphi} sin^{2n}!varphi mathrm{d}varphi}$
$$begin{aligned}
& int_{0}^{infty} {e^{-zvarphi} sin^{2n}!varphi mathrm{d}varphi}\
= & -frac{e^{-zvarphi}}{z} sin^{2n}!varphi bigr|_{varphi=0}^{infty} + frac{2n}{z} int_{0}^{infty} {e^{-zvarphi} sin^{2n-1}!varphi cosvarphi mathrm{d}varphi}\
= & -frac{2n}{z^{2}} sin^{2n-1}!varphi cosvarphi bigr|_{varphi=0}^{infty} + frac{2n(2n-1)}{z^{2}} int_{0}^{infty} {e^{-zvarphi} sin^{2n-2}!varphi cos^{2}!varphi mathrm{d}varphi} - frac{2n}{z^{2}} int_{0}^{infty} {e^{-zvarphi} sin^{2n}!varphi mathrm{d}varphi}\
= & frac{2n(2n-1)}{z^{2}} int_{0}^{infty} {e^{-zvarphi} sin^{2n-2}!varphi mathrm{d}varphi} - frac{(2n)^2}{z^{2}} int_{0}^{infty} {e^{-zvarphi} sin^{2n}!varphi mathrm{d}varphi}\
end{aligned}$$
which deduces the recurrence relation
$$I_{n}(z) = frac{2n(2n-1)}{z^{2}+(2n)^{2}}I_{n-1}(z)$$
with $I_{0}(z)=1/z$
$$I_{n}(z) = frac{(2n)!}{prod_{k=1}^{n} {(z^{2}+(2n)^{2})}} I_{0}(z) = frac{(2n)!}{zprod_{k=1}^{n} {(z^{2}+(2k)^{2})}}$$
thus we find this integral
$$begin{aligned}
int_{0}^{infty} {e^{-zvarphi} cos(zoperatorname{arsinh}(sinvarphi)) mathrm{d}varphi}
& = int_{0}^{infty} {e^{-zvarphi} sum_{n=0}^{infty}{a_{n}sin^{2n}!varphi} mathrm{d}varphi}\
& = sum_{n=0}^{infty} {a_{n} int_{0}^{infty} {e^{-zvarphi} sin^{2n}!varphi mathrm{d}varphi}}\
& = sum_{n=0}^{infty} {a_{n} I_{n}(z)}\
& = sum_{n=0}^{infty} {(-1)^{n}frac{z}{z^{2}+(2n)^{2}}}\
& = frac1{z} + frac1{4} sum_{n=1}^{infty} {(-1)^{n}frac{z}{(z/2)^{2}+(n)^{2}}}
end{aligned}$$
since (actually, this identity can be proven with the Herglotz trick, also suggested in other answers)
$$frac{pi}{sinpi z} = frac1{z} + sum_{n=1}^{infty} {(-1)^{n}frac{2z}{z^{2}-n^{2}}}$$
let $zto iz/2$ with $sinh(z)=isin(iz)$ we have
$$int_{0}^{infty} {e^{-zvarphi} cos(zoperatorname{arsinh}(sinvarphi)) mathrm{d}varphi} = frac1{2z} + frac1{4}frac{pi}{sinh(pi z/2)}$$
the last step is to write
$$begin{aligned}
int_{0}^{infty} {e^{-zvarphi} cos(zoperatorname{arsinh}(sinvarphi)) mathrm{d}varphi}
& = sum_{k=0}^{infty} {int_{kpi}^{(k+1)pi} {e^{-zvarphi} cos(zoperatorname{arsinh}(sinvarphi)) mathrm{d}varphi}}\
& = sum_{k=0}^{infty} {e^{-kpi z} int_{0}^{pi} {e^{-zvarphi} cos(zoperatorname{arsinh}(sinvarphi)) mathrm{d}varphi}}\
& = frac1{1-e^{-pi z}} int_{0}^{pi} {e^{-zvarphi} cos(zoperatorname{arsinh}(sinvarphi)) mathrm{d}varphi}\
& = frac{e^{pi z/2}}{2sinh(pi z/2)} int_{0}^{pi} {e^{-zvarphi} cos(zoperatorname{arsinh}(sinvarphi)) mathrm{d}varphi}
end{aligned}$$
where the original integral can be proven by times $sinh(pi z/2)$ in both side of the identity above
answered Jan 20 at 20:49
NanayajitzukiNanayajitzuki
3185
edited Jan 20 at 21:06
answered Jan 20 at 20:49
NanayajitzukiNanayajitzuki
3185
answered Jan 20 at 20:49
NanayajitzukiNanayajitzuki
3185
answered Jan 20 at 20:49
NanayajitzukiNanayajitzuki
3185
3185
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5
$begingroup$
Do you have some reason to think it's true? Did you derive it yourself or find it in some book or what?
$endgroup$
– Michael Hardy
Aug 24 '17 at 17:05
$begingroup$
By substituting $x=costheta$ we are left with $$ int_{0}^{pi/2}cosh(atheta)cosleft(alog(costheta+sqrt{1+cos^2theta})right),dtheta $$ with by parity equals $$ frac{1}{2}int_{-pi/2}^{pi/2}cosh(atheta),underbrace{cosleft(alog(costheta+sqrt{1+cos^2theta})right)}_{text{has the same value at }theta,,theta+pi,,theta+2pitext{ and so on}.},dtheta $$ Maybe this is relevant.
$endgroup$
– Jack D'Aurizio
Aug 24 '17 at 19:27
$begingroup$
@JackD'Aurizio I was too hasty and there was a serious flaw in my now-deleted-post. Apology for posting without checking. I'm pursuing another tact.
$endgroup$
– Mark Viola
Aug 24 '17 at 19:31
3
$begingroup$
@MarkViola: no need for apologies. For the OP: a bit of extra context would not be bad. Where does this integral come from?
$endgroup$
– Jack D'Aurizio
Aug 24 '17 at 20:53
$begingroup$
Hey, that's an integral I first posted on a site that was then called "integrals and series" in 2016: tapatalk.com/groups/integralsandseries/… It's even written in exactly the same format as I wrote it there.
$endgroup$
– nospoon
Jun 5 '18 at 19:49