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Integration Contest - Season 3

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Post Fri Aug 30, 2013 4:30 pm
galactus User avatar
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Random Variable wrote:
No one seems to have an evaluation of problem 44.

So here is another one.



Problem 45



\(\displaystyle \int_{0}^{\infty} \frac{x^{2}}{\cosh x + \cos a} \ dx = \frac{a(\pi^{2}-a^{2})}{3 \sin a} \ \ 0 < a < \pi\)



Hey G:

I think I have an eval for this one.

I started with:

\(\displaystyle \int_{0}^{\infty}\frac{e^{-sx}}{\cosh(x)+\cos(a)}dx=\frac{2}{\sin(a)}\sum_{k=1}^{\infty}\frac{(-1)^{k-1}\sin(ak)}{s+k}\)....[1]

To show this, write \(\displaystyle \frac{e^{-sx}}{\cosh(x)+\cos(a)}=\frac{e^{-sx}}{\sin(a)}\left(\frac{1}{1+e^{-(x+ia)}}-\frac{1}{1+e^{-(x-ia)}}\right)\)

\(\displaystyle =\frac{2e^{sx}}{\sin(a)}\sum_{k=0}^{\infty}(-1)^{k-1}e^{-xk}\sin(ak)\)

Integrating gives the above formula in [1].

Now, take [1] and diff twice w.r.t s, then let s=0:

This gives:

\(\displaystyle \int_{0}^{\infty}\frac{x^{2}}{\cosh(x)+\cos(a)}dx=\frac{4}{\sin(a)}\sum_{k=1}^{\infty}\frac{(-1)^{k-1}\sin(ka)}{k^{3}}\)

If the sin(ak) on the right is written in terms of its exponential, then we have a trilog series.

\(\displaystyle \int_{0}^{\infty}\frac{x^{2}}{\cosh(x)+\cos(a)}dx=\frac{2i}{\sin(a)}\left[Li_{3}(-e^{ia})-Li_{3}(-e^{-ia})\right]\)

This is the same as \(\displaystyle \frac{a(\pi^{2}-a^{2})}{3\sin(a)}\) by virtue of the identity

\(\displaystyle Li_{3}(z)-Li_{3}(1/z)=-1/6log^{3}(-z)+\frac{\pi^{2}}{6}log(-z)\)

\(\displaystyle \frac{2i}{\sin(a)}\left(\frac{-1}{6}log^{3}(e^{ia})+\frac{\pi^{2}}{6}log(e^{-ia})\right)\)

\(\displaystyle =\frac{a(\pi^{2}-a^{2})}{3\sin(a)}\)

Post Fri Aug 30, 2013 5:00 pm
Random Variable Integration Guru
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You could use the Fourier series for \(\displaystyle f(x) = x\), that is \(\displaystyle x = 2 \sum_{k=1}^{\infty} (-1)^{n-1} \frac{\sin kx}{k}\), and integrate twice.

Post Fri Aug 30, 2013 5:03 pm
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Yeah, there's is usually many ways to go about something. Fourier can be mighty handy.

Problem 44 remains undone. Heck, G, I figured you would have went through that one like Grant through Richmond using contours :):)

I dabbled with it a little, made a sub and got it down to something with a \(\displaystyle x^{2}+4\) and a plain cos(x), but then got a little stuck, got too busy, and then forgot about it.

Make the sub \(\displaystyle t=x-1/x\) gives:

\(\displaystyle \int_{-\infty}^{\infty}\frac{1}{(t^{2}+4)(1-a\cos(t)+a^{2})}dt\)

I tried the classic \(\displaystyle \cos(z)=\frac{z+z^{-1}}{2}\), then kind of drifted away.

I played around a little and arrived at some residues with e^2 terms as are in the solution. May be a viable approach afterall.

Post Fri Aug 30, 2013 7:17 pm
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problem 46:

\(\displaystyle \int_{0}^{\infty}\frac{\sinh(ax)\cosh(bx)}{\sinh(cx)}dx=\frac{\pi}{2c}\left(\frac{\sin(\pi a/c)}{\cos(\pi a/c)+\cos(\pi b/c)}\right)\)

Post Fri Aug 30, 2013 9:00 pm
Random Variable Integration Guru
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\(\displaystyle \sinh(ax+bx) = \sinh(ax) \cosh(bx) + \cosh(ax) \sinh(bx)\)


So \(\displaystyle \int_{0}^{\infty} \frac{\sinh (ax) \cosh(bx)}{\sinh(cx)} \ dx = \frac{1}{2} \int_{0}^{\infty} \frac{\sinh \big((a+b)x \big) + \sinh \big((a-b )x \big)}{\sinh (cx)} \ dx\)

\(\displaystyle = \frac{\pi}{2c} \Big[ \tan \Big(\frac{\pi(a+b)}{2c} \Big) + \tan \Big( \frac{ \pi(a-b)}{2c} \Big) \Big]\)


I tried to get my answer in that form, but I gave up after about 15 minutes.

Post Fri Aug 30, 2013 10:15 pm
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I got the same thing, G. I went about it differently, though. Ultimately, I used \(\displaystyle \psi(1/2+x)-\psi(1/2-x)=\pi\tan(\pi x)\)

Maybe note that \(\displaystyle \tan\left(\frac{\pi (a\pm b)}{2c}\right)=\frac{\sin \left(\frac{\pi (a\pm b)}{c}\right)}{1+\cos\left(\frac{\pi (a\pm b)}{c}\right)}\)

It looked encouraging, but then I got tired. :cry:

Post Sat Aug 31, 2013 7:37 pm
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zaidalyafey wrote:
Hi RV , nice solution :) . Can we prove that \(\displaystyle \zeta(-2n)=0\) ?


Hey Z, I just noticed this.

I think one way to show is to use ol' Riemann's functional equation.

\(\displaystyle \zeta(1-s)=2(2\pi)^{-s}\Gamma(s)\cos\left(\frac{\pi s}{2}\right)\zeta(s)\)

Now, let \(\displaystyle s=2n+1\)

\(\displaystyle \zeta(-2n)=2(2\pi)^{-(2n+1)}\Gamma(2n+1)\cos(\frac{\pi}{2}(2n+1))\zeta(2n+1)\)

Note that \(\displaystyle \cos(\frac{\pi}{2}(2n+1))=-\sin(\pi n)=0\), and so \(\displaystyle \zeta(-2n)=0\)

Post Sat Aug 31, 2013 7:44 pm
zaidalyafey Global Moderator
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Hey C . Yes I know that but I meant using the following integral representation


Random Variable wrote:
Problem 40(b)


\(\displaystyle \zeta(s) = \frac{2^{s-1}}{1-2^{1-s}} \int_{0}^{\infty} \frac{\cos (s \arctan t)}{(1+t^{2})^{s/2} \cosh \left( \frac{\pi t}{2} \right)} \ dt\)

Wanna learn what we discuss , see Book

Post Sat Aug 31, 2013 10:42 pm
Random Variable Integration Guru
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Problem 47


\(\displaystyle \int_{0}^{1} K(k) \ dk = 2G\) where \(\displaystyle K(k)\) is the complete elliptical integral of the first kind and \(\displaystyle G\) is Catalan's constant

Post Sun Sep 01, 2013 3:10 am
Shobhit Site Admin
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Random Variable wrote:
Problem 47


\(\displaystyle \int_{0}^{1} K(k) \ dk = 2G\) where \(\displaystyle K(k)\) is the complete elliptical integral of the first kind and \(\displaystyle G\) is Catalan's constant


Nice problem! These integrals are called "Moments of Elliptic Integrals".

We have

\(\displaystyle \begin{align*}
I &=\int_0^1 \int_0^{\frac{\pi}{2}} \frac{1}{\sqrt{1-k^2 \sin^2 \theta}}d\theta \ dk \\
&= \int_0^{\frac{\pi}{2}} \int_0^1 \frac{1}{\sqrt{1-k^2 \sin^2 \theta}}\ dk d\theta \\
&= \int_0^{\frac{\pi}{2}} \frac{\theta}{\sin \theta} d\theta \\
&= -\int_0^{\frac{\pi}{2}}\log \tan \frac{\theta}{2} \ d\theta \\
&= -2 \int_0^{\frac{\pi}{4}} \log \tan \theta \ d\theta \\
&= -2 \int_0^1 \frac{\log x}{1+x^2}dx \quad x=\tan \theta \\
&= -2 \sum_{n=0}^\infty (-1)^n \int_0^1 x^{2n}\log(x) \ dx \\
&= 2\sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)^2} \\
&= 2G
\end{align*}\)

The following closed form is also known to exist:

\(\displaystyle \int_0^1 K'(k)^3 \ dk = \frac{\Gamma^8 \left(\frac{1}{4}\right)}{128\pi^2}\)

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