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How to prove that $int_{0}^{infty}ln^2(x)sin(x^2)dx=frac{1}{32}sqrt{frac{pi}{2}}(2gamma-pi+ln16)^2$
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Wolfram Alpha provides
$$int_{0}^{infty}ln^2(x)sin(x^2)dx=frac{1}{32}sqrt{frac{pi}{2}}(2gamma-pi+ln16)^2tag{1}$$
But I haven't figured out the way to verify this result.
I know Frullani's Integral
$$ln(x)= int_{0}^{infty}frac{e^{-t}-e^{-xt}}{t}dt$$
I also know
$$int_{0}^{infty}sin(x^2)~dx=frac{1}{2}int_{0}^{infty}x^{-1/2}sin(x)~dx$$
Then,
$$begin{align}
int_{0}^{infty}ln^2(x)sin(x^2)dx&=int_{0}^{infty}left(int_{0}^{infty}frac{e^{-t}-e^{-xt}}{t}dtright)left(int_{0}^{infty}frac{e^{-n}-e^{-xn}}{n}dnright)sin(x^2)~dx\
&=frac{1}{2}int_{0}^{infty}left(int_{0}^{infty}frac{e^{-t}-e^{-xt}}{t}dtright)left(int_{0}^{infty}frac{e^{-n}-e^{-xn}}{n}dnright)frac{sin(x)}{sqrt{x}}dx\
&=frac{1}{2}int_{0}^{infty}int_{0}^{infty}int_{0}^{infty}frac{e^{-t}-e^{-xt}}{t}frac{e^{-n}-e^{-xn}}{n}frac{sin(x)}{sqrt{x}}~dx~dn~dt\
&=frac{1}{2}int_{0}^{infty}frac{1}{t}int_{0}^{infty}frac{1}{n}int_{0}^{infty}(e^{-t}-e^{-xt})(e^{-n}-e^{-xn})frac{sin(x)}{sqrt{x}}~dx~dn~dt\
&=frac{1}{2}int_{0}^{infty}frac{1}{t}int_{0}^{infty}frac{1}{n}int_{0}^{infty}(e^{-t-n}-e^{-xn-t}-e^{-xt-n}+e^{-xt-xn})frac{sin(x)}{sqrt{x}}~dx~dn~dt
end{align}$$
What should I do next?
There is also a general case
$$int_{0}^{infty}ln^2(x^a)sin(x^2)dx=frac{a^2}{32}sqrt{frac{pi}{2}}(2gamma-pi+ln16)^2tag{2}$$
But I think $(2)$ becomes easy to prove if we can prove $(1)$.
calculus integration
edited Jan 26 at 20:18
user
5,65111031
asked Jan 26 at 19:48
LarryLarry
2,53531131
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show 2 more comments
$begingroup$
Wolfram Alpha provides
$$int_{0}^{infty}ln^2(x)sin(x^2)dx=frac{1}{32}sqrt{frac{pi}{2}}(2gamma-pi+ln16)^2tag{1}$$
But I haven't figured out the way to verify this result.
I know Frullani's Integral
$$ln(x)= int_{0}^{infty}frac{e^{-t}-e^{-xt}}{t}dt$$
I also know
$$int_{0}^{infty}sin(x^2)~dx=frac{1}{2}int_{0}^{infty}x^{-1/2}sin(x)~dx$$
Then,
$$begin{align}
int_{0}^{infty}ln^2(x)sin(x^2)dx&=int_{0}^{infty}left(int_{0}^{infty}frac{e^{-t}-e^{-xt}}{t}dtright)left(int_{0}^{infty}frac{e^{-n}-e^{-xn}}{n}dnright)sin(x^2)~dx\
&=frac{1}{2}int_{0}^{infty}left(int_{0}^{infty}frac{e^{-t}-e^{-xt}}{t}dtright)left(int_{0}^{infty}frac{e^{-n}-e^{-xn}}{n}dnright)frac{sin(x)}{sqrt{x}}dx\
&=frac{1}{2}int_{0}^{infty}int_{0}^{infty}int_{0}^{infty}frac{e^{-t}-e^{-xt}}{t}frac{e^{-n}-e^{-xn}}{n}frac{sin(x)}{sqrt{x}}~dx~dn~dt\
&=frac{1}{2}int_{0}^{infty}frac{1}{t}int_{0}^{infty}frac{1}{n}int_{0}^{infty}(e^{-t}-e^{-xt})(e^{-n}-e^{-xn})frac{sin(x)}{sqrt{x}}~dx~dn~dt\
&=frac{1}{2}int_{0}^{infty}frac{1}{t}int_{0}^{infty}frac{1}{n}int_{0}^{infty}(e^{-t-n}-e^{-xn-t}-e^{-xt-n}+e^{-xt-xn})frac{sin(x)}{sqrt{x}}~dx~dn~dt
end{align}$$
What should I do next?
There is also a general case
$$int_{0}^{infty}ln^2(x^a)sin(x^2)dx=frac{a^2}{32}sqrt{frac{pi}{2}}(2gamma-pi+ln16)^2tag{2}$$
But I think $(2)$ becomes easy to prove if we can prove $(1)$.
calculus integration
edited Jan 26 at 20:18
user
5,65111031
asked Jan 26 at 19:48
LarryLarry
2,53531131
$endgroup$
$begingroup$
Computer algebra solves this instantly: $frac{1}{32} sqrt{frac{pi }{2}} (2 gamma -pi +log (16))^2$. Is there really a need to perform this by hand? ted.com/talks/…
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– David G. Stork
Jan 26 at 20:10
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It is an interesting video. "Hand calculating procedures teach understanding."
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– Larry
Jan 26 at 20:16
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Did you actually watch the full video? One of Wolfram's key points was exposing the fallacy in "hand calculating procedures teach understanding." Very convincing.
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– David G. Stork
Jan 26 at 20:29
1
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@DavidG.Stork I entirely disagree. I personally am interested in knowing how to preform such manual calculations as they teach me new techniques which I can apply to other problems. Also, what is the point of knowing that $int_0^infty ln^2(x)sin(x^2)mathrm dx$ has a closed form (or exact evaluation instead of a decimal expansion) if one cannot prove it? Without a proof it is just a useless fact.
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– clathratus
Jan 26 at 20:51
2
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Ok, I see. That is actually a good point. I will keep that in mind.
$endgroup$
– Larry
Jan 26 at 21:12
|
show 2 more comments
$begingroup$
Wolfram Alpha provides
$$int_{0}^{infty}ln^2(x)sin(x^2)dx=frac{1}{32}sqrt{frac{pi}{2}}(2gamma-pi+ln16)^2tag{1}$$
But I haven't figured out the way to verify this result.
I know Frullani's Integral
$$ln(x)= int_{0}^{infty}frac{e^{-t}-e^{-xt}}{t}dt$$
I also know
$$int_{0}^{infty}sin(x^2)~dx=frac{1}{2}int_{0}^{infty}x^{-1/2}sin(x)~dx$$
Then,
$$begin{align}
int_{0}^{infty}ln^2(x)sin(x^2)dx&=int_{0}^{infty}left(int_{0}^{infty}frac{e^{-t}-e^{-xt}}{t}dtright)left(int_{0}^{infty}frac{e^{-n}-e^{-xn}}{n}dnright)sin(x^2)~dx\
&=frac{1}{2}int_{0}^{infty}left(int_{0}^{infty}frac{e^{-t}-e^{-xt}}{t}dtright)left(int_{0}^{infty}frac{e^{-n}-e^{-xn}}{n}dnright)frac{sin(x)}{sqrt{x}}dx\
&=frac{1}{2}int_{0}^{infty}int_{0}^{infty}int_{0}^{infty}frac{e^{-t}-e^{-xt}}{t}frac{e^{-n}-e^{-xn}}{n}frac{sin(x)}{sqrt{x}}~dx~dn~dt\
&=frac{1}{2}int_{0}^{infty}frac{1}{t}int_{0}^{infty}frac{1}{n}int_{0}^{infty}(e^{-t}-e^{-xt})(e^{-n}-e^{-xn})frac{sin(x)}{sqrt{x}}~dx~dn~dt\
&=frac{1}{2}int_{0}^{infty}frac{1}{t}int_{0}^{infty}frac{1}{n}int_{0}^{infty}(e^{-t-n}-e^{-xn-t}-e^{-xt-n}+e^{-xt-xn})frac{sin(x)}{sqrt{x}}~dx~dn~dt
end{align}$$
What should I do next?
There is also a general case
$$int_{0}^{infty}ln^2(x^a)sin(x^2)dx=frac{a^2}{32}sqrt{frac{pi}{2}}(2gamma-pi+ln16)^2tag{2}$$
But I think $(2)$ becomes easy to prove if we can prove $(1)$.
calculus integration
edited Jan 26 at 20:18
user
5,65111031
asked Jan 26 at 19:48
LarryLarry
2,53531131
$endgroup$
Wolfram Alpha provides
$$int_{0}^{infty}ln^2(x)sin(x^2)dx=frac{1}{32}sqrt{frac{pi}{2}}(2gamma-pi+ln16)^2tag{1}$$
But I haven't figured out the way to verify this result.
I know Frullani's Integral
$$ln(x)= int_{0}^{infty}frac{e^{-t}-e^{-xt}}{t}dt$$
I also know
$$int_{0}^{infty}sin(x^2)~dx=frac{1}{2}int_{0}^{infty}x^{-1/2}sin(x)~dx$$
Then,
$$begin{align}
int_{0}^{infty}ln^2(x)sin(x^2)dx&=int_{0}^{infty}left(int_{0}^{infty}frac{e^{-t}-e^{-xt}}{t}dtright)left(int_{0}^{infty}frac{e^{-n}-e^{-xn}}{n}dnright)sin(x^2)~dx\
&=frac{1}{2}int_{0}^{infty}left(int_{0}^{infty}frac{e^{-t}-e^{-xt}}{t}dtright)left(int_{0}^{infty}frac{e^{-n}-e^{-xn}}{n}dnright)frac{sin(x)}{sqrt{x}}dx\
&=frac{1}{2}int_{0}^{infty}int_{0}^{infty}int_{0}^{infty}frac{e^{-t}-e^{-xt}}{t}frac{e^{-n}-e^{-xn}}{n}frac{sin(x)}{sqrt{x}}~dx~dn~dt\
&=frac{1}{2}int_{0}^{infty}frac{1}{t}int_{0}^{infty}frac{1}{n}int_{0}^{infty}(e^{-t}-e^{-xt})(e^{-n}-e^{-xn})frac{sin(x)}{sqrt{x}}~dx~dn~dt\
&=frac{1}{2}int_{0}^{infty}frac{1}{t}int_{0}^{infty}frac{1}{n}int_{0}^{infty}(e^{-t-n}-e^{-xn-t}-e^{-xt-n}+e^{-xt-xn})frac{sin(x)}{sqrt{x}}~dx~dn~dt
end{align}$$
What should I do next?
There is also a general case
$$int_{0}^{infty}ln^2(x^a)sin(x^2)dx=frac{a^2}{32}sqrt{frac{pi}{2}}(2gamma-pi+ln16)^2tag{2}$$
But I think $(2)$ becomes easy to prove if we can prove $(1)$.
calculus integration
calculus integration
edited Jan 26 at 20:18
user
5,65111031
asked Jan 26 at 19:48
LarryLarry
2,53531131
edited Jan 26 at 20:18
user
5,65111031
asked Jan 26 at 19:48
LarryLarry
2,53531131
edited Jan 26 at 20:18
user
5,65111031
edited Jan 26 at 20:18
user
5,65111031
edited Jan 26 at 20:18
user
5,65111031
5,65111031
asked Jan 26 at 19:48
LarryLarry
2,53531131
asked Jan 26 at 19:48
LarryLarry
2,53531131
asked Jan 26 at 19:48
LarryLarry
2,53531131
2,53531131
$begingroup$
Computer algebra solves this instantly: $frac{1}{32} sqrt{frac{pi }{2}} (2 gamma -pi +log (16))^2$. Is there really a need to perform this by hand? ted.com/talks/…
$endgroup$
– David G. Stork
Jan 26 at 20:10
$begingroup$
It is an interesting video. "Hand calculating procedures teach understanding."
$endgroup$
– Larry
Jan 26 at 20:16
$begingroup$
Did you actually watch the full video? One of Wolfram's key points was exposing the fallacy in "hand calculating procedures teach understanding." Very convincing.
$endgroup$
– David G. Stork
Jan 26 at 20:29
1
$begingroup$
@DavidG.Stork I entirely disagree. I personally am interested in knowing how to preform such manual calculations as they teach me new techniques which I can apply to other problems. Also, what is the point of knowing that $int_0^infty ln^2(x)sin(x^2)mathrm dx$ has a closed form (or exact evaluation instead of a decimal expansion) if one cannot prove it? Without a proof it is just a useless fact.
$endgroup$
– clathratus
Jan 26 at 20:51
2
$begingroup$
Ok, I see. That is actually a good point. I will keep that in mind.
$endgroup$
– Larry
Jan 26 at 21:12
|
show 2 more comments
$begingroup$
Computer algebra solves this instantly: $frac{1}{32} sqrt{frac{pi }{2}} (2 gamma -pi +log (16))^2$. Is there really a need to perform this by hand? ted.com/talks/…
$endgroup$
– David G. Stork
Jan 26 at 20:10
$begingroup$
It is an interesting video. "Hand calculating procedures teach understanding."
$endgroup$
– Larry
Jan 26 at 20:16
$begingroup$
Did you actually watch the full video? One of Wolfram's key points was exposing the fallacy in "hand calculating procedures teach understanding." Very convincing.
$endgroup$
– David G. Stork
Jan 26 at 20:29
1
$begingroup$
@DavidG.Stork I entirely disagree. I personally am interested in knowing how to preform such manual calculations as they teach me new techniques which I can apply to other problems. Also, what is the point of knowing that $int_0^infty ln^2(x)sin(x^2)mathrm dx$ has a closed form (or exact evaluation instead of a decimal expansion) if one cannot prove it? Without a proof it is just a useless fact.
$endgroup$
– clathratus
Jan 26 at 20:51
2
$begingroup$
Ok, I see. That is actually a good point. I will keep that in mind.
$endgroup$
– Larry
Jan 26 at 21:12
$begingroup$
Computer algebra solves this instantly: $frac{1}{32} sqrt{frac{pi }{2}} (2 gamma -pi +log (16))^2$. Is there really a need to perform this by hand? ted.com/talks/…
$endgroup$
– David G. Stork
Jan 26 at 20:10
$begingroup$
Computer algebra solves this instantly: $frac{1}{32} sqrt{frac{pi }{2}} (2 gamma -pi +log (16))^2$. Is there really a need to perform this by hand? ted.com/talks/…
$endgroup$
– David G. Stork
Jan 26 at 20:10
$begingroup$
It is an interesting video. "Hand calculating procedures teach understanding."
$endgroup$
– Larry
Jan 26 at 20:16
$begingroup$
It is an interesting video. "Hand calculating procedures teach understanding."
$endgroup$
– Larry
Jan 26 at 20:16
$begingroup$
Did you actually watch the full video? One of Wolfram's key points was exposing the fallacy in "hand calculating procedures teach understanding." Very convincing.
$endgroup$
– David G. Stork
Jan 26 at 20:29
$begingroup$
Did you actually watch the full video? One of Wolfram's key points was exposing the fallacy in "hand calculating procedures teach understanding." Very convincing.
$endgroup$
– David G. Stork
Jan 26 at 20:29
1
1
$begingroup$
@DavidG.Stork I entirely disagree. I personally am interested in knowing how to preform such manual calculations as they teach me new techniques which I can apply to other problems. Also, what is the point of knowing that $int_0^infty ln^2(x)sin(x^2)mathrm dx$ has a closed form (or exact evaluation instead of a decimal expansion) if one cannot prove it? Without a proof it is just a useless fact.
$endgroup$
– clathratus
Jan 26 at 20:51
$begingroup$
@DavidG.Stork I entirely disagree. I personally am interested in knowing how to preform such manual calculations as they teach me new techniques which I can apply to other problems. Also, what is the point of knowing that $int_0^infty ln^2(x)sin(x^2)mathrm dx$ has a closed form (or exact evaluation instead of a decimal expansion) if one cannot prove it? Without a proof it is just a useless fact.
$endgroup$
– clathratus
Jan 26 at 20:51
2
2
$begingroup$
Ok, I see. That is actually a good point. I will keep that in mind.
$endgroup$
– Larry
Jan 26 at 21:12
$begingroup$
Ok, I see. That is actually a good point. I will keep that in mind.
$endgroup$
– Larry
Jan 26 at 21:12
|
show 2 more comments
5 Answers
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$$I=int_{0}^{infty}ln^2(x)sin(x^2)dx overset{x^2=t}=int_0^infty frac{1}{2sqrt t} ln^2 (sqrt t) sin t dt =frac18 int_0^infty t^{-1/2}sin t ln^2 t ,dt$$
Note that the last integral is the Mellin transform in $s=frac12 $ of the sine after being differentiated twice.
See for example here a proof for:
$$int_0^infty x^{s-1}sin x dx= Gamma(s) sinleft(frac{pi s}{2}right)$$
$$Rightarrow I=frac18frac{d^2}{ds^2}Gamma(s) sinleft(frac{pi s}{2}right)bigg|_{s=frac12}$$
It's not the end of the world to differentiate that twice since the digamma function comes in our help.
From the wiki page we have:
$Gamma'(x)=Gamma(x)psi(x)$
$$Rightarrow frac{d}{ds}Gamma(s) sinleft(frac{pi s}{2}right)=Gamma(s)psi(s)sinleft(frac{pi s}{2}right) +frac{pi}{2}Gamma(s)cosleft(frac{pi s}{2}right)$$
$$Rightarrow frac{d^2}{ds^2}Gamma(s) sinleft(frac{pi s}{2}right)=frac{d}{ds}Gamma(s)left(psi(s)sinleft(frac{pi s}{2}right)+frac{pi}{2}cosleft(frac{pi s}{2}right)right)$$
$$=Gamma(x)psi(x)left(psi(s)sinleft(frac{pi s}{2}right)+frac{pi}{2}cosleft(frac{pi s}{2}right)right)+Gamma(s)left(psi_1(x)sinleft(frac{pi s}{2}right)+frac{pi}{2}Gamma(s)cosleft(frac{pi s}{2}right)-frac{pi^2}{4}sinleft(frac{pi s}{2}right)right)$$
And now setting $s=frac12$ we get using $Gammaleft(frac12right)=sqrt{pi}$, $psileft(frac12 right)=-gamma -2ln 2 $,$ psi_1left(frac12right)=frac{pi^2}{2}$ the result.
answered Jan 26 at 20:24
ZackyZacky
7,89511061
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2
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Note that the identity used by @Zacky applied is a direct application of Ramanujan's Master Theorem: en.wikipedia.org/wiki/Ramanujan%27s_master_theorem
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– DavidG
Jan 28 at 9:09
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We have
$$ F(alpha)=int_{0}^{+infty} x^alpha sin(x^2),dx = frac{1}{2}int_{0}^{+infty} x^{alpha/2-1}sin(x),dx\=frac{1}{2Gamma(1-alpha/2)}int_{0}^{+infty} frac{ds}{s^{alpha/2}(s^2+1)} $$
by the properties of the Laplace transform. The last integral can be computed through the Beta and Gamma functions, producing
$$ F(alpha) = frac{1}{2},Gammaleft(frac{1+alpha}{2}right)sinleft(frac{pi}{4}(1+alpha)right) $$
for any $alpha$ such that $text{Re}(alpha)in(-3,1)$. In order to prove the claim, it is enough to apply $lim_{alphato 0}frac{d^2}{dalpha^2}$ to both sides of the last identity and recall the special values of $Gamma,psi$ and $psi'$ at $frac{1}{2}$.
answered Jan 26 at 20:23
Jack D'AurizioJack D'Aurizio
291k33284669
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Let us rewrite your integral as
$$int_0^infty ln^2(x)sin(x^2)dx=frac{1}{8}int_0^infty frac{ln^2(x)sin(x)}{sqrt{x}}dx$$
To solve this integral, you can employ the following identity, which holds for any $pin (0,1)$:
$$int_0^infty x^{p-1}sin(x)dx=Gamma(p)sin(pi p/2)$$
The value of your integral can be obtained from this by differentiating both sides of this equation twice with respect to $p$, moving the derivative inside of the definite integral on the LHS, and making use of the known special values of the Digamma function.
This can be done by hand, but it requires a lot of algebra and would be best left to a CAS, as suggested in the comments.
answered Jan 26 at 20:17
FrpzzdFrpzzd
23k841110
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add a comment |
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Just a generalization of @Zacky's answer
$$F(a)=int_0^{infty}log^2(x^a)sin(x^2)mathrm dx$$
Since $log(x^a)=log(e^{alog x})=alog x$,
$$F(a)=a^2int_0^{infty}log^2(x)sin(x^2)mathrm dx$$
$$F(a)=a^2F(1)$$
And as @Zacky showed,
$$F(1)=frac18mathrm{D}^2_{s=frac12}Gamma(s)sinfrac{pi s}{2}=frac1{32}sqrt{fracpi2}(2gamma-pi+log16)^2$$
So
$$F(a)=frac{a^2}{32}sqrt{fracpi2}(2gamma-pi+log16)^2$$
I will edit my answer to include a proof of my own once I find one.
answered Jan 26 at 21:05
clathratusclathratus
5,1701338
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Nice, I forgot the fact that $log(x^a)=alog(x)$.
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– Larry
Jan 26 at 21:14
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@Larry I legitimately can't tell if you're being sarcastic or not
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– clathratus
Jan 26 at 21:15
1
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I am not being sarcastic.
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– Larry
Jan 26 at 21:18
1
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@Larry Thank you for the clarification. You were very right when you mentioned that $(2)$ would be easy to prove once one had proven $(1)$.
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– clathratus
Jan 26 at 21:20
add a comment |
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An alternative approach is to employ Feynman's Trick and Laplace Transforms to solve:
begin{equation}
I = int_0^inftyln^2(x)sinleft(x^2right):dx
end{equation}
We first observe that:
begin{equation}
I = int_0^inftyln^2(x)sinleft(x^2right):dx = lim_{krightarrow 0^+} frac{d^2}{dk^2}int_0^infty x^ksinleft(x^2right):dx = lim_{krightarrow 0^+} frac{d^2}{dk^2} H(k)
end{equation}
We proceed by solving $H(k)$. To do so, we introduce a new parameter $'t'$:
begin{equation}
J(t; k) = int_0^infty x^ksinleft(tx^2right):dx
end{equation}
(This is allowable through the Dominated Convergence Theorem). Thus:
begin{equation}
H(k) = lim_{trightarrow 1^+} J(t; k)
end{equation}
Using Fubini's Theorem we now take the Laplace Transform with respect to '$t$'
begin{align}
mathscr{L}_tleft[J(t;k) right] &= int_0^infty x^kmathscr{L}_tleft[sinleft(tx^2right)right]:dx = int_0^infty frac{x^{k + 2}}{s^2 + x^4}:dx
end{align}
As I address here we find this becomes:
begin{align}
mathscr{L}_tleft[J(t;k) right] &= frac{1}{4}cdot left(s^2right)^{frac{k + 2 + 1}{2} - 1} cdot Bleft(1 - frac{k + 2 + 1 }{4}, frac{k + 2 + 1 }{4} right) = frac{1}{4} s^{frac{k - 1}{2}} Bleft(1 - frac{k + 3}{4} , frac{k + 3}{4}right)
end{align}
Using the relationship between the Gamma and Beta Function we find:
begin{equation}
mathscr{L}_tleft[J(t;k) right] = frac{1}{4} s^{frac{k - 1}{2}} Gammaleft(1 - frac{k + 3}{4}right) Gammaleft( frac{k + 3}{4}right)
end{equation}
Using Euler's Reflection Formula we find:
begin{equation}
mathscr{L}_tleft[J(t;k) right] = frac{1}{4} s^{frac{k - 1}{2}} frac{pi}{sinleft(pileft(frac{k + 3}{4}right) right)}
end{equation}
Taking the inverse Laplace Transforms is rather tricky here. To evaluate recall that:
begin{equation}
I = lim_{krightarrow 0^+} frac{d^2}{dk^2} H(k) = lim_{krightarrow 0^+} frac{d^2}{dk^2}left[ lim_{trightarrow 1^+} J(t;k)right]
end{equation}
In this process we solve for $H(k)$ using
begin{equation}
H(k) = lim_{trightarrow 1^+} mathscr{L}_s^{-1}left[mathscr{L}_tleft[J(t; k)right]right]
end{equation}
Thus, our definition of $I$ becomes:
begin{align}
I &= lim_{krightarrow 0^+} frac{d^2}{dk^2} H(k) = lim_{krightarrow 0^+} frac{d^2}{dk^2}left[ lim_{trightarrow 1^+} J(t;k)right] = lim_{krightarrow 0^+} frac{d^2}{dk^2}left[ lim_{trightarrow 1^+} mathscr{L}_s^{-1}left[mathscr{L}_tleft[J(t; k)right]right]right] \
&= lim_{trightarrow 1^+} mathscr{L}_s^{-1}left[ lim_{krightarrow 0^+} frac{d^2}{dk^2}mathscr{L}_tleft[J(t; k)right]right] = lim_{trightarrow 1^+} mathscr{L}_s^{-1}left[ lim_{krightarrow 0^+} frac{d^2}{dk^2}left[ frac{1}{4} s^{frac{k - 1}{2}} frac{pi}{sinleft(pileft(frac{k + 3}{4}right) right)}right]right]
end{align}
Because I'm lazy, I used Wolframalpha to evaluate the second derivate at $0$:
begin{align}
I &= lim_{trightarrow 1^+} mathscr{L}_s^{-1}left[ lim_{krightarrow 0^+} frac{d^2}{dk^2}left[ frac{1}{4} s^{frac{k - 1}{2}} frac{pi}{sinleft(pileft(frac{k + 3}{4}right) right)}right]right] = lim_{trightarrow 1^+} mathscr{L}_s^{-1}left[ frac{pi}{4}left( frac{3pi^2}{8sqrt{2}sqrt{s}} + frac{ln^2(s)}{2sqrt{2}sqrt{s}} + frac{piln(s)}{2sqrt{2}sqrt{s}}right)right] \
&= lim_{trightarrow 1^+} left[ frac{3pi^3}{32sqrt{2}} mathscr{L}_s^{-1}left[ frac{1}{sqrt{s}}right] + frac{pi}{8sqrt{2}} mathscr{L}_s^{-1}left[ frac{ln^2(s)}{sqrt{s}}right]+ frac{pi^2}{8sqrt{2}} mathscr{L}_s^{-1}left[ frac{ln(s)}{sqrt{s}}right]right] \
&= lim_{trightarrow 1^+} left[ frac{3pi^3}{32sqrt{2}} left[ frac{1}{sqrt{pi}sqrt{t}}right] + frac{pi}{32sqrt{2}} left[ frac{
left(psi^{(0)}left(frac{1}{2}right)-ln(t)right)^2 -frac{pi^2}{2}}{sqrt{pi}sqrt{t}}right]+ frac{pi^2}{16sqrt{2}} left[ frac{
psi^{(0)}left(frac{1}{2}right)-ln(t)}{sqrt{pi}sqrt{t}}right]right] \
&= frac{3pi^3}{32sqrt{2}} left[ frac{1}{sqrt{pi}}right] + frac{pi}{32sqrt{2}} left[ frac{
psi^{(0)}left(frac{1}{2}right)^2 -frac{pi^2}{2}}{sqrt{pi}}right]+ frac{pi^2}{16sqrt{2}} left[ frac{
psi^{(0)}left(frac{1}{2}right)}{sqrt{pi}}right]
end{align}
Noting that
begin{equation}
psi^{(0)}left(frac{1}{2}right) = -gamma - 2ln(2)
end{equation}
Where $gamma$ is the Euler–Mascheroni constant.
Thus,
begin{align}
I = frac{3pi^3}{32sqrt{2}} left[ frac{1}{sqrt{pi}}right] + frac{pi}{32sqrt{2}} left[ frac{
left(gamma + 2ln(2)right)^2 -frac{pi^2}{2}}{sqrt{pi}}right]+ frac{pi^2}{16sqrt{2}} left[ frac{
gamma - 2ln(2)}{sqrt{pi}}right] = frac{1}{32}sqrt{frac{pi}{2}}(2gamma-pi+4ln2)^2
end{align}
answered Jan 28 at 12:51
DavidGDavidG
1
$endgroup$
$begingroup$
Very nice solution! I am a fan of your method.
$endgroup$
– clathratus
Jan 28 at 16:07
$begingroup$
Do you use a similar technique to evaluate $$F(n)=int_0^infty cos(x^n)mathrm dx$$
$endgroup$
– clathratus
Feb 6 at 20:58
$begingroup$
@clathratus You definitely can
$endgroup$
– DavidG
Feb 7 at 0:09
add a comment |
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5 Answers
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$begingroup$
$$I=int_{0}^{infty}ln^2(x)sin(x^2)dx overset{x^2=t}=int_0^infty frac{1}{2sqrt t} ln^2 (sqrt t) sin t dt =frac18 int_0^infty t^{-1/2}sin t ln^2 t ,dt$$
Note that the last integral is the Mellin transform in $s=frac12 $ of the sine after being differentiated twice.
See for example here a proof for:
$$int_0^infty x^{s-1}sin x dx= Gamma(s) sinleft(frac{pi s}{2}right)$$
$$Rightarrow I=frac18frac{d^2}{ds^2}Gamma(s) sinleft(frac{pi s}{2}right)bigg|_{s=frac12}$$
It's not the end of the world to differentiate that twice since the digamma function comes in our help.
From the wiki page we have:
$Gamma'(x)=Gamma(x)psi(x)$
$$Rightarrow frac{d}{ds}Gamma(s) sinleft(frac{pi s}{2}right)=Gamma(s)psi(s)sinleft(frac{pi s}{2}right) +frac{pi}{2}Gamma(s)cosleft(frac{pi s}{2}right)$$
$$Rightarrow frac{d^2}{ds^2}Gamma(s) sinleft(frac{pi s}{2}right)=frac{d}{ds}Gamma(s)left(psi(s)sinleft(frac{pi s}{2}right)+frac{pi}{2}cosleft(frac{pi s}{2}right)right)$$
$$=Gamma(x)psi(x)left(psi(s)sinleft(frac{pi s}{2}right)+frac{pi}{2}cosleft(frac{pi s}{2}right)right)+Gamma(s)left(psi_1(x)sinleft(frac{pi s}{2}right)+frac{pi}{2}Gamma(s)cosleft(frac{pi s}{2}right)-frac{pi^2}{4}sinleft(frac{pi s}{2}right)right)$$
And now setting $s=frac12$ we get using $Gammaleft(frac12right)=sqrt{pi}$, $psileft(frac12 right)=-gamma -2ln 2 $,$ psi_1left(frac12right)=frac{pi^2}{2}$ the result.
answered Jan 26 at 20:24
ZackyZacky
7,89511061
$endgroup$
2
$begingroup$
Note that the identity used by @Zacky applied is a direct application of Ramanujan's Master Theorem: en.wikipedia.org/wiki/Ramanujan%27s_master_theorem
$endgroup$
– DavidG
Jan 28 at 9:09
add a comment |
$begingroup$
$$I=int_{0}^{infty}ln^2(x)sin(x^2)dx overset{x^2=t}=int_0^infty frac{1}{2sqrt t} ln^2 (sqrt t) sin t dt =frac18 int_0^infty t^{-1/2}sin t ln^2 t ,dt$$
Note that the last integral is the Mellin transform in $s=frac12 $ of the sine after being differentiated twice.
See for example here a proof for:
$$int_0^infty x^{s-1}sin x dx= Gamma(s) sinleft(frac{pi s}{2}right)$$
$$Rightarrow I=frac18frac{d^2}{ds^2}Gamma(s) sinleft(frac{pi s}{2}right)bigg|_{s=frac12}$$
It's not the end of the world to differentiate that twice since the digamma function comes in our help.
From the wiki page we have:
$Gamma'(x)=Gamma(x)psi(x)$
$$Rightarrow frac{d}{ds}Gamma(s) sinleft(frac{pi s}{2}right)=Gamma(s)psi(s)sinleft(frac{pi s}{2}right) +frac{pi}{2}Gamma(s)cosleft(frac{pi s}{2}right)$$
$$Rightarrow frac{d^2}{ds^2}Gamma(s) sinleft(frac{pi s}{2}right)=frac{d}{ds}Gamma(s)left(psi(s)sinleft(frac{pi s}{2}right)+frac{pi}{2}cosleft(frac{pi s}{2}right)right)$$
$$=Gamma(x)psi(x)left(psi(s)sinleft(frac{pi s}{2}right)+frac{pi}{2}cosleft(frac{pi s}{2}right)right)+Gamma(s)left(psi_1(x)sinleft(frac{pi s}{2}right)+frac{pi}{2}Gamma(s)cosleft(frac{pi s}{2}right)-frac{pi^2}{4}sinleft(frac{pi s}{2}right)right)$$
And now setting $s=frac12$ we get using $Gammaleft(frac12right)=sqrt{pi}$, $psileft(frac12 right)=-gamma -2ln 2 $,$ psi_1left(frac12right)=frac{pi^2}{2}$ the result.
answered Jan 26 at 20:24
ZackyZacky
7,89511061
$endgroup$
2
$begingroup$
Note that the identity used by @Zacky applied is a direct application of Ramanujan's Master Theorem: en.wikipedia.org/wiki/Ramanujan%27s_master_theorem
$endgroup$
– DavidG
Jan 28 at 9:09
add a comment |
$begingroup$
$$I=int_{0}^{infty}ln^2(x)sin(x^2)dx overset{x^2=t}=int_0^infty frac{1}{2sqrt t} ln^2 (sqrt t) sin t dt =frac18 int_0^infty t^{-1/2}sin t ln^2 t ,dt$$
Note that the last integral is the Mellin transform in $s=frac12 $ of the sine after being differentiated twice.
See for example here a proof for:
$$int_0^infty x^{s-1}sin x dx= Gamma(s) sinleft(frac{pi s}{2}right)$$
$$Rightarrow I=frac18frac{d^2}{ds^2}Gamma(s) sinleft(frac{pi s}{2}right)bigg|_{s=frac12}$$
It's not the end of the world to differentiate that twice since the digamma function comes in our help.
From the wiki page we have:
$Gamma'(x)=Gamma(x)psi(x)$
$$Rightarrow frac{d}{ds}Gamma(s) sinleft(frac{pi s}{2}right)=Gamma(s)psi(s)sinleft(frac{pi s}{2}right) +frac{pi}{2}Gamma(s)cosleft(frac{pi s}{2}right)$$
$$Rightarrow frac{d^2}{ds^2}Gamma(s) sinleft(frac{pi s}{2}right)=frac{d}{ds}Gamma(s)left(psi(s)sinleft(frac{pi s}{2}right)+frac{pi}{2}cosleft(frac{pi s}{2}right)right)$$
$$=Gamma(x)psi(x)left(psi(s)sinleft(frac{pi s}{2}right)+frac{pi}{2}cosleft(frac{pi s}{2}right)right)+Gamma(s)left(psi_1(x)sinleft(frac{pi s}{2}right)+frac{pi}{2}Gamma(s)cosleft(frac{pi s}{2}right)-frac{pi^2}{4}sinleft(frac{pi s}{2}right)right)$$
And now setting $s=frac12$ we get using $Gammaleft(frac12right)=sqrt{pi}$, $psileft(frac12 right)=-gamma -2ln 2 $,$ psi_1left(frac12right)=frac{pi^2}{2}$ the result.
answered Jan 26 at 20:24
ZackyZacky
7,89511061
$endgroup$
$$I=int_{0}^{infty}ln^2(x)sin(x^2)dx overset{x^2=t}=int_0^infty frac{1}{2sqrt t} ln^2 (sqrt t) sin t dt =frac18 int_0^infty t^{-1/2}sin t ln^2 t ,dt$$
Note that the last integral is the Mellin transform in $s=frac12 $ of the sine after being differentiated twice.
See for example here a proof for:
$$int_0^infty x^{s-1}sin x dx= Gamma(s) sinleft(frac{pi s}{2}right)$$
$$Rightarrow I=frac18frac{d^2}{ds^2}Gamma(s) sinleft(frac{pi s}{2}right)bigg|_{s=frac12}$$
It's not the end of the world to differentiate that twice since the digamma function comes in our help.
From the wiki page we have:
$Gamma'(x)=Gamma(x)psi(x)$
$$Rightarrow frac{d}{ds}Gamma(s) sinleft(frac{pi s}{2}right)=Gamma(s)psi(s)sinleft(frac{pi s}{2}right) +frac{pi}{2}Gamma(s)cosleft(frac{pi s}{2}right)$$
$$Rightarrow frac{d^2}{ds^2}Gamma(s) sinleft(frac{pi s}{2}right)=frac{d}{ds}Gamma(s)left(psi(s)sinleft(frac{pi s}{2}right)+frac{pi}{2}cosleft(frac{pi s}{2}right)right)$$
$$=Gamma(x)psi(x)left(psi(s)sinleft(frac{pi s}{2}right)+frac{pi}{2}cosleft(frac{pi s}{2}right)right)+Gamma(s)left(psi_1(x)sinleft(frac{pi s}{2}right)+frac{pi}{2}Gamma(s)cosleft(frac{pi s}{2}right)-frac{pi^2}{4}sinleft(frac{pi s}{2}right)right)$$
And now setting $s=frac12$ we get using $Gammaleft(frac12right)=sqrt{pi}$, $psileft(frac12 right)=-gamma -2ln 2 $,$ psi_1left(frac12right)=frac{pi^2}{2}$ the result.
answered Jan 26 at 20:24
ZackyZacky
7,89511061
answered Jan 26 at 20:24
ZackyZacky
7,89511061
answered Jan 26 at 20:24
ZackyZacky
7,89511061
answered Jan 26 at 20:24
ZackyZacky
7,89511061
7,89511061
2
$begingroup$
Note that the identity used by @Zacky applied is a direct application of Ramanujan's Master Theorem: en.wikipedia.org/wiki/Ramanujan%27s_master_theorem
$endgroup$
– DavidG
Jan 28 at 9:09
add a comment |
2
$begingroup$
Note that the identity used by @Zacky applied is a direct application of Ramanujan's Master Theorem: en.wikipedia.org/wiki/Ramanujan%27s_master_theorem
$endgroup$
– DavidG
Jan 28 at 9:09
2
2
$begingroup$
Note that the identity used by @Zacky applied is a direct application of Ramanujan's Master Theorem: en.wikipedia.org/wiki/Ramanujan%27s_master_theorem
$endgroup$
– DavidG
Jan 28 at 9:09
$begingroup$
Note that the identity used by @Zacky applied is a direct application of Ramanujan's Master Theorem: en.wikipedia.org/wiki/Ramanujan%27s_master_theorem
$endgroup$
– DavidG
Jan 28 at 9:09
add a comment |
$begingroup$
We have
$$ F(alpha)=int_{0}^{+infty} x^alpha sin(x^2),dx = frac{1}{2}int_{0}^{+infty} x^{alpha/2-1}sin(x),dx\=frac{1}{2Gamma(1-alpha/2)}int_{0}^{+infty} frac{ds}{s^{alpha/2}(s^2+1)} $$
by the properties of the Laplace transform. The last integral can be computed through the Beta and Gamma functions, producing
$$ F(alpha) = frac{1}{2},Gammaleft(frac{1+alpha}{2}right)sinleft(frac{pi}{4}(1+alpha)right) $$
for any $alpha$ such that $text{Re}(alpha)in(-3,1)$. In order to prove the claim, it is enough to apply $lim_{alphato 0}frac{d^2}{dalpha^2}$ to both sides of the last identity and recall the special values of $Gamma,psi$ and $psi'$ at $frac{1}{2}$.
answered Jan 26 at 20:23
Jack D'AurizioJack D'Aurizio
291k33284669
$endgroup$
add a comment |
$begingroup$
We have
$$ F(alpha)=int_{0}^{+infty} x^alpha sin(x^2),dx = frac{1}{2}int_{0}^{+infty} x^{alpha/2-1}sin(x),dx\=frac{1}{2Gamma(1-alpha/2)}int_{0}^{+infty} frac{ds}{s^{alpha/2}(s^2+1)} $$
by the properties of the Laplace transform. The last integral can be computed through the Beta and Gamma functions, producing
$$ F(alpha) = frac{1}{2},Gammaleft(frac{1+alpha}{2}right)sinleft(frac{pi}{4}(1+alpha)right) $$
for any $alpha$ such that $text{Re}(alpha)in(-3,1)$. In order to prove the claim, it is enough to apply $lim_{alphato 0}frac{d^2}{dalpha^2}$ to both sides of the last identity and recall the special values of $Gamma,psi$ and $psi'$ at $frac{1}{2}$.
answered Jan 26 at 20:23
Jack D'AurizioJack D'Aurizio
291k33284669
$endgroup$
add a comment |
$begingroup$
We have
$$ F(alpha)=int_{0}^{+infty} x^alpha sin(x^2),dx = frac{1}{2}int_{0}^{+infty} x^{alpha/2-1}sin(x),dx\=frac{1}{2Gamma(1-alpha/2)}int_{0}^{+infty} frac{ds}{s^{alpha/2}(s^2+1)} $$
by the properties of the Laplace transform. The last integral can be computed through the Beta and Gamma functions, producing
$$ F(alpha) = frac{1}{2},Gammaleft(frac{1+alpha}{2}right)sinleft(frac{pi}{4}(1+alpha)right) $$
for any $alpha$ such that $text{Re}(alpha)in(-3,1)$. In order to prove the claim, it is enough to apply $lim_{alphato 0}frac{d^2}{dalpha^2}$ to both sides of the last identity and recall the special values of $Gamma,psi$ and $psi'$ at $frac{1}{2}$.
answered Jan 26 at 20:23
Jack D'AurizioJack D'Aurizio
291k33284669
$endgroup$
We have
$$ F(alpha)=int_{0}^{+infty} x^alpha sin(x^2),dx = frac{1}{2}int_{0}^{+infty} x^{alpha/2-1}sin(x),dx\=frac{1}{2Gamma(1-alpha/2)}int_{0}^{+infty} frac{ds}{s^{alpha/2}(s^2+1)} $$
by the properties of the Laplace transform. The last integral can be computed through the Beta and Gamma functions, producing
$$ F(alpha) = frac{1}{2},Gammaleft(frac{1+alpha}{2}right)sinleft(frac{pi}{4}(1+alpha)right) $$
for any $alpha$ such that $text{Re}(alpha)in(-3,1)$. In order to prove the claim, it is enough to apply $lim_{alphato 0}frac{d^2}{dalpha^2}$ to both sides of the last identity and recall the special values of $Gamma,psi$ and $psi'$ at $frac{1}{2}$.
answered Jan 26 at 20:23
Jack D'AurizioJack D'Aurizio
291k33284669
answered Jan 26 at 20:23
Jack D'AurizioJack D'Aurizio
291k33284669
answered Jan 26 at 20:23
Jack D'AurizioJack D'Aurizio
291k33284669
answered Jan 26 at 20:23
Jack D'AurizioJack D'Aurizio
291k33284669
291k33284669
add a comment |
add a comment |
$begingroup$
Let us rewrite your integral as
$$int_0^infty ln^2(x)sin(x^2)dx=frac{1}{8}int_0^infty frac{ln^2(x)sin(x)}{sqrt{x}}dx$$
To solve this integral, you can employ the following identity, which holds for any $pin (0,1)$:
$$int_0^infty x^{p-1}sin(x)dx=Gamma(p)sin(pi p/2)$$
The value of your integral can be obtained from this by differentiating both sides of this equation twice with respect to $p$, moving the derivative inside of the definite integral on the LHS, and making use of the known special values of the Digamma function.
This can be done by hand, but it requires a lot of algebra and would be best left to a CAS, as suggested in the comments.
answered Jan 26 at 20:17
FrpzzdFrpzzd
23k841110
$endgroup$
add a comment |
$begingroup$
Let us rewrite your integral as
$$int_0^infty ln^2(x)sin(x^2)dx=frac{1}{8}int_0^infty frac{ln^2(x)sin(x)}{sqrt{x}}dx$$
To solve this integral, you can employ the following identity, which holds for any $pin (0,1)$:
$$int_0^infty x^{p-1}sin(x)dx=Gamma(p)sin(pi p/2)$$
The value of your integral can be obtained from this by differentiating both sides of this equation twice with respect to $p$, moving the derivative inside of the definite integral on the LHS, and making use of the known special values of the Digamma function.
This can be done by hand, but it requires a lot of algebra and would be best left to a CAS, as suggested in the comments.
answered Jan 26 at 20:17
FrpzzdFrpzzd
23k841110
$endgroup$
add a comment |
$begingroup$
Let us rewrite your integral as
$$int_0^infty ln^2(x)sin(x^2)dx=frac{1}{8}int_0^infty frac{ln^2(x)sin(x)}{sqrt{x}}dx$$
To solve this integral, you can employ the following identity, which holds for any $pin (0,1)$:
$$int_0^infty x^{p-1}sin(x)dx=Gamma(p)sin(pi p/2)$$
The value of your integral can be obtained from this by differentiating both sides of this equation twice with respect to $p$, moving the derivative inside of the definite integral on the LHS, and making use of the known special values of the Digamma function.
This can be done by hand, but it requires a lot of algebra and would be best left to a CAS, as suggested in the comments.
answered Jan 26 at 20:17
FrpzzdFrpzzd
23k841110
$endgroup$
Let us rewrite your integral as
$$int_0^infty ln^2(x)sin(x^2)dx=frac{1}{8}int_0^infty frac{ln^2(x)sin(x)}{sqrt{x}}dx$$
To solve this integral, you can employ the following identity, which holds for any $pin (0,1)$:
$$int_0^infty x^{p-1}sin(x)dx=Gamma(p)sin(pi p/2)$$
The value of your integral can be obtained from this by differentiating both sides of this equation twice with respect to $p$, moving the derivative inside of the definite integral on the LHS, and making use of the known special values of the Digamma function.
This can be done by hand, but it requires a lot of algebra and would be best left to a CAS, as suggested in the comments.
answered Jan 26 at 20:17
FrpzzdFrpzzd
23k841110
answered Jan 26 at 20:17
FrpzzdFrpzzd
23k841110
answered Jan 26 at 20:17
FrpzzdFrpzzd
23k841110
answered Jan 26 at 20:17
FrpzzdFrpzzd
23k841110
23k841110
add a comment |
add a comment |
$begingroup$
Just a generalization of @Zacky's answer
$$F(a)=int_0^{infty}log^2(x^a)sin(x^2)mathrm dx$$
Since $log(x^a)=log(e^{alog x})=alog x$,
$$F(a)=a^2int_0^{infty}log^2(x)sin(x^2)mathrm dx$$
$$F(a)=a^2F(1)$$
And as @Zacky showed,
$$F(1)=frac18mathrm{D}^2_{s=frac12}Gamma(s)sinfrac{pi s}{2}=frac1{32}sqrt{fracpi2}(2gamma-pi+log16)^2$$
So
$$F(a)=frac{a^2}{32}sqrt{fracpi2}(2gamma-pi+log16)^2$$
I will edit my answer to include a proof of my own once I find one.
answered Jan 26 at 21:05
clathratusclathratus
5,1701338
$endgroup$
$begingroup$
Nice, I forgot the fact that $log(x^a)=alog(x)$.
$endgroup$
– Larry
Jan 26 at 21:14
$begingroup$
@Larry I legitimately can't tell if you're being sarcastic or not
$endgroup$
– clathratus
Jan 26 at 21:15
1
$begingroup$
I am not being sarcastic.
$endgroup$
– Larry
Jan 26 at 21:18
1
$begingroup$
@Larry Thank you for the clarification. You were very right when you mentioned that $(2)$ would be easy to prove once one had proven $(1)$.
$endgroup$
– clathratus
Jan 26 at 21:20
add a comment |
$begingroup$
Just a generalization of @Zacky's answer
$$F(a)=int_0^{infty}log^2(x^a)sin(x^2)mathrm dx$$
Since $log(x^a)=log(e^{alog x})=alog x$,
$$F(a)=a^2int_0^{infty}log^2(x)sin(x^2)mathrm dx$$
$$F(a)=a^2F(1)$$
And as @Zacky showed,
$$F(1)=frac18mathrm{D}^2_{s=frac12}Gamma(s)sinfrac{pi s}{2}=frac1{32}sqrt{fracpi2}(2gamma-pi+log16)^2$$
So
$$F(a)=frac{a^2}{32}sqrt{fracpi2}(2gamma-pi+log16)^2$$
I will edit my answer to include a proof of my own once I find one.
answered Jan 26 at 21:05
clathratusclathratus
5,1701338
$endgroup$
$begingroup$
Nice, I forgot the fact that $log(x^a)=alog(x)$.
$endgroup$
– Larry
Jan 26 at 21:14
$begingroup$
@Larry I legitimately can't tell if you're being sarcastic or not
$endgroup$
– clathratus
Jan 26 at 21:15
1
$begingroup$
I am not being sarcastic.
$endgroup$
– Larry
Jan 26 at 21:18
1
$begingroup$
@Larry Thank you for the clarification. You were very right when you mentioned that $(2)$ would be easy to prove once one had proven $(1)$.
$endgroup$
– clathratus
Jan 26 at 21:20
add a comment |
$begingroup$
Just a generalization of @Zacky's answer
$$F(a)=int_0^{infty}log^2(x^a)sin(x^2)mathrm dx$$
Since $log(x^a)=log(e^{alog x})=alog x$,
$$F(a)=a^2int_0^{infty}log^2(x)sin(x^2)mathrm dx$$
$$F(a)=a^2F(1)$$
And as @Zacky showed,
$$F(1)=frac18mathrm{D}^2_{s=frac12}Gamma(s)sinfrac{pi s}{2}=frac1{32}sqrt{fracpi2}(2gamma-pi+log16)^2$$
So
$$F(a)=frac{a^2}{32}sqrt{fracpi2}(2gamma-pi+log16)^2$$
I will edit my answer to include a proof of my own once I find one.
answered Jan 26 at 21:05
clathratusclathratus
5,1701338
$endgroup$
Just a generalization of @Zacky's answer
$$F(a)=int_0^{infty}log^2(x^a)sin(x^2)mathrm dx$$
Since $log(x^a)=log(e^{alog x})=alog x$,
$$F(a)=a^2int_0^{infty}log^2(x)sin(x^2)mathrm dx$$
$$F(a)=a^2F(1)$$
And as @Zacky showed,
$$F(1)=frac18mathrm{D}^2_{s=frac12}Gamma(s)sinfrac{pi s}{2}=frac1{32}sqrt{fracpi2}(2gamma-pi+log16)^2$$
So
$$F(a)=frac{a^2}{32}sqrt{fracpi2}(2gamma-pi+log16)^2$$
I will edit my answer to include a proof of my own once I find one.
answered Jan 26 at 21:05
clathratusclathratus
5,1701338
answered Jan 26 at 21:05
clathratusclathratus
5,1701338
answered Jan 26 at 21:05
clathratusclathratus
5,1701338
answered Jan 26 at 21:05
clathratusclathratus
5,1701338
5,1701338
$begingroup$
Nice, I forgot the fact that $log(x^a)=alog(x)$.
$endgroup$
– Larry
Jan 26 at 21:14
$begingroup$
@Larry I legitimately can't tell if you're being sarcastic or not
$endgroup$
– clathratus
Jan 26 at 21:15
1
$begingroup$
I am not being sarcastic.
$endgroup$
– Larry
Jan 26 at 21:18
1
$begingroup$
@Larry Thank you for the clarification. You were very right when you mentioned that $(2)$ would be easy to prove once one had proven $(1)$.
$endgroup$
– clathratus
Jan 26 at 21:20
add a comment |
$begingroup$
Nice, I forgot the fact that $log(x^a)=alog(x)$.
$endgroup$
– Larry
Jan 26 at 21:14
$begingroup$
@Larry I legitimately can't tell if you're being sarcastic or not
$endgroup$
– clathratus
Jan 26 at 21:15
1
$begingroup$
I am not being sarcastic.
$endgroup$
– Larry
Jan 26 at 21:18
1
$begingroup$
@Larry Thank you for the clarification. You were very right when you mentioned that $(2)$ would be easy to prove once one had proven $(1)$.
$endgroup$
– clathratus
Jan 26 at 21:20
$begingroup$
Nice, I forgot the fact that $log(x^a)=alog(x)$.
$endgroup$
– Larry
Jan 26 at 21:14
$begingroup$
Nice, I forgot the fact that $log(x^a)=alog(x)$.
$endgroup$
– Larry
Jan 26 at 21:14
$begingroup$
@Larry I legitimately can't tell if you're being sarcastic or not
$endgroup$
– clathratus
Jan 26 at 21:15
$begingroup$
@Larry I legitimately can't tell if you're being sarcastic or not
$endgroup$
– clathratus
Jan 26 at 21:15
1
1
$begingroup$
I am not being sarcastic.
$endgroup$
– Larry
Jan 26 at 21:18
$begingroup$
I am not being sarcastic.
$endgroup$
– Larry
Jan 26 at 21:18
1
1
$begingroup$
@Larry Thank you for the clarification. You were very right when you mentioned that $(2)$ would be easy to prove once one had proven $(1)$.
$endgroup$
– clathratus
Jan 26 at 21:20
$begingroup$
@Larry Thank you for the clarification. You were very right when you mentioned that $(2)$ would be easy to prove once one had proven $(1)$.
$endgroup$
– clathratus
Jan 26 at 21:20
add a comment |
$begingroup$
An alternative approach is to employ Feynman's Trick and Laplace Transforms to solve:
begin{equation}
I = int_0^inftyln^2(x)sinleft(x^2right):dx
end{equation}
We first observe that:
begin{equation}
I = int_0^inftyln^2(x)sinleft(x^2right):dx = lim_{krightarrow 0^+} frac{d^2}{dk^2}int_0^infty x^ksinleft(x^2right):dx = lim_{krightarrow 0^+} frac{d^2}{dk^2} H(k)
end{equation}
We proceed by solving $H(k)$. To do so, we introduce a new parameter $'t'$:
begin{equation}
J(t; k) = int_0^infty x^ksinleft(tx^2right):dx
end{equation}
(This is allowable through the Dominated Convergence Theorem). Thus:
begin{equation}
H(k) = lim_{trightarrow 1^+} J(t; k)
end{equation}
Using Fubini's Theorem we now take the Laplace Transform with respect to '$t$'
begin{align}
mathscr{L}_tleft[J(t;k) right] &= int_0^infty x^kmathscr{L}_tleft[sinleft(tx^2right)right]:dx = int_0^infty frac{x^{k + 2}}{s^2 + x^4}:dx
end{align}
As I address here we find this becomes:
begin{align}
mathscr{L}_tleft[J(t;k) right] &= frac{1}{4}cdot left(s^2right)^{frac{k + 2 + 1}{2} - 1} cdot Bleft(1 - frac{k + 2 + 1 }{4}, frac{k + 2 + 1 }{4} right) = frac{1}{4} s^{frac{k - 1}{2}} Bleft(1 - frac{k + 3}{4} , frac{k + 3}{4}right)
end{align}
Using the relationship between the Gamma and Beta Function we find:
begin{equation}
mathscr{L}_tleft[J(t;k) right] = frac{1}{4} s^{frac{k - 1}{2}} Gammaleft(1 - frac{k + 3}{4}right) Gammaleft( frac{k + 3}{4}right)
end{equation}
Using Euler's Reflection Formula we find:
begin{equation}
mathscr{L}_tleft[J(t;k) right] = frac{1}{4} s^{frac{k - 1}{2}} frac{pi}{sinleft(pileft(frac{k + 3}{4}right) right)}
end{equation}
Taking the inverse Laplace Transforms is rather tricky here. To evaluate recall that:
begin{equation}
I = lim_{krightarrow 0^+} frac{d^2}{dk^2} H(k) = lim_{krightarrow 0^+} frac{d^2}{dk^2}left[ lim_{trightarrow 1^+} J(t;k)right]
end{equation}
In this process we solve for $H(k)$ using
begin{equation}
H(k) = lim_{trightarrow 1^+} mathscr{L}_s^{-1}left[mathscr{L}_tleft[J(t; k)right]right]
end{equation}
Thus, our definition of $I$ becomes:
begin{align}
I &= lim_{krightarrow 0^+} frac{d^2}{dk^2} H(k) = lim_{krightarrow 0^+} frac{d^2}{dk^2}left[ lim_{trightarrow 1^+} J(t;k)right] = lim_{krightarrow 0^+} frac{d^2}{dk^2}left[ lim_{trightarrow 1^+} mathscr{L}_s^{-1}left[mathscr{L}_tleft[J(t; k)right]right]right] \
&= lim_{trightarrow 1^+} mathscr{L}_s^{-1}left[ lim_{krightarrow 0^+} frac{d^2}{dk^2}mathscr{L}_tleft[J(t; k)right]right] = lim_{trightarrow 1^+} mathscr{L}_s^{-1}left[ lim_{krightarrow 0^+} frac{d^2}{dk^2}left[ frac{1}{4} s^{frac{k - 1}{2}} frac{pi}{sinleft(pileft(frac{k + 3}{4}right) right)}right]right]
end{align}
Because I'm lazy, I used Wolframalpha to evaluate the second derivate at $0$:
begin{align}
I &= lim_{trightarrow 1^+} mathscr{L}_s^{-1}left[ lim_{krightarrow 0^+} frac{d^2}{dk^2}left[ frac{1}{4} s^{frac{k - 1}{2}} frac{pi}{sinleft(pileft(frac{k + 3}{4}right) right)}right]right] = lim_{trightarrow 1^+} mathscr{L}_s^{-1}left[ frac{pi}{4}left( frac{3pi^2}{8sqrt{2}sqrt{s}} + frac{ln^2(s)}{2sqrt{2}sqrt{s}} + frac{piln(s)}{2sqrt{2}sqrt{s}}right)right] \
&= lim_{trightarrow 1^+} left[ frac{3pi^3}{32sqrt{2}} mathscr{L}_s^{-1}left[ frac{1}{sqrt{s}}right] + frac{pi}{8sqrt{2}} mathscr{L}_s^{-1}left[ frac{ln^2(s)}{sqrt{s}}right]+ frac{pi^2}{8sqrt{2}} mathscr{L}_s^{-1}left[ frac{ln(s)}{sqrt{s}}right]right] \
&= lim_{trightarrow 1^+} left[ frac{3pi^3}{32sqrt{2}} left[ frac{1}{sqrt{pi}sqrt{t}}right] + frac{pi}{32sqrt{2}} left[ frac{
left(psi^{(0)}left(frac{1}{2}right)-ln(t)right)^2 -frac{pi^2}{2}}{sqrt{pi}sqrt{t}}right]+ frac{pi^2}{16sqrt{2}} left[ frac{
psi^{(0)}left(frac{1}{2}right)-ln(t)}{sqrt{pi}sqrt{t}}right]right] \
&= frac{3pi^3}{32sqrt{2}} left[ frac{1}{sqrt{pi}}right] + frac{pi}{32sqrt{2}} left[ frac{
psi^{(0)}left(frac{1}{2}right)^2 -frac{pi^2}{2}}{sqrt{pi}}right]+ frac{pi^2}{16sqrt{2}} left[ frac{
psi^{(0)}left(frac{1}{2}right)}{sqrt{pi}}right]
end{align}
Noting that
begin{equation}
psi^{(0)}left(frac{1}{2}right) = -gamma - 2ln(2)
end{equation}
Where $gamma$ is the Euler–Mascheroni constant.
Thus,
begin{align}
I = frac{3pi^3}{32sqrt{2}} left[ frac{1}{sqrt{pi}}right] + frac{pi}{32sqrt{2}} left[ frac{
left(gamma + 2ln(2)right)^2 -frac{pi^2}{2}}{sqrt{pi}}right]+ frac{pi^2}{16sqrt{2}} left[ frac{
gamma - 2ln(2)}{sqrt{pi}}right] = frac{1}{32}sqrt{frac{pi}{2}}(2gamma-pi+4ln2)^2
end{align}
answered Jan 28 at 12:51
DavidGDavidG
1
$endgroup$
$begingroup$
Very nice solution! I am a fan of your method.
$endgroup$
– clathratus
Jan 28 at 16:07
$begingroup$
Do you use a similar technique to evaluate $$F(n)=int_0^infty cos(x^n)mathrm dx$$
$endgroup$
– clathratus
Feb 6 at 20:58
$begingroup$
@clathratus You definitely can
$endgroup$
– DavidG
Feb 7 at 0:09
add a comment |
$begingroup$
An alternative approach is to employ Feynman's Trick and Laplace Transforms to solve:
begin{equation}
I = int_0^inftyln^2(x)sinleft(x^2right):dx
end{equation}
We first observe that:
begin{equation}
I = int_0^inftyln^2(x)sinleft(x^2right):dx = lim_{krightarrow 0^+} frac{d^2}{dk^2}int_0^infty x^ksinleft(x^2right):dx = lim_{krightarrow 0^+} frac{d^2}{dk^2} H(k)
end{equation}
We proceed by solving $H(k)$. To do so, we introduce a new parameter $'t'$:
begin{equation}
J(t; k) = int_0^infty x^ksinleft(tx^2right):dx
end{equation}
(This is allowable through the Dominated Convergence Theorem). Thus:
begin{equation}
H(k) = lim_{trightarrow 1^+} J(t; k)
end{equation}
Using Fubini's Theorem we now take the Laplace Transform with respect to '$t$'
begin{align}
mathscr{L}_tleft[J(t;k) right] &= int_0^infty x^kmathscr{L}_tleft[sinleft(tx^2right)right]:dx = int_0^infty frac{x^{k + 2}}{s^2 + x^4}:dx
end{align}
As I address here we find this becomes:
begin{align}
mathscr{L}_tleft[J(t;k) right] &= frac{1}{4}cdot left(s^2right)^{frac{k + 2 + 1}{2} - 1} cdot Bleft(1 - frac{k + 2 + 1 }{4}, frac{k + 2 + 1 }{4} right) = frac{1}{4} s^{frac{k - 1}{2}} Bleft(1 - frac{k + 3}{4} , frac{k + 3}{4}right)
end{align}
Using the relationship between the Gamma and Beta Function we find:
begin{equation}
mathscr{L}_tleft[J(t;k) right] = frac{1}{4} s^{frac{k - 1}{2}} Gammaleft(1 - frac{k + 3}{4}right) Gammaleft( frac{k + 3}{4}right)
end{equation}
Using Euler's Reflection Formula we find:
begin{equation}
mathscr{L}_tleft[J(t;k) right] = frac{1}{4} s^{frac{k - 1}{2}} frac{pi}{sinleft(pileft(frac{k + 3}{4}right) right)}
end{equation}
Taking the inverse Laplace Transforms is rather tricky here. To evaluate recall that:
begin{equation}
I = lim_{krightarrow 0^+} frac{d^2}{dk^2} H(k) = lim_{krightarrow 0^+} frac{d^2}{dk^2}left[ lim_{trightarrow 1^+} J(t;k)right]
end{equation}
In this process we solve for $H(k)$ using
begin{equation}
H(k) = lim_{trightarrow 1^+} mathscr{L}_s^{-1}left[mathscr{L}_tleft[J(t; k)right]right]
end{equation}
Thus, our definition of $I$ becomes:
begin{align}
I &= lim_{krightarrow 0^+} frac{d^2}{dk^2} H(k) = lim_{krightarrow 0^+} frac{d^2}{dk^2}left[ lim_{trightarrow 1^+} J(t;k)right] = lim_{krightarrow 0^+} frac{d^2}{dk^2}left[ lim_{trightarrow 1^+} mathscr{L}_s^{-1}left[mathscr{L}_tleft[J(t; k)right]right]right] \
&= lim_{trightarrow 1^+} mathscr{L}_s^{-1}left[ lim_{krightarrow 0^+} frac{d^2}{dk^2}mathscr{L}_tleft[J(t; k)right]right] = lim_{trightarrow 1^+} mathscr{L}_s^{-1}left[ lim_{krightarrow 0^+} frac{d^2}{dk^2}left[ frac{1}{4} s^{frac{k - 1}{2}} frac{pi}{sinleft(pileft(frac{k + 3}{4}right) right)}right]right]
end{align}
Because I'm lazy, I used Wolframalpha to evaluate the second derivate at $0$:
begin{align}
I &= lim_{trightarrow 1^+} mathscr{L}_s^{-1}left[ lim_{krightarrow 0^+} frac{d^2}{dk^2}left[ frac{1}{4} s^{frac{k - 1}{2}} frac{pi}{sinleft(pileft(frac{k + 3}{4}right) right)}right]right] = lim_{trightarrow 1^+} mathscr{L}_s^{-1}left[ frac{pi}{4}left( frac{3pi^2}{8sqrt{2}sqrt{s}} + frac{ln^2(s)}{2sqrt{2}sqrt{s}} + frac{piln(s)}{2sqrt{2}sqrt{s}}right)right] \
&= lim_{trightarrow 1^+} left[ frac{3pi^3}{32sqrt{2}} mathscr{L}_s^{-1}left[ frac{1}{sqrt{s}}right] + frac{pi}{8sqrt{2}} mathscr{L}_s^{-1}left[ frac{ln^2(s)}{sqrt{s}}right]+ frac{pi^2}{8sqrt{2}} mathscr{L}_s^{-1}left[ frac{ln(s)}{sqrt{s}}right]right] \
&= lim_{trightarrow 1^+} left[ frac{3pi^3}{32sqrt{2}} left[ frac{1}{sqrt{pi}sqrt{t}}right] + frac{pi}{32sqrt{2}} left[ frac{
left(psi^{(0)}left(frac{1}{2}right)-ln(t)right)^2 -frac{pi^2}{2}}{sqrt{pi}sqrt{t}}right]+ frac{pi^2}{16sqrt{2}} left[ frac{
psi^{(0)}left(frac{1}{2}right)-ln(t)}{sqrt{pi}sqrt{t}}right]right] \
&= frac{3pi^3}{32sqrt{2}} left[ frac{1}{sqrt{pi}}right] + frac{pi}{32sqrt{2}} left[ frac{
psi^{(0)}left(frac{1}{2}right)^2 -frac{pi^2}{2}}{sqrt{pi}}right]+ frac{pi^2}{16sqrt{2}} left[ frac{
psi^{(0)}left(frac{1}{2}right)}{sqrt{pi}}right]
end{align}
Noting that
begin{equation}
psi^{(0)}left(frac{1}{2}right) = -gamma - 2ln(2)
end{equation}
Where $gamma$ is the Euler–Mascheroni constant.
Thus,
begin{align}
I = frac{3pi^3}{32sqrt{2}} left[ frac{1}{sqrt{pi}}right] + frac{pi}{32sqrt{2}} left[ frac{
left(gamma + 2ln(2)right)^2 -frac{pi^2}{2}}{sqrt{pi}}right]+ frac{pi^2}{16sqrt{2}} left[ frac{
gamma - 2ln(2)}{sqrt{pi}}right] = frac{1}{32}sqrt{frac{pi}{2}}(2gamma-pi+4ln2)^2
end{align}
answered Jan 28 at 12:51
DavidGDavidG
1
$endgroup$
$begingroup$
Very nice solution! I am a fan of your method.
$endgroup$
– clathratus
Jan 28 at 16:07
$begingroup$
Do you use a similar technique to evaluate $$F(n)=int_0^infty cos(x^n)mathrm dx$$
$endgroup$
– clathratus
Feb 6 at 20:58
$begingroup$
@clathratus You definitely can
$endgroup$
– DavidG
Feb 7 at 0:09
add a comment |
$begingroup$
An alternative approach is to employ Feynman's Trick and Laplace Transforms to solve:
begin{equation}
I = int_0^inftyln^2(x)sinleft(x^2right):dx
end{equation}
We first observe that:
begin{equation}
I = int_0^inftyln^2(x)sinleft(x^2right):dx = lim_{krightarrow 0^+} frac{d^2}{dk^2}int_0^infty x^ksinleft(x^2right):dx = lim_{krightarrow 0^+} frac{d^2}{dk^2} H(k)
end{equation}
We proceed by solving $H(k)$. To do so, we introduce a new parameter $'t'$:
begin{equation}
J(t; k) = int_0^infty x^ksinleft(tx^2right):dx
end{equation}
(This is allowable through the Dominated Convergence Theorem). Thus:
begin{equation}
H(k) = lim_{trightarrow 1^+} J(t; k)
end{equation}
Using Fubini's Theorem we now take the Laplace Transform with respect to '$t$'
begin{align}
mathscr{L}_tleft[J(t;k) right] &= int_0^infty x^kmathscr{L}_tleft[sinleft(tx^2right)right]:dx = int_0^infty frac{x^{k + 2}}{s^2 + x^4}:dx
end{align}
As I address here we find this becomes:
begin{align}
mathscr{L}_tleft[J(t;k) right] &= frac{1}{4}cdot left(s^2right)^{frac{k + 2 + 1}{2} - 1} cdot Bleft(1 - frac{k + 2 + 1 }{4}, frac{k + 2 + 1 }{4} right) = frac{1}{4} s^{frac{k - 1}{2}} Bleft(1 - frac{k + 3}{4} , frac{k + 3}{4}right)
end{align}
Using the relationship between the Gamma and Beta Function we find:
begin{equation}
mathscr{L}_tleft[J(t;k) right] = frac{1}{4} s^{frac{k - 1}{2}} Gammaleft(1 - frac{k + 3}{4}right) Gammaleft( frac{k + 3}{4}right)
end{equation}
Using Euler's Reflection Formula we find:
begin{equation}
mathscr{L}_tleft[J(t;k) right] = frac{1}{4} s^{frac{k - 1}{2}} frac{pi}{sinleft(pileft(frac{k + 3}{4}right) right)}
end{equation}
Taking the inverse Laplace Transforms is rather tricky here. To evaluate recall that:
begin{equation}
I = lim_{krightarrow 0^+} frac{d^2}{dk^2} H(k) = lim_{krightarrow 0^+} frac{d^2}{dk^2}left[ lim_{trightarrow 1^+} J(t;k)right]
end{equation}
In this process we solve for $H(k)$ using
begin{equation}
H(k) = lim_{trightarrow 1^+} mathscr{L}_s^{-1}left[mathscr{L}_tleft[J(t; k)right]right]
end{equation}
Thus, our definition of $I$ becomes:
begin{align}
I &= lim_{krightarrow 0^+} frac{d^2}{dk^2} H(k) = lim_{krightarrow 0^+} frac{d^2}{dk^2}left[ lim_{trightarrow 1^+} J(t;k)right] = lim_{krightarrow 0^+} frac{d^2}{dk^2}left[ lim_{trightarrow 1^+} mathscr{L}_s^{-1}left[mathscr{L}_tleft[J(t; k)right]right]right] \
&= lim_{trightarrow 1^+} mathscr{L}_s^{-1}left[ lim_{krightarrow 0^+} frac{d^2}{dk^2}mathscr{L}_tleft[J(t; k)right]right] = lim_{trightarrow 1^+} mathscr{L}_s^{-1}left[ lim_{krightarrow 0^+} frac{d^2}{dk^2}left[ frac{1}{4} s^{frac{k - 1}{2}} frac{pi}{sinleft(pileft(frac{k + 3}{4}right) right)}right]right]
end{align}
Because I'm lazy, I used Wolframalpha to evaluate the second derivate at $0$:
begin{align}
I &= lim_{trightarrow 1^+} mathscr{L}_s^{-1}left[ lim_{krightarrow 0^+} frac{d^2}{dk^2}left[ frac{1}{4} s^{frac{k - 1}{2}} frac{pi}{sinleft(pileft(frac{k + 3}{4}right) right)}right]right] = lim_{trightarrow 1^+} mathscr{L}_s^{-1}left[ frac{pi}{4}left( frac{3pi^2}{8sqrt{2}sqrt{s}} + frac{ln^2(s)}{2sqrt{2}sqrt{s}} + frac{piln(s)}{2sqrt{2}sqrt{s}}right)right] \
&= lim_{trightarrow 1^+} left[ frac{3pi^3}{32sqrt{2}} mathscr{L}_s^{-1}left[ frac{1}{sqrt{s}}right] + frac{pi}{8sqrt{2}} mathscr{L}_s^{-1}left[ frac{ln^2(s)}{sqrt{s}}right]+ frac{pi^2}{8sqrt{2}} mathscr{L}_s^{-1}left[ frac{ln(s)}{sqrt{s}}right]right] \
&= lim_{trightarrow 1^+} left[ frac{3pi^3}{32sqrt{2}} left[ frac{1}{sqrt{pi}sqrt{t}}right] + frac{pi}{32sqrt{2}} left[ frac{
left(psi^{(0)}left(frac{1}{2}right)-ln(t)right)^2 -frac{pi^2}{2}}{sqrt{pi}sqrt{t}}right]+ frac{pi^2}{16sqrt{2}} left[ frac{
psi^{(0)}left(frac{1}{2}right)-ln(t)}{sqrt{pi}sqrt{t}}right]right] \
&= frac{3pi^3}{32sqrt{2}} left[ frac{1}{sqrt{pi}}right] + frac{pi}{32sqrt{2}} left[ frac{
psi^{(0)}left(frac{1}{2}right)^2 -frac{pi^2}{2}}{sqrt{pi}}right]+ frac{pi^2}{16sqrt{2}} left[ frac{
psi^{(0)}left(frac{1}{2}right)}{sqrt{pi}}right]
end{align}
Noting that
begin{equation}
psi^{(0)}left(frac{1}{2}right) = -gamma - 2ln(2)
end{equation}
Where $gamma$ is the Euler–Mascheroni constant.
Thus,
begin{align}
I = frac{3pi^3}{32sqrt{2}} left[ frac{1}{sqrt{pi}}right] + frac{pi}{32sqrt{2}} left[ frac{
left(gamma + 2ln(2)right)^2 -frac{pi^2}{2}}{sqrt{pi}}right]+ frac{pi^2}{16sqrt{2}} left[ frac{
gamma - 2ln(2)}{sqrt{pi}}right] = frac{1}{32}sqrt{frac{pi}{2}}(2gamma-pi+4ln2)^2
end{align}
answered Jan 28 at 12:51
DavidGDavidG
1
$endgroup$
An alternative approach is to employ Feynman's Trick and Laplace Transforms to solve:
begin{equation}
I = int_0^inftyln^2(x)sinleft(x^2right):dx
end{equation}
We first observe that:
begin{equation}
I = int_0^inftyln^2(x)sinleft(x^2right):dx = lim_{krightarrow 0^+} frac{d^2}{dk^2}int_0^infty x^ksinleft(x^2right):dx = lim_{krightarrow 0^+} frac{d^2}{dk^2} H(k)
end{equation}
We proceed by solving $H(k)$. To do so, we introduce a new parameter $'t'$:
begin{equation}
J(t; k) = int_0^infty x^ksinleft(tx^2right):dx
end{equation}
(This is allowable through the Dominated Convergence Theorem). Thus:
begin{equation}
H(k) = lim_{trightarrow 1^+} J(t; k)
end{equation}
Using Fubini's Theorem we now take the Laplace Transform with respect to '$t$'
begin{align}
mathscr{L}_tleft[J(t;k) right] &= int_0^infty x^kmathscr{L}_tleft[sinleft(tx^2right)right]:dx = int_0^infty frac{x^{k + 2}}{s^2 + x^4}:dx
end{align}
As I address here we find this becomes:
begin{align}
mathscr{L}_tleft[J(t;k) right] &= frac{1}{4}cdot left(s^2right)^{frac{k + 2 + 1}{2} - 1} cdot Bleft(1 - frac{k + 2 + 1 }{4}, frac{k + 2 + 1 }{4} right) = frac{1}{4} s^{frac{k - 1}{2}} Bleft(1 - frac{k + 3}{4} , frac{k + 3}{4}right)
end{align}
Using the relationship between the Gamma and Beta Function we find:
begin{equation}
mathscr{L}_tleft[J(t;k) right] = frac{1}{4} s^{frac{k - 1}{2}} Gammaleft(1 - frac{k + 3}{4}right) Gammaleft( frac{k + 3}{4}right)
end{equation}
Using Euler's Reflection Formula we find:
begin{equation}
mathscr{L}_tleft[J(t;k) right] = frac{1}{4} s^{frac{k - 1}{2}} frac{pi}{sinleft(pileft(frac{k + 3}{4}right) right)}
end{equation}
Taking the inverse Laplace Transforms is rather tricky here. To evaluate recall that:
begin{equation}
I = lim_{krightarrow 0^+} frac{d^2}{dk^2} H(k) = lim_{krightarrow 0^+} frac{d^2}{dk^2}left[ lim_{trightarrow 1^+} J(t;k)right]
end{equation}
In this process we solve for $H(k)$ using
begin{equation}
H(k) = lim_{trightarrow 1^+} mathscr{L}_s^{-1}left[mathscr{L}_tleft[J(t; k)right]right]
end{equation}
Thus, our definition of $I$ becomes:
begin{align}
I &= lim_{krightarrow 0^+} frac{d^2}{dk^2} H(k) = lim_{krightarrow 0^+} frac{d^2}{dk^2}left[ lim_{trightarrow 1^+} J(t;k)right] = lim_{krightarrow 0^+} frac{d^2}{dk^2}left[ lim_{trightarrow 1^+} mathscr{L}_s^{-1}left[mathscr{L}_tleft[J(t; k)right]right]right] \
&= lim_{trightarrow 1^+} mathscr{L}_s^{-1}left[ lim_{krightarrow 0^+} frac{d^2}{dk^2}mathscr{L}_tleft[J(t; k)right]right] = lim_{trightarrow 1^+} mathscr{L}_s^{-1}left[ lim_{krightarrow 0^+} frac{d^2}{dk^2}left[ frac{1}{4} s^{frac{k - 1}{2}} frac{pi}{sinleft(pileft(frac{k + 3}{4}right) right)}right]right]
end{align}
Because I'm lazy, I used Wolframalpha to evaluate the second derivate at $0$:
begin{align}
I &= lim_{trightarrow 1^+} mathscr{L}_s^{-1}left[ lim_{krightarrow 0^+} frac{d^2}{dk^2}left[ frac{1}{4} s^{frac{k - 1}{2}} frac{pi}{sinleft(pileft(frac{k + 3}{4}right) right)}right]right] = lim_{trightarrow 1^+} mathscr{L}_s^{-1}left[ frac{pi}{4}left( frac{3pi^2}{8sqrt{2}sqrt{s}} + frac{ln^2(s)}{2sqrt{2}sqrt{s}} + frac{piln(s)}{2sqrt{2}sqrt{s}}right)right] \
&= lim_{trightarrow 1^+} left[ frac{3pi^3}{32sqrt{2}} mathscr{L}_s^{-1}left[ frac{1}{sqrt{s}}right] + frac{pi}{8sqrt{2}} mathscr{L}_s^{-1}left[ frac{ln^2(s)}{sqrt{s}}right]+ frac{pi^2}{8sqrt{2}} mathscr{L}_s^{-1}left[ frac{ln(s)}{sqrt{s}}right]right] \
&= lim_{trightarrow 1^+} left[ frac{3pi^3}{32sqrt{2}} left[ frac{1}{sqrt{pi}sqrt{t}}right] + frac{pi}{32sqrt{2}} left[ frac{
left(psi^{(0)}left(frac{1}{2}right)-ln(t)right)^2 -frac{pi^2}{2}}{sqrt{pi}sqrt{t}}right]+ frac{pi^2}{16sqrt{2}} left[ frac{
psi^{(0)}left(frac{1}{2}right)-ln(t)}{sqrt{pi}sqrt{t}}right]right] \
&= frac{3pi^3}{32sqrt{2}} left[ frac{1}{sqrt{pi}}right] + frac{pi}{32sqrt{2}} left[ frac{
psi^{(0)}left(frac{1}{2}right)^2 -frac{pi^2}{2}}{sqrt{pi}}right]+ frac{pi^2}{16sqrt{2}} left[ frac{
psi^{(0)}left(frac{1}{2}right)}{sqrt{pi}}right]
end{align}
Noting that
begin{equation}
psi^{(0)}left(frac{1}{2}right) = -gamma - 2ln(2)
end{equation}
Where $gamma$ is the Euler–Mascheroni constant.
Thus,
begin{align}
I = frac{3pi^3}{32sqrt{2}} left[ frac{1}{sqrt{pi}}right] + frac{pi}{32sqrt{2}} left[ frac{
left(gamma + 2ln(2)right)^2 -frac{pi^2}{2}}{sqrt{pi}}right]+ frac{pi^2}{16sqrt{2}} left[ frac{
gamma - 2ln(2)}{sqrt{pi}}right] = frac{1}{32}sqrt{frac{pi}{2}}(2gamma-pi+4ln2)^2
end{align}
answered Jan 28 at 12:51
DavidGDavidG
1
edited Jan 31 at 12:44
answered Jan 28 at 12:51
DavidGDavidG
1
answered Jan 28 at 12:51
DavidGDavidG
1
answered Jan 28 at 12:51
DavidGDavidG
1
1
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Very nice solution! I am a fan of your method.
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– clathratus
Jan 28 at 16:07
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Do you use a similar technique to evaluate $$F(n)=int_0^infty cos(x^n)mathrm dx$$
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– clathratus
Feb 6 at 20:58
$begingroup$
@clathratus You definitely can
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– DavidG
Feb 7 at 0:09
add a comment |
$begingroup$
Very nice solution! I am a fan of your method.
$endgroup$
– clathratus
Jan 28 at 16:07
$begingroup$
Do you use a similar technique to evaluate $$F(n)=int_0^infty cos(x^n)mathrm dx$$
$endgroup$
– clathratus
Feb 6 at 20:58
$begingroup$
@clathratus You definitely can
$endgroup$
– DavidG
Feb 7 at 0:09
$begingroup$
Very nice solution! I am a fan of your method.
$endgroup$
– clathratus
Jan 28 at 16:07
$begingroup$
Very nice solution! I am a fan of your method.
$endgroup$
– clathratus
Jan 28 at 16:07
$begingroup$
Do you use a similar technique to evaluate $$F(n)=int_0^infty cos(x^n)mathrm dx$$
$endgroup$
– clathratus
Feb 6 at 20:58
$begingroup$
Do you use a similar technique to evaluate $$F(n)=int_0^infty cos(x^n)mathrm dx$$
$endgroup$
– clathratus
Feb 6 at 20:58
$begingroup$
@clathratus You definitely can
$endgroup$
– DavidG
Feb 7 at 0:09
$begingroup$
@clathratus You definitely can
$endgroup$
– DavidG
Feb 7 at 0:09
add a comment |
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$begingroup$
Computer algebra solves this instantly: $frac{1}{32} sqrt{frac{pi }{2}} (2 gamma -pi +log (16))^2$. Is there really a need to perform this by hand? ted.com/talks/…
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– David G. Stork
Jan 26 at 20:10
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It is an interesting video. "Hand calculating procedures teach understanding."
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– Larry
Jan 26 at 20:16
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Did you actually watch the full video? One of Wolfram's key points was exposing the fallacy in "hand calculating procedures teach understanding." Very convincing.
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– David G. Stork
Jan 26 at 20:29
1
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@DavidG.Stork I entirely disagree. I personally am interested in knowing how to preform such manual calculations as they teach me new techniques which I can apply to other problems. Also, what is the point of knowing that $int_0^infty ln^2(x)sin(x^2)mathrm dx$ has a closed form (or exact evaluation instead of a decimal expansion) if one cannot prove it? Without a proof it is just a useless fact.
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– clathratus
Jan 26 at 20:51
2
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Ok, I see. That is actually a good point. I will keep that in mind.
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– Larry
Jan 26 at 21:12