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Thread: Fun Integrals!

  1. #1
    Member sbhatnagar's Avatar
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    Fun Integrals!

    These are some of my favorites.

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

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

    3. $\displaystyle \int_{-\infty}^{\infty}\frac{e^{r\arctan(ax)}+e^{-r\arctan(ax)}}{1+x^2}\cos \left(\frac{r}{2}\log(1+a^2 x^2) \right)dx$ where $\displaystyle a\in \mathbb{R}^+$ and $\displaystyle r \in \mathbb{R}$
    Last edited by sbhatnagar; Feb 22nd 2013 at 01:09 AM.
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  2. #2
    Forum Admin topsquark's Avatar
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    Re: Fun Integrals!

    Quote Originally Posted by sbhatnagar View Post
    These are some of my favorites.

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

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

    3. $\displaystyle \int_{-\infty}^{\infty}\frac{e^{r\arctan(ax)}+e^{-r\arctan(ax)}}{1+x^2}\cos \left(\frac{r}{2}\log(1+a^2 x^2) \right)dx$ where $\displaystyle a\in \mathbb{R}^+$ and $\displaystyle r \in \mathbb{R}$
    Jeez, I can't even come up with a good comment to make. What circle of Hades did these come from?

    -Dan
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  3. #3
    Member sbhatnagar's Avatar
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    Re: Fun Integrals!

    I will start posting the answers now.

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

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

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

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

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

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

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

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