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Integration Contest - Season 4

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Post Thu Jan 02, 2014 4:55 pm
Shobhit Site Admin
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Integration Contest - Season 4


Contests have played a vital role in giving shape to the ideology of Integrals and Series Forum. In fact the community of Integrals and Series originated from a contest! :)

Our new members can take some ideas from the previous contests:

  • I will start by posting the first problem. If the problem is solved, anyone can post a new problem. If the problem remains unsolved for more than one day, we move on to the next problem.
  • 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 kind.
  • There is no winner or loser. The point is to simply to learn from one another!

Post Thu Jan 02, 2014 5:17 pm
Shobhit Site Admin
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Posts: 852
Location: Jaipur, India

Here is the first problem:

Problem 1

\(\displaystyle \int_0^\infty \frac{x(1-x^2)}{(x^2+a^2)^2}\log \Big|\frac{x+1}{x-1} \Big|\; dx=\pi \left( \frac{1}{2a}-\tan^{-1}\frac{1}{a}\right)\)

Post Fri Jan 03, 2014 5:47 am
Shobhit Site Admin
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Posts: 852
Location: Jaipur, India

Here is an identity that may help: \(\displaystyle \frac{x(1-x^2)}{(x^2+a^2)^2}=(1+a^2) \frac{x}{(x^2+a^2)^2}-\frac{x}{x^2+a^2}\)

Post Fri Jan 03, 2014 11:51 am
Shobhit Site Admin
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Posts: 852
Location: Jaipur, India

After working for some time, I came up with the following solution. I hope it is correct.

Let \(\displaystyle f(z)=\frac{z(1-z^2)}{(z^2+a^2)^2} \log \left(\frac{1+z}{1-z} \right)\) and consider the contour integral

\[\oint_{C} \frac{z(1-z^2)}{(z^2+a^2)^2} \log \left(\frac{1+z}{1-z}\right) \; dz\]
where \(C\) is the following contour
Image

1. \( \gamma \) is a large semicircle of radius \(R\) centered at the origin.

2. \(\alpha\) and \(\beta\) are quarter circle indents around the branch points \(-1,1\) respectively of radius \(\epsilon \).

3. There are 3 line segments joining

  • \(-\sqrt{R^2-\epsilon^2}+i \epsilon\) and \(-1+i \epsilon\)
  • \(-1+\epsilon \) and \(1-\epsilon \)
  • \(1+i \epsilon\) and \(\sqrt{R^2-\epsilon^2}+i \epsilon\)

For the line segments above the branch cuts, we have

\(\displaystyle \int_{-\sqrt{R^2-\epsilon^2}+i\epsilon}^{-1+i\epsilon} f(z) \; dz +\int_{1+i\epsilon}^{\sqrt{R^2-\epsilon^2}+i\epsilon} f(z)\; dz\)

\(\displaystyle \begin{align*}
\; &= \int_{-\sqrt{R^2-\epsilon^2}+i\epsilon}^{-1+i\epsilon} \frac{z(1-z^2)}{(z^2+a^2)^2}\left\{ \log \Bigg( \frac{|1+z|}{1-z} \Bigg)+i\pi \right\}dz + \int_{1+i\epsilon}^{\sqrt{R^2-\epsilon^2}+i\epsilon} \frac{z(1-z^2)}{(z^2+a^2)^2}\left\{ \log \Bigg( \frac{1+z}{|1-z|} \Bigg) -i\pi \right\}dz \\
&\;
\\
&\to 2 \int_1^\infty \frac{x(1-x^2)}{(x^2+a^2)^2}\log \left( \frac{x+1}{x-1} \right) dx \\ &\quad \text{as } R\to \infty \text{ and }\epsilon\to 0^+
\end{align*}\)

Similarly for the other segment,

\(\displaystyle \int_{-1+\epsilon}^{1+\epsilon}f(z) \; dz \to 2 \int_0^1 \frac{x(1-x^2)}{(x^2+a^2)^2}\log \left( \frac{1+x}{1-x} \right)dx \quad \text{as }\epsilon \to 0^+\)

And also,

\(\displaystyle \int_\gamma f(z)\; dz \to \pi^2 \text{ as }R\to \infty\)

\(\displaystyle \int_\alpha f(z)\; dz \to 0 \quad , \int_\beta f(z)\; dz \to 0 \text{ as }\epsilon \to 0^+\)

So by the residue theorem, we get

\(\displaystyle \begin{align*}
\lim_{\begin{matrix}R\to \infty \\ \epsilon \to 0^+\end{matrix}}\oint_{C} f(z)\; dz &=2\pi i \text{Res}_{z=i a} \frac{z(1-z^2)}{(z^2+a^2)^2} \log \left(\frac{1+z}{1-z}\right) \\
&\;
\\ \implies 2\int_0^1 \frac{x(1-x^2)}{(x^2+a^2)^2}\log \left( \frac{1+x}{1-x} \right) dx+2 \int_1^\infty \frac{x(1-x^2)}{(x^2+a^2)^2}\log \left( \frac{x+1}{x-1} \right) dx +\pi^2 &= 2\pi i \left( -\frac{i}{2a}-\frac{1}{2} \log \left( \frac{i-a}{i+a}\right)\right) \\
&\; \\
\implies 2 \int_0^\infty \frac{x(1-x^2)}{(x^2+a^2)^2}\log \Bigg| \frac{1+x}{1-x} \Bigg| dx +\pi^2 &= \frac{\pi}{a}-2\pi \tan^{-1}\frac{1}{a}+\pi^2 \\
&\; \\
\implies \int_0^\infty \frac{x(1-x^2)}{(x^2+a^2)^2}\log \Bigg| \frac{1+x}{1-x} \Bigg| dx &= \boxed{\displaystyle \pi \left(\frac{1}{2a}- \tan^{-1}\frac{1}{a}\right)}
\end{align*}\)

EDIT

I just noticed that this problem is much easier if we choose \(\displaystyle f(z)=\frac{z(1-z^2)}{(z^2+a^2)^2} \log \left(\frac{1+z}{\color{red}{z-1}} \right)\) instead of \(\displaystyle f(z)=\frac{z(1-z^2)}{(z^2+a^2)^2} \log \left(\frac{1+z}{1-z} \right)\).

Post Fri Jan 03, 2014 3:14 pm
Shobhit Site Admin
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Posts: 852
Location: Jaipur, India

Problem 2

\(\displaystyle \int_0^\infty \frac{\cos(x)}{x}\left(\int_0^x \frac{\sin t}{t} dt\right)^2 dx = -\frac{7}{6}\zeta(3)\)

I have not tried this one but I think it is interesting.

Post Fri Jan 03, 2014 3:15 pm
galactus User avatar
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That is very cool S. That is a different kind of contour. Ain't seen one like that before. One can do about anything if one can find the right contour.

Post Fri Jan 03, 2014 3:48 pm
Shobhit Site Admin
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Posts: 852
Location: Jaipur, India

galactus wrote:
That is very cool S. That is a different kind of contour. Ain't seen one like that before. One can do about anything if one can find the right contour.


Thanks! :) Any idea about problem 2? It looks extremely rough. It is just the kind of integral SOS loves.

Post Fri Jan 03, 2014 4:00 pm
Random Variable Integration Guru
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Posts: 381
Is the contour used for integral 1 from the book The Cauchy Method of Residues?

A similar integral came up before.

Post Fri Jan 03, 2014 4:51 pm
Shobhit Site Admin
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Posts: 852
Location: Jaipur, India

Random Variable wrote:
Is the contour used for integral 1 from the book The Cauchy Method of Residues?

A similar integral came up before.


The integral was taken from 'The Cauchy Method of Residue' but they use a different approach to solve the problem.

They have even proved a general formula for these kind of integrals.

If \(f(-z)=-f(z)\) and \(\displaystyle f(z)=o\left(\frac{1}{z\log|z|}\right)+\frac{c}{z}\) then we have

\(\displaystyle \int_0^\infty f(x) \log \Bigg|\frac{x+1}{x-1} \Bigg|dx = \frac{\pi^2 c}{2}-2\pi \Im \sum_{k=1}^n \text{Res}_{z=z_k} f(z) \log(1+z)\)

where \(z_k\) are the poles of \(f(z)\) in the upper half plane.

Post Fri Jan 03, 2014 5:04 pm
Shobhit Site Admin
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Posts: 852
Location: Jaipur, India

Here's another one in case no wants to do Problem 2 :

Alternate Problem 2

\(\displaystyle \int_0^\infty \frac{\left(\tan^{-1}x \right)^2 \log(x)}{x^{3/2}}dx =\pi \sqrt{2}\left( 4 G-\pi+\frac{\pi^2}{4}-2-\frac{\pi}{2}\log(2)\right)\)

where \(G\) is Catlan's Constant.

Or you can post any other problem if you feel like it.

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