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Integration Contest - Season 3

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Post Wed Jul 03, 2013 7:50 am
Shobhit Site Admin
Site Admin

Posts: 850
Location: Jaipur, India

Welcome! I announce the Beginning of another epic Integration Contest (INC3)! The aim of the Integration Contest is to improve skills in the computation of integrals and to learn from each other as much as possible.


You can check out the previous integration contests -
The rules are as before.


  • I will start by posting the first problem. If anybody solves it, he is given the ability to post a new one.
  • You may post only one problem at a time.
  • The Scope of questions is only computation of integrals. You are allowed to post complex as well as multiple integrals.
  • The final answer can contain special functions of any kinds.
  • If someone solves a problem and he has no another to post then he should indicate that.
  • This time we do not have any points for an answer.

Post Thu Jul 04, 2013 2:17 pm
Shobhit Site Admin
Site Admin

Posts: 850
Location: Jaipur, India

Let us Begin NOW! :)

  • Problem 1

Prove that

\(\displaystyle \int_0^\infty e^{-ax} \sin (ax) \left(\frac{1}{\tan x}+\frac{1}{\tanh x} \right)dx = \frac{\pi}{2}\frac{\sinh(\pi a)}{\cosh(\pi a)-\cos(\pi a)} \quad a>0\)

This integral was Ramanujan's discovery.

Post Fri Jul 05, 2013 4:32 pm
Shobhit Site Admin
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Posts: 850
Location: Jaipur, India

Should I post the solution? Are you interested? :|

Post Fri Jul 05, 2013 8:06 pm
zaidalyafey Global Moderator
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Posts: 357
I didn't know that the contest is already on. Please give us more time I had internet problems last two days .

You said the contest will start on Monday, next week!
Wanna learn what we discuss , see Book

Post Sat Jul 06, 2013 11:35 am
galactus User avatar
Global Moderator
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Posts: 902
I initially attempted this with contours. But, then I tried series.

The coth is a wee bit easier using series because we can write coth as

\(\displaystyle \coth(x)=\frac{e^{2x}+1}{e^{2x}-1}\)

So, we have:

\(\displaystyle \int_{0}^{\infty}e^{-ax}\sin(ax)\left(\frac{e^{2x}+1}{e^{2x}-1}\right)dx\)

\(\displaystyle =\int_{0}^{\infty}e^{-ax}\sin(ax)dx+2\int_{0}^{\infty}e^{-ax}\sin(ax)\cdot \frac{1}{e^{2x}-1}dx\)

\(\displaystyle =\frac{1}{2a}+2\int_{0}^{\infty}\sin(ax)\sum_{n=1}^{\infty}e^{-(a+2n)x}dx\)

\(\displaystyle =\frac{1}{2a}+2\sum_{n=1}^{\infty}\int_{0}^{\infty}e^{-(a+2n)x}\sin(ax)dx\)

\(\displaystyle =\frac{1}{2a}+\sum_{n=1}^{\infty}\frac{a}{a^{2}+2an+2n^{2}}\)

Now, the cot one requires some sort of identity that'll work.

I do know a closed form sum for this, but have not derived it yet.

\(\displaystyle \sum_{n=1}^{\infty}\frac{a}{2a^{2}-4an+4a^{2}}-\sum_{n=1}^{\infty}\frac{a}{2a^{2}+4an+4a^{2}}\)

Now, add them and get:

\(\displaystyle =\frac{1}{2a}+\frac{1}{2}\sum_{n=1}^{\infty}\frac{a}{a^{2}+2an+2n^{2}}+\frac{1}{2}\sum_{n=1}^{\infty}\frac{a}{a^{2}-2an+2n^{2}}\)

\(\displaystyle =\frac{\pi}{2}\cdot \frac{\sinh(\pi a)}{\cosh(\pi a)-\cos(\pi a)}\)

This evaluates to the required result, but as I said, I have not worked it out by hand.

E, you can go ahead and post your method. It is, no doubt, better than mine.

I managed to make some headway on this using contours. But, not a nice, full solution.

The poles are infinite and the resulting series depend on whether 'a' is even or odd.

So, the above solution can be reduced down to tanh and coth terms depending on odd or even 'a'.


EDIT: I found the solution to this on that German site. Here, they use a sin identity for the cot term.

http://de.wikibooks.org/wiki/Formelsamm ... ot,coth%29

Post Sun Jul 07, 2013 9:01 am
zaidalyafey Global Moderator
Global Moderator

Posts: 357
I officially give up , no more Ramanujan's discoveries :lol: .
Wanna learn what we discuss , see Book

Post Sun Jul 07, 2013 9:40 am
galactus User avatar
Global Moderator
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Posts: 902
No doubt. We have to remember that Ramanujan was a gifted fellow.

Anyway, I was interested in a contour method of doing this one.

Here is a contour solution that Ron Gordon posted on SE. It's cool and clever.

http://math.stackexchange.com/questions ... ng-contour

Post Sun Jul 07, 2013 10:34 am
galactus User avatar
Global Moderator
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Posts: 902
Well, anyway, enough of that one. It was a goodie, though.

Here is an indefinite one that is kind of curious.

Problem #2:

\(\displaystyle \int\left(\frac{\tan^{-1}(x)}{x-\tan^{-1}(x)}\right)^{2}dx=\frac{1+x\tan^{-1}(x)}{\tan^{-1}(x)-x}+C\)

Post Sun Jul 07, 2013 5:23 pm
sos440 User avatar
Integration Guru
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Posts: 124
Location: California, US

With the substitution \(\displaystyle x = \tan t\), we have
\begin{align*}
\int \left( \frac{\arctan x}{x - \arctan x} \right)^{2} \, dx
&= \int \frac{t^2}{(\tan t - t)^{2}} \sec^{2} t \, dt
= \int \frac{t^2}{(\sin t - t \cos t)^{2}} \, dt \\
&= \int \left(- \frac{t}{\sin t} \right) \left( - \frac{t \sin t}{(\sin t - t \cos t)^{2}} \right) \, dt \\
&= - \frac{t}{\sin t} \frac{1}{\sin t - t \cos t} + \int \frac{dt}{\sin^{2} t} \\
&= - \frac{(1 + \tan^{2} t) t}{\tan t ( \tan t - t )} - \frac{1}{\tan t} + C \\
&= - \frac{(1 + x^{2}) \arctan x}{x ( x - \arctan x )} - \frac{1}{x} + C \\
&= - \frac{1+x \arctan x}{x - \arctan x} + C.
\end{align*}
Every integral harbors a story as fascinating as a comic book!

Post Sun Jul 07, 2013 5:33 pm
sos440 User avatar
Integration Guru
Integration Guru

Posts: 124
Location: California, US

Here is an easy one.

Problem #3. Show that \begin{equation*}
\int_{0}^{\frac{\pi}{2}} \frac{\sin (2x/3)}{\tan x} \, dx = \frac{\sqrt{3}}{4} ( 3 - \log 4 ).
\end{equation*} Similarly, show that \begin{equation*}
\int_{0}^{\frac{\pi}{2}} \frac{x \cos (2x/3)}{\sin x} \, dx = 5G - \frac{3\pi}{4} \log (2 + \sqrt{3}),
\end{equation*} where \(\displaystyle G\) denotes the Catalan constant.
Every integral harbors a story as fascinating as a comic book!

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