1. ## Integral

Hi,

Like before, I've come up with this integral (with several difficulties though...) by a probabilistic way. I'll show it later. But I ask you to find it ^^

$\displaystyle \int_{\mathbb{R}} \frac{\ln(z^2)}{z^2-1} ~dz=\pi^2$

Yes, I can see Tessy coming with complex analysis and actually, I can do it too ^^'

2. let $\displaystyle f(t)=\frac{\ln t^{2}}{t^{2}-1}.$

we have $\displaystyle \int_{0}^{1}{t^{2j}\ln t^{2}\,dt}=-\int_{0}^{1}{\int_{t^{2}}^{1}{\frac{t^{2j}}{u}\,du }\,dt}=-\int_{0}^{1}{\int_{0}^{\sqrt{u}}{\frac{t^{2j}}{u}\ ,dt}\,du}=-\frac{2}{(2j+1)^{2}},$ thus $\displaystyle \int_{0}^{1}{\frac{\ln t^{2}}{t^{2}-1}\,dt}=\sum\limits_{j=0}^{\infty }{\frac{2}{(2j+1)^{2}}}=\frac{\pi ^{2}}{4},$ finally $\displaystyle \int_{\mathbb{R}}{f}=2\int_{0}^{\infty }{f}=4\int_{0}^{1}{f}=\pi ^{2},$ as required. $\displaystyle \blacksquare$

3. Okay, actually, it can be done without mentioning "probability"

Consider $\displaystyle I=\int_{\mathbb{R}} \frac{dx}{1+x^2}=\pi$

Hence $\displaystyle \pi^2=\iint_{\mathbb{R}^2} \frac{1}{1+x^2}\cdot\frac{1}{1+y^2} ~dx ~ dy$ (Fubini)

Consider the following transformation :
\displaystyle \begin{aligned}T ~:~& \mathbb{R}^*\times\mathbb{R}\to \mathbb{R}^*\times\mathbb{R}^* \\ & (x,y) \mapsto (u,z)=(x,xy)\end{aligned}
We have $\displaystyle \begin{cases}x=u \\y=\frac zu\end{cases}$

$\displaystyle J_T=\begin{pmatrix} 1&0\\y&x\end{pmatrix}\Rightarrow |\det J_T|=|x|=|u|$

The integral is now :
\displaystyle \begin{aligned}\pi^2&=\iint_{\mathbb{R}^2}\frac{1} {|u|}\cdot\frac{1}{1+(z/u)^2}\cdot\frac{1}{1+u^2} ~du ~dz \\ &=\iint_{\mathbb{R}^2} \frac{|u|}{z^2+u^2}\cdot\frac{1}{1+u^2}~du ~dz \\ &=2\int_{\mathbb{R}}\int_{\mathbb{R}_+}\frac{u}{z^ 2+u^2}\cdot\frac{1}{1+u^2} ~du~dz \end{aligned}

Substitute $\displaystyle t=u^2$ :

\displaystyle \begin{aligned}\pi^2&=\int_{\mathbb{R}}\int_0^\inf ty\frac{1}{t+z^2}\cdot\frac{1}{1+t}~dt ~dz \\ &=\int_{\mathbb{R}}\int_0^\infty\frac{1}{1-z^2}\cdot\frac{1+t-t-z^2}{(1+t)(t+z^2)} ~dt~dz \\ &=\int_{\mathbb{R}}\frac{1}{1-z^2}\int_0^\infty \frac{1}{t+z^2}-\frac{1}{1+t} ~dt~dz \\ &=\int_{\mathbb{R}}\frac{1}{1-z^2} \cdot\left.\ln\left(\frac{t+z^2}{1+t}\right)\right |_0^\infty ~dt~dz \end{aligned}

$\displaystyle \Rightarrow \boxed{\pi^2=\int_{\mathbb{R}}\frac{\ln(z^2)}{z^2-1} ~dz}$

4. Originally Posted by Moo
Hi,

Like before, I've come up with this integral (with several difficulties though...) by a probabilistic way. I'll show it later. But I ask you to find it ^^

$\displaystyle \int_{\mathbb{R}} \frac{\ln(z^2)}{z^2-1} ~dz=\pi^2$

Yes, I can see Tessy coming with complex analysis and actually, I can do it too ^^'
a consequence of your integral: $\displaystyle \sum_{m=1}^{\infty} \frac{H_m}{m2^m}=\frac{\pi^2}{12},$ where $\displaystyle H_m=1+\frac{1}{2} + \cdots + \frac{1}{m}.$ the idea of the proof is to start with $\displaystyle \int_0^1 \frac{\ln x}{x^2-1} \ dx=\frac{\pi^2}{8},$ which is a result of your integral.

then substitute $\displaystyle x \to 1-x$ to get: $\displaystyle \int_0^1 \frac{\ln(1-x)}{-x(2-x)} \ dx = \frac{\pi^2}{8}.$ but we have $\displaystyle \ln(1-x)=-\sum_{n=0}^{\infty} \frac{x^{n+1}}{n+1}$ and $\displaystyle \frac{1}{2-x}= \frac{1}{2}\sum_{n=0}^{\infty} \frac{x^n}{2^n}.$ thus: $\displaystyle \frac{\pi^2}{4}=\int_0^1\left(\sum_{n=0}^{\infty} \frac{x^n}{n+1} \right) \left(\sum_{n=0}^{\infty} \frac{x^n}{2^n} \right) \ dx,$

which gives us: $\displaystyle \frac{\pi^2}{4}=\sum_{n=0}^{\infty} \sum_{m=0}^n \frac{1}{(n+1)(n+1-m)2^m}.$ now change the order of summation to get: $\displaystyle \frac{\pi^2}{4}=\sum_{m=0}^{\infty} \frac{1}{2^m} \sum_{n=m}^{\infty} \frac{1}{(n+1)(n+1-m)}=\frac{\pi^2}{6} + \sum_{m=1}^{\infty} \frac{1}{2^m} \sum_{n=m}^{\infty} \frac{1}{(n+1)(n+1-m)}.$

but it's easy to see that $\displaystyle \sum_{n=m}^{\infty} \frac{1}{(n+1)(n+1-m)}=\frac{H_m}{m}$ and the result follows. i'm pretty sure there are easier ways to prove this result. maybe some of you want to try it?