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## Integration Contest - Season 4

This is not an ordinary forum. This is the place for the most brilliant ideas and techniques.

Moderators: Random Variable, sos440

### Integration Contest - Season 4

Thu Jan 02, 2014 4:55 pm

Posts: 852
Location: Jaipur, India

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!

### Re: Integration Contest - Season 4

Thu Jan 02, 2014 5:17 pm

Posts: 852
Location: Jaipur, India

Here is the first problem:

Problem 1

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

### Re: Integration Contest - Season 4

Fri Jan 03, 2014 5:47 am

Posts: 852
Location: Jaipur, India

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

### Re: Integration Contest - Season 4

Fri Jan 03, 2014 11:51 am

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

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 $\displaystyle f(z)=\frac{z(1-z^2)}{(z^2+a^2)^2} \log \left(\frac{1+z}{1-z} \right)$.

### Re: Integration Contest - Season 4

Fri Jan 03, 2014 3:14 pm

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.

### Re: Integration Contest - Season 4

Fri Jan 03, 2014 3:15 pm
galactus
Global Moderator

Posts: 902
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.

### Re: Integration Contest - Season 4

Fri Jan 03, 2014 3:48 pm

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.

### Re: Integration Contest - Season 4

Fri Jan 03, 2014 4:00 pm
Random Variable Integration Guru

Posts: 381
Is the contour used for integral 1 from the book The Cauchy Method of Residues?

A similar integral came up before.

### Re: Integration Contest - Season 4

Fri Jan 03, 2014 4:51 pm

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.

### Re: Integration Contest - Season 4

Fri Jan 03, 2014 5:04 pm

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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