Fun Integrals!

**sbhatnagar** · February 22nd 2013, 01:07 AM

These are some of my favorites.

1. $\int_0^1 \frac{1}{1+ _2F_1 \left( \frac{1}{n},x;\frac{1}{n};\frac{1}{n}}\right)}dx$ where $_2F_1 (a,b;c;z)$ is the Gaussian Hypergeometric Function.

2. $\int_{-\infty}^{\infty}\frac{\cos(s\arctan (ax))}{(1+x^2)(1+a^2x^2)^{s/2}}\log(1+a^2x^2)dx$ where $a,s \in \mathbb{R}^+$

3. $\int_{-\infty}^{\infty}\frac{e^{r\arctan(ax)}+e^{-r\arctan(ax)}}{1+x^2}\cos \left(\frac{r}{2}\log(1+a^2 x^2) \right)dx$ where $a\in \mathbb{R}^+$ and $r \in \mathbb{R}$

**topsquark** · February 23rd 2013, 06:13 AM

Originally Posted by sbhatnagar

These are some of my favorites.

1. $\int_0^1 \frac{1}{1+ _2F_1 \left( \frac{1}{n},x;\frac{1}{n};\frac{1}{n}}\right)}dx$ where $_2F_1 (a,b;c;z)$ is the Gaussian Hypergeometric Function.

2. $\int_{-\infty}^{\infty}\frac{\cos(s\arctan (ax))}{(1+x^2)(1+a^2x^2)^{s/2}}\log(1+a^2x^2)dx$ where $a,s \in \mathbb{R}^+$

3. $\int_{-\infty}^{\infty}\frac{e^{r\arctan(ax)}+e^{-r\arctan(ax)}}{1+x^2}\cos \left(\frac{r}{2}\log(1+a^2 x^2) \right)dx$ where $a\in \mathbb{R}^+$ and $r \in \mathbb{R}$

Jeez, I can't even come up with a good comment to make. What circle of Hades did these come from?

-Dan

**sbhatnagar** · March 5th 2013, 01:45 AM

I will start posting the answers now.

1. $I=\int_0^1 \frac{1}{1+ {}_2F_1 \left(\frac{1}{n},x;\frac{1}{n};\frac{1}{n} \right)}dx$

First note that $_2F_1\left( \frac{1}{n},x;\frac{1}{n};\frac{1}{n} \right) =\left(1-\frac{1}{n}\right)^{-x}$

$\begin{aligned}I &= \int_0^1 \frac{1}{1+\left(1-\frac{1}{n}\right)^{-x}}dx \\ &= \int_0^1 \frac{1}{1+\left(\frac{n}{n-1}\right)^x}dx \end{aligned}$

For simplicity let $e^k = \frac{n}{n-1}$ . We need to evaluate

$I=\int_0^1 \frac{1}{1+e^{kx}}dx$

Substitute $t=1+e^{kx}$ and $dt=k(t-1)dx$ :

$\begin{aligned}I &=\int_2^{1+e^k}\frac{1}{k}\frac{1}{t(t-1)}dt \\ &= \frac{1}{k}\int_2^{1+e^k}\left( \frac{1}{t-1}-\frac{1}{t}\right)dt \\ &= \frac{1}{k}\left( k-\log(1+e^k)+\log(2)\right) \\&=1- \frac{\log(2n-1)-\log(n-1)-\log(2)}{\log(n)-\log(n-1)} \\ &= \boxed{\displaystyle \frac{\log \left( \frac{2n}{2n-1}\right)}{\log \left( \frac{n}{n-1}\right)}}\end{aligned}$

I will post the answers to the other two when I get time.

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