Mixing
Proving $Imoperatorname{Li}_2(sqrt i(sqrt 2-1))=frac34G+frac18piln(sqrt2-1)$
$begingroup$
$newcommand{Li}{operatorname{Li}_2}$
I found, numerically, that $$ImLi(sqrt i(sqrt 2-1))=frac34G+frac18piln(sqrt2-1).$$
How can we prove it?
My attempt of proving this equation:
Using identity $$Li(x)=int_0^1frac{x}{xt-1}ln tdt,$$
we can deduce$$begin{align}ImLi(sqrt i(sqrt 2-1))&=frac1{2i}int_0^1left(frac{sqrt i(sqrt2-1)}{sqrt i(sqrt2-1)t-1}-frac{sqrt {-i}(sqrt2-1)}{sqrt {-i}(sqrt2-1)t-1}right)ln tdt\
&=int_0^1frac{2-sqrt{2}}{left(4 sqrt{2}-6right) t^2-2 left(sqrt{2}-2right) t-2}ln tdt\
&=int_0^{2-sqrt2}-frac{1}{u^2-2u+2}lnfrac u{2-sqrt 2}du\
&=frac18piln(2-sqrt2)-int_{-1}^{1-sqrt2}frac{ln(v+1)}{v^2+1}dv\
&=frac18piln(2-sqrt2)-int_{pi/8}^{pi/4}ln(1-tan x)dx\
&=frac18piln(sqrt2-1)-int_{pi/8}^{pi/4}lnsec x+lnsinleft(fracpi4-xright)dx\
end{align}$$
I have no idea how to deal with the log-trig integral.
calculus definite-integrals polylogarithm
asked Dec 7 '18 at 11:17
Kemono ChenKemono Chen
2,9701739
$endgroup$
|
show 1 more comment
$begingroup$
$newcommand{Li}{operatorname{Li}_2}$
I found, numerically, that $$ImLi(sqrt i(sqrt 2-1))=frac34G+frac18piln(sqrt2-1).$$
How can we prove it?
My attempt of proving this equation:
Using identity $$Li(x)=int_0^1frac{x}{xt-1}ln tdt,$$
we can deduce$$begin{align}ImLi(sqrt i(sqrt 2-1))&=frac1{2i}int_0^1left(frac{sqrt i(sqrt2-1)}{sqrt i(sqrt2-1)t-1}-frac{sqrt {-i}(sqrt2-1)}{sqrt {-i}(sqrt2-1)t-1}right)ln tdt\
&=int_0^1frac{2-sqrt{2}}{left(4 sqrt{2}-6right) t^2-2 left(sqrt{2}-2right) t-2}ln tdt\
&=int_0^{2-sqrt2}-frac{1}{u^2-2u+2}lnfrac u{2-sqrt 2}du\
&=frac18piln(2-sqrt2)-int_{-1}^{1-sqrt2}frac{ln(v+1)}{v^2+1}dv\
&=frac18piln(2-sqrt2)-int_{pi/8}^{pi/4}ln(1-tan x)dx\
&=frac18piln(sqrt2-1)-int_{pi/8}^{pi/4}lnsec x+lnsinleft(fracpi4-xright)dx\
end{align}$$
I have no idea how to deal with the log-trig integral.
calculus definite-integrals polylogarithm
asked Dec 7 '18 at 11:17
Kemono ChenKemono Chen
2,9701739
$endgroup$
$begingroup$
Try reducing the lower limit of your integral to zero and then use the identity ; $$G = int_0^{π/4}log cot theta dtheta $$
$endgroup$
– Awe Kumar Jha
Dec 7 '18 at 11:49
$begingroup$
$logsin x$ and $logcos x$ have well-known Fourier series, and you just have to perform a termwise integration of them.
$endgroup$
– Jack D'Aurizio
Dec 7 '18 at 17:06
$begingroup$
What is $sqrt{i}$?
$endgroup$
– FDP
Dec 7 '18 at 18:14
$begingroup$
@FDP $sqrt i= e^{pi i/4}=frac{1+i}{sqrt 2}$.
$endgroup$
– Kemono Chen
Dec 8 '18 at 0:55
$begingroup$
why not begin{align}sqrt i= -e^{pi i/4}?end{align} if $x_0^2=a$ then $(-x_0)^2=a$.
$endgroup$
– FDP
Dec 8 '18 at 11:30
|
show 1 more comment
$begingroup$
$newcommand{Li}{operatorname{Li}_2}$
I found, numerically, that $$ImLi(sqrt i(sqrt 2-1))=frac34G+frac18piln(sqrt2-1).$$
How can we prove it?
My attempt of proving this equation:
Using identity $$Li(x)=int_0^1frac{x}{xt-1}ln tdt,$$
we can deduce$$begin{align}ImLi(sqrt i(sqrt 2-1))&=frac1{2i}int_0^1left(frac{sqrt i(sqrt2-1)}{sqrt i(sqrt2-1)t-1}-frac{sqrt {-i}(sqrt2-1)}{sqrt {-i}(sqrt2-1)t-1}right)ln tdt\
&=int_0^1frac{2-sqrt{2}}{left(4 sqrt{2}-6right) t^2-2 left(sqrt{2}-2right) t-2}ln tdt\
&=int_0^{2-sqrt2}-frac{1}{u^2-2u+2}lnfrac u{2-sqrt 2}du\
&=frac18piln(2-sqrt2)-int_{-1}^{1-sqrt2}frac{ln(v+1)}{v^2+1}dv\
&=frac18piln(2-sqrt2)-int_{pi/8}^{pi/4}ln(1-tan x)dx\
&=frac18piln(sqrt2-1)-int_{pi/8}^{pi/4}lnsec x+lnsinleft(fracpi4-xright)dx\
end{align}$$
I have no idea how to deal with the log-trig integral.
calculus definite-integrals polylogarithm
asked Dec 7 '18 at 11:17
Kemono ChenKemono Chen
2,9701739
$endgroup$
$newcommand{Li}{operatorname{Li}_2}$
I found, numerically, that $$ImLi(sqrt i(sqrt 2-1))=frac34G+frac18piln(sqrt2-1).$$
How can we prove it?
My attempt of proving this equation:
Using identity $$Li(x)=int_0^1frac{x}{xt-1}ln tdt,$$
we can deduce$$begin{align}ImLi(sqrt i(sqrt 2-1))&=frac1{2i}int_0^1left(frac{sqrt i(sqrt2-1)}{sqrt i(sqrt2-1)t-1}-frac{sqrt {-i}(sqrt2-1)}{sqrt {-i}(sqrt2-1)t-1}right)ln tdt\
&=int_0^1frac{2-sqrt{2}}{left(4 sqrt{2}-6right) t^2-2 left(sqrt{2}-2right) t-2}ln tdt\
&=int_0^{2-sqrt2}-frac{1}{u^2-2u+2}lnfrac u{2-sqrt 2}du\
&=frac18piln(2-sqrt2)-int_{-1}^{1-sqrt2}frac{ln(v+1)}{v^2+1}dv\
&=frac18piln(2-sqrt2)-int_{pi/8}^{pi/4}ln(1-tan x)dx\
&=frac18piln(sqrt2-1)-int_{pi/8}^{pi/4}lnsec x+lnsinleft(fracpi4-xright)dx\
end{align}$$
I have no idea how to deal with the log-trig integral.
calculus definite-integrals polylogarithm
calculus definite-integrals polylogarithm
asked Dec 7 '18 at 11:17
Kemono ChenKemono Chen
2,9701739
asked Dec 7 '18 at 11:17
Kemono ChenKemono Chen
2,9701739
asked Dec 7 '18 at 11:17
Kemono ChenKemono Chen
2,9701739
asked Dec 7 '18 at 11:17
Kemono ChenKemono Chen
2,9701739
asked Dec 7 '18 at 11:17
Kemono ChenKemono Chen
2,9701739
2,9701739
$begingroup$
Try reducing the lower limit of your integral to zero and then use the identity ; $$G = int_0^{π/4}log cot theta dtheta $$
$endgroup$
– Awe Kumar Jha
Dec 7 '18 at 11:49
$begingroup$
$logsin x$ and $logcos x$ have well-known Fourier series, and you just have to perform a termwise integration of them.
$endgroup$
– Jack D'Aurizio
Dec 7 '18 at 17:06
$begingroup$
What is $sqrt{i}$?
$endgroup$
– FDP
Dec 7 '18 at 18:14
$begingroup$
@FDP $sqrt i= e^{pi i/4}=frac{1+i}{sqrt 2}$.
$endgroup$
– Kemono Chen
Dec 8 '18 at 0:55
$begingroup$
why not begin{align}sqrt i= -e^{pi i/4}?end{align} if $x_0^2=a$ then $(-x_0)^2=a$.
$endgroup$
– FDP
Dec 8 '18 at 11:30
|
show 1 more comment
$begingroup$
Try reducing the lower limit of your integral to zero and then use the identity ; $$G = int_0^{π/4}log cot theta dtheta $$
$endgroup$
– Awe Kumar Jha
Dec 7 '18 at 11:49
$begingroup$
$logsin x$ and $logcos x$ have well-known Fourier series, and you just have to perform a termwise integration of them.
$endgroup$
– Jack D'Aurizio
Dec 7 '18 at 17:06
$begingroup$
What is $sqrt{i}$?
$endgroup$
– FDP
Dec 7 '18 at 18:14
$begingroup$
@FDP $sqrt i= e^{pi i/4}=frac{1+i}{sqrt 2}$.
$endgroup$
– Kemono Chen
Dec 8 '18 at 0:55
$begingroup$
why not begin{align}sqrt i= -e^{pi i/4}?end{align} if $x_0^2=a$ then $(-x_0)^2=a$.
$endgroup$
– FDP
Dec 8 '18 at 11:30
$begingroup$
Try reducing the lower limit of your integral to zero and then use the identity ; $$G = int_0^{π/4}log cot theta dtheta $$
$endgroup$
– Awe Kumar Jha
Dec 7 '18 at 11:49
$begingroup$
Try reducing the lower limit of your integral to zero and then use the identity ; $$G = int_0^{π/4}log cot theta dtheta $$
$endgroup$
– Awe Kumar Jha
Dec 7 '18 at 11:49
$begingroup$
$logsin x$ and $logcos x$ have well-known Fourier series, and you just have to perform a termwise integration of them.
$endgroup$
– Jack D'Aurizio
Dec 7 '18 at 17:06
$begingroup$
$logsin x$ and $logcos x$ have well-known Fourier series, and you just have to perform a termwise integration of them.
$endgroup$
– Jack D'Aurizio
Dec 7 '18 at 17:06
$begingroup$
What is $sqrt{i}$?
$endgroup$
– FDP
Dec 7 '18 at 18:14
$begingroup$
What is $sqrt{i}$?
$endgroup$
– FDP
Dec 7 '18 at 18:14
$begingroup$
@FDP $sqrt i= e^{pi i/4}=frac{1+i}{sqrt 2}$.
$endgroup$
– Kemono Chen
Dec 8 '18 at 0:55
$begingroup$
@FDP $sqrt i= e^{pi i/4}=frac{1+i}{sqrt 2}$.
$endgroup$
– Kemono Chen
Dec 8 '18 at 0:55
$begingroup$
why not begin{align}sqrt i= -e^{pi i/4}?end{align} if $x_0^2=a$ then $(-x_0)^2=a$.
$endgroup$
– FDP
Dec 8 '18 at 11:30
$begingroup$
why not begin{align}sqrt i= -e^{pi i/4}?end{align} if $x_0^2=a$ then $(-x_0)^2=a$.
$endgroup$
– FDP
Dec 8 '18 at 11:30
|
show 1 more comment
1 Answer
1
active
oldest
votes
$begingroup$
$newcommand{Cl}{operatorname{Cl}}$In order to dodge the extensive usage of the Fourier Series Expansion I will use the Clausen Function $Cl_2(z)$ to shorten things up; nevertheless the result will remain the same as one could get going along the long way. Anyway the only difficulty that remains after your attempt is the evaluation of the following integral
$$mathfrak{I}~=~-int_{pi/8}^{pi/4}log(sec x)+logleft(sin fracpi4-xright)mathrm{d}xtag1$$
First of all I will do a bit of reshaping to actually apply useful formulae involving the Clausen Function. Therefore split up the first integral and do the substitution $x+fracpi4to x$ within the second integral to get
$$begin{align*}
mathfrak{I}&=-int_{pi/8}^{pi/4}log(sec x)+logleft(sin fracpi4-xright)mathrm{d}x\
&=int_0^{pi/4}log(cos x)mathrm{d}x-int_0^{pi/8}log(cos x)mathrm{d}x-int_{pi/8}^{pi/4}logleft(cos x+frac pi4right)mathrm{d}x\
&=int_0^{pi/4}log(cos x)mathrm{d}x-int_0^{pi/8}log(cos x)mathrm{d}x-int_{3pi/8}^{pi/2}log(cos x)mathrm{d}x\
&=int_0^{pi/4}log(cos x)mathrm{d}x-int_0^{pi/8}log(cos x)mathrm{d}x-int_0^{pi/2}log(cos x)mathrm{d}x+int_0^{3pi/8}log(cos x)mathrm{d}x
end{align*}$$
Now it is time to apply the first useful formula of the Clausen Function, namely
$$int_0^t log(cos x)mathrm{d}x~=~frac12Cl_2(pi-2t)-tlog(2)tag2$$
Formula $(2)$ can be shown rather easy be utilizing the well-know Fourier Series Expansion of $log(cos x)$ combined with the series representation of the function $Cl_2(z)$. However, with this knowledge we can rewrite the integrals from above in terms of the Clausen Function to get
$$begin{align*}
mathfrak{I}&=frac12Cl_2left(pi-2fracpi4right)-frac12Cl_2left(pi-2fracpi8right)-frac12Cl_2left(pi-2fracpi2right)+frac12Cl_2left(pi-2frac{3pi}8right)\
&~~~underbrace{-fracpi4log(2)+fracpi8log(2)+fracpi2log(2)-frac{3pi}8log(2)}_{=0}\
&=frac12left[Cl_2left(fracpi2right)-Cl_2left(0right)+Cl_2left(fracpi4right)-Cl_2left(frac{3pi}4right)right]
end{align*}$$
We are almost done! It is time to throw to more important formulae in: firstly the already mentioned series representation and secondly the so-called Duplication Formula
$$begin{align*}
Cl_2(z)~&=~sum_{n=1}^infty frac{sin(nz)}{n^2}tag3\
Cl_2(2z)~&=~2Cl_2(z)-2Cl_2(pi-z)tag4
end{align*}$$
From $(3)$ we can direcetly conclude that $Cl_2left(fracpi2right)=G$ and that $Cl_2(0)=0$ where $G$ denotes Catalan's Constant. Using the $(4)$ with $Z=fracpi4$ we get a representation for the other terms from above. Putting this all together gives us the final value
$$begin{align*}
mathfrak{I}&=frac12left[underbrace{Cl_2left(fracpi2right)}_{=G}-underbrace{Cl_2left(0right)}_{=0}+underbrace{Cl_2left(fracpi4right)-Cl_2left(frac{3pi}4right)}_{=frac G2}right]\
&=frac12left[G+frac G2right]
end{align*}$$
$$therefore~mathfrak{I}~=~-int_{pi/8}^{pi/4}log(sec x)+logleft(sin fracpi4-xright)mathrm{d}x~=~frac34 G$$
I recommend to study the Clausen Function hence it reduces the number of caculations needed for linear logarithmo-trigonometric integrals tremendously. If you are feeling uncomfortable with a part of the proof let me know and I will try to clear your doubts.
answered Jan 4 at 19:21
mrtaurhomrtaurho
4,16121234
$endgroup$
add a comment |
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1 Answer
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1 Answer
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$begingroup$
$newcommand{Cl}{operatorname{Cl}}$In order to dodge the extensive usage of the Fourier Series Expansion I will use the Clausen Function $Cl_2(z)$ to shorten things up; nevertheless the result will remain the same as one could get going along the long way. Anyway the only difficulty that remains after your attempt is the evaluation of the following integral
$$mathfrak{I}~=~-int_{pi/8}^{pi/4}log(sec x)+logleft(sin fracpi4-xright)mathrm{d}xtag1$$
First of all I will do a bit of reshaping to actually apply useful formulae involving the Clausen Function. Therefore split up the first integral and do the substitution $x+fracpi4to x$ within the second integral to get
$$begin{align*}
mathfrak{I}&=-int_{pi/8}^{pi/4}log(sec x)+logleft(sin fracpi4-xright)mathrm{d}x\
&=int_0^{pi/4}log(cos x)mathrm{d}x-int_0^{pi/8}log(cos x)mathrm{d}x-int_{pi/8}^{pi/4}logleft(cos x+frac pi4right)mathrm{d}x\
&=int_0^{pi/4}log(cos x)mathrm{d}x-int_0^{pi/8}log(cos x)mathrm{d}x-int_{3pi/8}^{pi/2}log(cos x)mathrm{d}x\
&=int_0^{pi/4}log(cos x)mathrm{d}x-int_0^{pi/8}log(cos x)mathrm{d}x-int_0^{pi/2}log(cos x)mathrm{d}x+int_0^{3pi/8}log(cos x)mathrm{d}x
end{align*}$$
Now it is time to apply the first useful formula of the Clausen Function, namely
$$int_0^t log(cos x)mathrm{d}x~=~frac12Cl_2(pi-2t)-tlog(2)tag2$$
Formula $(2)$ can be shown rather easy be utilizing the well-know Fourier Series Expansion of $log(cos x)$ combined with the series representation of the function $Cl_2(z)$. However, with this knowledge we can rewrite the integrals from above in terms of the Clausen Function to get
$$begin{align*}
mathfrak{I}&=frac12Cl_2left(pi-2fracpi4right)-frac12Cl_2left(pi-2fracpi8right)-frac12Cl_2left(pi-2fracpi2right)+frac12Cl_2left(pi-2frac{3pi}8right)\
&~~~underbrace{-fracpi4log(2)+fracpi8log(2)+fracpi2log(2)-frac{3pi}8log(2)}_{=0}\
&=frac12left[Cl_2left(fracpi2right)-Cl_2left(0right)+Cl_2left(fracpi4right)-Cl_2left(frac{3pi}4right)right]
end{align*}$$
We are almost done! It is time to throw to more important formulae in: firstly the already mentioned series representation and secondly the so-called Duplication Formula
$$begin{align*}
Cl_2(z)~&=~sum_{n=1}^infty frac{sin(nz)}{n^2}tag3\
Cl_2(2z)~&=~2Cl_2(z)-2Cl_2(pi-z)tag4
end{align*}$$
From $(3)$ we can direcetly conclude that $Cl_2left(fracpi2right)=G$ and that $Cl_2(0)=0$ where $G$ denotes Catalan's Constant. Using the $(4)$ with $Z=fracpi4$ we get a representation for the other terms from above. Putting this all together gives us the final value
$$begin{align*}
mathfrak{I}&=frac12left[underbrace{Cl_2left(fracpi2right)}_{=G}-underbrace{Cl_2left(0right)}_{=0}+underbrace{Cl_2left(fracpi4right)-Cl_2left(frac{3pi}4right)}_{=frac G2}right]\
&=frac12left[G+frac G2right]
end{align*}$$
$$therefore~mathfrak{I}~=~-int_{pi/8}^{pi/4}log(sec x)+logleft(sin fracpi4-xright)mathrm{d}x~=~frac34 G$$
I recommend to study the Clausen Function hence it reduces the number of caculations needed for linear logarithmo-trigonometric integrals tremendously. If you are feeling uncomfortable with a part of the proof let me know and I will try to clear your doubts.
answered Jan 4 at 19:21
mrtaurhomrtaurho
4,16121234
$endgroup$
add a comment |
$begingroup$
$newcommand{Cl}{operatorname{Cl}}$In order to dodge the extensive usage of the Fourier Series Expansion I will use the Clausen Function $Cl_2(z)$ to shorten things up; nevertheless the result will remain the same as one could get going along the long way. Anyway the only difficulty that remains after your attempt is the evaluation of the following integral
$$mathfrak{I}~=~-int_{pi/8}^{pi/4}log(sec x)+logleft(sin fracpi4-xright)mathrm{d}xtag1$$
First of all I will do a bit of reshaping to actually apply useful formulae involving the Clausen Function. Therefore split up the first integral and do the substitution $x+fracpi4to x$ within the second integral to get
$$begin{align*}
mathfrak{I}&=-int_{pi/8}^{pi/4}log(sec x)+logleft(sin fracpi4-xright)mathrm{d}x\
&=int_0^{pi/4}log(cos x)mathrm{d}x-int_0^{pi/8}log(cos x)mathrm{d}x-int_{pi/8}^{pi/4}logleft(cos x+frac pi4right)mathrm{d}x\
&=int_0^{pi/4}log(cos x)mathrm{d}x-int_0^{pi/8}log(cos x)mathrm{d}x-int_{3pi/8}^{pi/2}log(cos x)mathrm{d}x\
&=int_0^{pi/4}log(cos x)mathrm{d}x-int_0^{pi/8}log(cos x)mathrm{d}x-int_0^{pi/2}log(cos x)mathrm{d}x+int_0^{3pi/8}log(cos x)mathrm{d}x
end{align*}$$
Now it is time to apply the first useful formula of the Clausen Function, namely
$$int_0^t log(cos x)mathrm{d}x~=~frac12Cl_2(pi-2t)-tlog(2)tag2$$
Formula $(2)$ can be shown rather easy be utilizing the well-know Fourier Series Expansion of $log(cos x)$ combined with the series representation of the function $Cl_2(z)$. However, with this knowledge we can rewrite the integrals from above in terms of the Clausen Function to get
$$begin{align*}
mathfrak{I}&=frac12Cl_2left(pi-2fracpi4right)-frac12Cl_2left(pi-2fracpi8right)-frac12Cl_2left(pi-2fracpi2right)+frac12Cl_2left(pi-2frac{3pi}8right)\
&~~~underbrace{-fracpi4log(2)+fracpi8log(2)+fracpi2log(2)-frac{3pi}8log(2)}_{=0}\
&=frac12left[Cl_2left(fracpi2right)-Cl_2left(0right)+Cl_2left(fracpi4right)-Cl_2left(frac{3pi}4right)right]
end{align*}$$
We are almost done! It is time to throw to more important formulae in: firstly the already mentioned series representation and secondly the so-called Duplication Formula
$$begin{align*}
Cl_2(z)~&=~sum_{n=1}^infty frac{sin(nz)}{n^2}tag3\
Cl_2(2z)~&=~2Cl_2(z)-2Cl_2(pi-z)tag4
end{align*}$$
From $(3)$ we can direcetly conclude that $Cl_2left(fracpi2right)=G$ and that $Cl_2(0)=0$ where $G$ denotes Catalan's Constant. Using the $(4)$ with $Z=fracpi4$ we get a representation for the other terms from above. Putting this all together gives us the final value
$$begin{align*}
mathfrak{I}&=frac12left[underbrace{Cl_2left(fracpi2right)}_{=G}-underbrace{Cl_2left(0right)}_{=0}+underbrace{Cl_2left(fracpi4right)-Cl_2left(frac{3pi}4right)}_{=frac G2}right]\
&=frac12left[G+frac G2right]
end{align*}$$
$$therefore~mathfrak{I}~=~-int_{pi/8}^{pi/4}log(sec x)+logleft(sin fracpi4-xright)mathrm{d}x~=~frac34 G$$
I recommend to study the Clausen Function hence it reduces the number of caculations needed for linear logarithmo-trigonometric integrals tremendously. If you are feeling uncomfortable with a part of the proof let me know and I will try to clear your doubts.
answered Jan 4 at 19:21
mrtaurhomrtaurho
4,16121234
$endgroup$
add a comment |
$begingroup$
$newcommand{Cl}{operatorname{Cl}}$In order to dodge the extensive usage of the Fourier Series Expansion I will use the Clausen Function $Cl_2(z)$ to shorten things up; nevertheless the result will remain the same as one could get going along the long way. Anyway the only difficulty that remains after your attempt is the evaluation of the following integral
$$mathfrak{I}~=~-int_{pi/8}^{pi/4}log(sec x)+logleft(sin fracpi4-xright)mathrm{d}xtag1$$
First of all I will do a bit of reshaping to actually apply useful formulae involving the Clausen Function. Therefore split up the first integral and do the substitution $x+fracpi4to x$ within the second integral to get
$$begin{align*}
mathfrak{I}&=-int_{pi/8}^{pi/4}log(sec x)+logleft(sin fracpi4-xright)mathrm{d}x\
&=int_0^{pi/4}log(cos x)mathrm{d}x-int_0^{pi/8}log(cos x)mathrm{d}x-int_{pi/8}^{pi/4}logleft(cos x+frac pi4right)mathrm{d}x\
&=int_0^{pi/4}log(cos x)mathrm{d}x-int_0^{pi/8}log(cos x)mathrm{d}x-int_{3pi/8}^{pi/2}log(cos x)mathrm{d}x\
&=int_0^{pi/4}log(cos x)mathrm{d}x-int_0^{pi/8}log(cos x)mathrm{d}x-int_0^{pi/2}log(cos x)mathrm{d}x+int_0^{3pi/8}log(cos x)mathrm{d}x
end{align*}$$
Now it is time to apply the first useful formula of the Clausen Function, namely
$$int_0^t log(cos x)mathrm{d}x~=~frac12Cl_2(pi-2t)-tlog(2)tag2$$
Formula $(2)$ can be shown rather easy be utilizing the well-know Fourier Series Expansion of $log(cos x)$ combined with the series representation of the function $Cl_2(z)$. However, with this knowledge we can rewrite the integrals from above in terms of the Clausen Function to get
$$begin{align*}
mathfrak{I}&=frac12Cl_2left(pi-2fracpi4right)-frac12Cl_2left(pi-2fracpi8right)-frac12Cl_2left(pi-2fracpi2right)+frac12Cl_2left(pi-2frac{3pi}8right)\
&~~~underbrace{-fracpi4log(2)+fracpi8log(2)+fracpi2log(2)-frac{3pi}8log(2)}_{=0}\
&=frac12left[Cl_2left(fracpi2right)-Cl_2left(0right)+Cl_2left(fracpi4right)-Cl_2left(frac{3pi}4right)right]
end{align*}$$
We are almost done! It is time to throw to more important formulae in: firstly the already mentioned series representation and secondly the so-called Duplication Formula
$$begin{align*}
Cl_2(z)~&=~sum_{n=1}^infty frac{sin(nz)}{n^2}tag3\
Cl_2(2z)~&=~2Cl_2(z)-2Cl_2(pi-z)tag4
end{align*}$$
From $(3)$ we can direcetly conclude that $Cl_2left(fracpi2right)=G$ and that $Cl_2(0)=0$ where $G$ denotes Catalan's Constant. Using the $(4)$ with $Z=fracpi4$ we get a representation for the other terms from above. Putting this all together gives us the final value
$$begin{align*}
mathfrak{I}&=frac12left[underbrace{Cl_2left(fracpi2right)}_{=G}-underbrace{Cl_2left(0right)}_{=0}+underbrace{Cl_2left(fracpi4right)-Cl_2left(frac{3pi}4right)}_{=frac G2}right]\
&=frac12left[G+frac G2right]
end{align*}$$
$$therefore~mathfrak{I}~=~-int_{pi/8}^{pi/4}log(sec x)+logleft(sin fracpi4-xright)mathrm{d}x~=~frac34 G$$
I recommend to study the Clausen Function hence it reduces the number of caculations needed for linear logarithmo-trigonometric integrals tremendously. If you are feeling uncomfortable with a part of the proof let me know and I will try to clear your doubts.
answered Jan 4 at 19:21
mrtaurhomrtaurho
4,16121234
$endgroup$
$newcommand{Cl}{operatorname{Cl}}$In order to dodge the extensive usage of the Fourier Series Expansion I will use the Clausen Function $Cl_2(z)$ to shorten things up; nevertheless the result will remain the same as one could get going along the long way. Anyway the only difficulty that remains after your attempt is the evaluation of the following integral
$$mathfrak{I}~=~-int_{pi/8}^{pi/4}log(sec x)+logleft(sin fracpi4-xright)mathrm{d}xtag1$$
First of all I will do a bit of reshaping to actually apply useful formulae involving the Clausen Function. Therefore split up the first integral and do the substitution $x+fracpi4to x$ within the second integral to get
$$begin{align*}
mathfrak{I}&=-int_{pi/8}^{pi/4}log(sec x)+logleft(sin fracpi4-xright)mathrm{d}x\
&=int_0^{pi/4}log(cos x)mathrm{d}x-int_0^{pi/8}log(cos x)mathrm{d}x-int_{pi/8}^{pi/4}logleft(cos x+frac pi4right)mathrm{d}x\
&=int_0^{pi/4}log(cos x)mathrm{d}x-int_0^{pi/8}log(cos x)mathrm{d}x-int_{3pi/8}^{pi/2}log(cos x)mathrm{d}x\
&=int_0^{pi/4}log(cos x)mathrm{d}x-int_0^{pi/8}log(cos x)mathrm{d}x-int_0^{pi/2}log(cos x)mathrm{d}x+int_0^{3pi/8}log(cos x)mathrm{d}x
end{align*}$$
Now it is time to apply the first useful formula of the Clausen Function, namely
$$int_0^t log(cos x)mathrm{d}x~=~frac12Cl_2(pi-2t)-tlog(2)tag2$$
Formula $(2)$ can be shown rather easy be utilizing the well-know Fourier Series Expansion of $log(cos x)$ combined with the series representation of the function $Cl_2(z)$. However, with this knowledge we can rewrite the integrals from above in terms of the Clausen Function to get
$$begin{align*}
mathfrak{I}&=frac12Cl_2left(pi-2fracpi4right)-frac12Cl_2left(pi-2fracpi8right)-frac12Cl_2left(pi-2fracpi2right)+frac12Cl_2left(pi-2frac{3pi}8right)\
&~~~underbrace{-fracpi4log(2)+fracpi8log(2)+fracpi2log(2)-frac{3pi}8log(2)}_{=0}\
&=frac12left[Cl_2left(fracpi2right)-Cl_2left(0right)+Cl_2left(fracpi4right)-Cl_2left(frac{3pi}4right)right]
end{align*}$$
We are almost done! It is time to throw to more important formulae in: firstly the already mentioned series representation and secondly the so-called Duplication Formula
$$begin{align*}
Cl_2(z)~&=~sum_{n=1}^infty frac{sin(nz)}{n^2}tag3\
Cl_2(2z)~&=~2Cl_2(z)-2Cl_2(pi-z)tag4
end{align*}$$
From $(3)$ we can direcetly conclude that $Cl_2left(fracpi2right)=G$ and that $Cl_2(0)=0$ where $G$ denotes Catalan's Constant. Using the $(4)$ with $Z=fracpi4$ we get a representation for the other terms from above. Putting this all together gives us the final value
$$begin{align*}
mathfrak{I}&=frac12left[underbrace{Cl_2left(fracpi2right)}_{=G}-underbrace{Cl_2left(0right)}_{=0}+underbrace{Cl_2left(fracpi4right)-Cl_2left(frac{3pi}4right)}_{=frac G2}right]\
&=frac12left[G+frac G2right]
end{align*}$$
$$therefore~mathfrak{I}~=~-int_{pi/8}^{pi/4}log(sec x)+logleft(sin fracpi4-xright)mathrm{d}x~=~frac34 G$$
I recommend to study the Clausen Function hence it reduces the number of caculations needed for linear logarithmo-trigonometric integrals tremendously. If you are feeling uncomfortable with a part of the proof let me know and I will try to clear your doubts.
answered Jan 4 at 19:21
mrtaurhomrtaurho
4,16121234
edited Jan 4 at 19:57
answered Jan 4 at 19:21
mrtaurhomrtaurho
4,16121234
answered Jan 4 at 19:21
mrtaurhomrtaurho
4,16121234
answered Jan 4 at 19:21
mrtaurhomrtaurho
4,16121234
4,16121234
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$begingroup$
Try reducing the lower limit of your integral to zero and then use the identity ; $$G = int_0^{π/4}log cot theta dtheta $$
$endgroup$
– Awe Kumar Jha
Dec 7 '18 at 11:49
$begingroup$
$logsin x$ and $logcos x$ have well-known Fourier series, and you just have to perform a termwise integration of them.
$endgroup$
– Jack D'Aurizio
Dec 7 '18 at 17:06
$begingroup$
What is $sqrt{i}$?
$endgroup$
– FDP
Dec 7 '18 at 18:14
$begingroup$
@FDP $sqrt i= e^{pi i/4}=frac{1+i}{sqrt 2}$.
$endgroup$
– Kemono Chen
Dec 8 '18 at 0:55
$begingroup$
why not begin{align}sqrt i= -e^{pi i/4}?end{align} if $x_0^2=a$ then $(-x_0)^2=a$.
$endgroup$
– FDP
Dec 8 '18 at 11:30