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## Strange Identities

Moderator: galactus

### Strange Identities

Thu Jan 09, 2014 12:49 pm

Posts: 850
Location: Jaipur, India

Hi all, I would like to share some curious/strange identities with you. Maybe in the future will be skilled enough to prove them?

This thread is not for proving identities but to simply state them. If you know any other crazy identities or want to comment don't hesitate to post.

1. $\displaystyle \sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)\cosh(10n+5)\pi}=\frac{1}{2} \sin^{-1} \left[\left\{ \left( 3-2\sqrt2\right)\left( 2+\sqrt5\right) \left( \sqrt{10}-3\right)\left( -\sqrt2+\sqrt[4]5\right)^2 \right\}^2 \right]$

2. $\displaystyle \sum_{n=0}^\infty (-1)^n \frac{(2n+1)^2}{\sinh (2n+1)\pi }=\frac{\sqrt{2}-1}{8}\frac{\pi^{3/2}}{\Gamma^6 \left(\frac{3}{4}\right)}$

3. $\displaystyle \prod_{k=1}^\infty \left(1+e^{-\frac{k\pi}{3}} \right)=\frac{e^{\frac{\pi}{72}}}{\sqrt[8]{2}}\sqrt[24]{\frac{\sqrt{2}+\sqrt[4]{3}}{(\sqrt{2}-\sqrt[4]{3})^5}}$

4. $\displaystyle \int_0^1 K(k)^3 \; dk = \frac{3}{1280\pi^2}\Gamma^8\left( \frac{1}{4}\right)$ where $K(k)$ is the complete elliptic integral of the first kind.

5. $\displaystyle \sum_{k=0}^\infty \frac{2k+1}{\left\{ 25+\frac{(2k+1)^4}{100}\right\}(1+e^{(2k+1)\pi}) }=\frac{4689}{11890}-\frac{\pi}{8}\text{coth}^2 \left(\frac{5\pi}{2} \right)$

6.$\displaystyle \int_0^\infty \frac{1}{\prod\limits_{n=0}^\infty (1+e^{-10 n \pi}x^2)}dx=\frac{\pi^{\frac{3}{4}}\Gamma \left(\frac{3}{4}\right)}{2e^{\frac{5\pi}{8}}}\sqrt{5}\sqrt[8]{2}\left( 1+\sqrt[4]{5}\right)\sqrt{\frac{1+\sqrt{5}}{2}}$

### Re: Strange Identities

Thu Jan 09, 2014 8:41 pm
galactus
Global Moderator

Posts: 902
I posted this one under that Si thread. I thought it was worth noting.

Since $\displaystyle \text{si}(x)=\int_{0}^{x}\frac{\sin(t)}{t}dt-\int_{0}^{\infty}\frac{\sin(t)}{t}dt=-\int_{x}^{\infty}\frac{\sin(t)}{t}dt$

we have $\displaystyle \text{Si}(x)-\frac{\pi}{2}=\text{si}(x)$

The curious identity is

7. $\displaystyle \text{si}(x)=-\int_{0}^{\frac{\pi}{2}}e^{-x\cos(t)}\cos(x\sin(t))dt$

In other words: $\displaystyle \text{Si}(x)=-\int_{0}^{\frac{\pi}{2}}e^{-x\cos(t)}\cos(x\sin(t))dt+\frac{\pi}{2}$

### Re: Strange Identities

Sat Jan 11, 2014 9:02 am
8. $\displaystyle \int_1^\infty\frac{\operatorname{arccot}\left(1+\frac{2\,\pi}{\operatorname{arcoth}x\,-\,\operatorname{arccsc}x}\right)}{\sqrt{x^2-1}}dx=\frac{\pi}{8}\log \left( \frac{\pi^2}{8}\right)$