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Math Help - Integral

  1. #1
    Moo
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    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 ^^

    \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 ^^'
    Last edited by Moo; May 22nd 2009 at 11:39 AM. Reason: minus sign...
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  2. #2
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    let f(t)=\frac{\ln t^{2}}{t^{2}-1}.

    we have \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 \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 \int_{\mathbb{R}}{f}=2\int_{0}^{\infty }{f}=4\int_{0}^{1}{f}=\pi ^{2}, as required. \blacksquare
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  3. #3
    Moo
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    Okay, actually, it can be done without mentioning "probability"


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

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

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

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

    The integral is now :
    \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 \\<br />
&=\iint_{\mathbb{R}^2} \frac{|u|}{z^2+u^2}\cdot\frac{1}{1+u^2}~du ~dz \\<br />
&=2\int_{\mathbb{R}}\int_{\mathbb{R}_+}\frac{u}{z^  2+u^2}\cdot\frac{1}{1+u^2} ~du~dz<br />
\end{aligned}

    Substitute t=u^2 :

    \begin{aligned}\pi^2&=\int_{\mathbb{R}}\int_0^\inf  ty\frac{1}{t+z^2}\cdot\frac{1}{1+t}~dt ~dz \\<br />
&=\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 \\<br />
&=\int_{\mathbb{R}}\frac{1}{1-z^2}\int_0^\infty \frac{1}{t+z^2}-\frac{1}{1+t} ~dt~dz \\<br />
&=\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}

    \Rightarrow \boxed{\pi^2=\int_{\mathbb{R}}\frac{\ln(z^2)}{z^2-1} ~dz}
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  4. #4
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    Quote Originally Posted by Moo View Post
    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 ^^

    \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: \sum_{m=1}^{\infty} \frac{H_m}{m2^m}=\frac{\pi^2}{12}, where H_m=1+\frac{1}{2} + \cdots + \frac{1}{m}. the idea of the proof is to start with \int_0^1 \frac{\ln x}{x^2-1} \ dx=\frac{\pi^2}{8}, which is a result of your integral.

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