integral

May 2009
959
362
\(\displaystyle \zeta (2) = \sum^{\infty}_{n=1} \frac{1}{n^{2}} = Li_{2} (1) = -\int^{1}_{0} \frac{\ln(1-x)}{x} \ dx \)

So how do I show that \(\displaystyle -\int^{1}_{0} \frac{\ln(1-x)}{x} \ dx = \frac{\pi^{2}}{6} \) ?

The only solution I've found is to write it as \(\displaystyle \int^{1}_{0} \int^{1}_{0} \frac{1}{1+xy} \ dy \ dx \) and then make a crazy change of variables that rotates the coordinate system by \(\displaystyle \frac{\pi}{4}\). But that gets really messy.
 
Last edited:
Apr 2010
384
153
Canada
\(\displaystyle \zeta (2) = \sum^{\infty}_{n=1} \frac{1}{n^{2}} = Li_{2} (1) = -\int^{1}_{0} \frac{\ln(1-t)}{t} \ dt \)

So how do I show that \(\displaystyle -\int^{1}_{0} \frac{\ln(1-x)}{x} \ dx = \frac{\pi^{2}}{6} \) ?

The only solution I've found is to write it as \(\displaystyle \int^{1}_{0} \int^{1}_{0} \frac{1}{1+xy} \ dy \ dx \) and then make a crazy change of variables that rotates the coordinate system by \(\displaystyle \frac{\pi}{4}\). But that gets really messy.

Hmmm...I believe you would have to use a power series representation here.

For \(\displaystyle 0 \le x \le 1 \)

\(\displaystyle \frac{ ln(1-x) }{ x} = \frac{ -(x + \frac{x^2}{2} + \frac{x^3}{3}+...) }{x} = -(1+ \frac{x}{2} + \frac{x^2}{3}+...)\)

If we take the integral of this,

\(\displaystyle -\int^{1}_{0} \frac{\ln(1-x)}{x} \ dx = \int_0^1 (1+ \frac{x}{2} + \frac{x^2}{3}+...) = (1 + \frac{1}{2^2} + \frac{1}{3^2}+...) \)

Note how this is the series,

\(\displaystyle \sum \frac{1}{n^2} = \frac{ ( \pi)^2 }{6} \)

So \(\displaystyle -\int^{1}_{0} \frac{\ln(1-x)}{x} \ dx = \frac{ ( \pi)^2 }{6} \)

Edit- Can you explain your way? I don't quite comprehend what you were trying to do but it seems like a cool exercise (so if you could, could you post your solution?)
 
May 2009
959
362
Hmmm...I believe you would have to use a power series representation here.

For \(\displaystyle 0 \le x \le 1 \)

\(\displaystyle \frac{ ln(1-x) }{ x} = \frac{ -(x + \frac{x^2}{2} + \frac{x^3}{3}+...) }{x} = -(1+ \frac{x}{2} + \frac{x^2}{3}+...)\)

If we take the integral of this,

\(\displaystyle -\int^{1}_{0} \frac{\ln(1-x)}{x} \ dx = \int_0^1 (1+ \frac{x}{2} + \frac{x^2}{3}+...) = (1 + \frac{1}{2^2} + \frac{1}{3^2}+...) \)

Note how this is the series,

\(\displaystyle \sum \frac{1}{n^2} = \frac{ ( \pi)^2 }{6} \)

So \(\displaystyle -\int^{1}_{0} \frac{\ln(1-x)}{x} \ dx = \frac{ ( \pi)^2 }{6} \)

Edit- Can you explain your way? I don't quite comprehend what you were trying to do but it seems like a cool exercise (so if you could, could you post your solution?)
Yes, that's definitely true. But I'm not trying to show that the integral is equal to \(\displaystyle \sum^{\infty}_{n=1} \frac{1}{n^{2}} \). I'm trying to evaluate the integral to show that \(\displaystyle \sum^{\infty}_{n=1}\frac{1}{n^{2}} = \frac{\pi^{2}}{6} \)

The rotation equations are \(\displaystyle x'= x \cos \theta - y \sin \theta \) and \(\displaystyle y' = x \sin \theta + y \cos \theta \)

so just let \(\displaystyle \theta = \frac{\pi}{4} \)
 
Apr 2010
384
153
Canada
Yes, that's definitely true. But I'm not trying to show that the integral is equal to \(\displaystyle \sum^{\infty}_{n=1} \frac{1}{n^{2}} \). I'm trying to evaluate the integral to show that \(\displaystyle \sum^{\infty}_{n=1}\frac{1}{n^{2}} = \frac{\pi^{2}}{6} \)

The rotation equations are \(\displaystyle x'= x \cos \theta - y \sin \theta \) and \(\displaystyle y' = x \sin \theta + y \cos \theta \)

so just let \(\displaystyle \theta = \frac{\pi}{4} \)
Yikes! I kind of jumped the gun on that one eh?

This should help: Math Forum - Ask Dr. Math

It would appear that you have to represent the integral in a completely differn't form.
 
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Drexel28

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Nov 2009
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\(\displaystyle \zeta (2) = \sum^{\infty}_{n=1} \frac{1}{n^{2}} = Li_{2} (1) = -\int^{1}_{0} \frac{\ln(1-x)}{x} \ dx \)

So how do I show that \(\displaystyle -\int^{1}_{0} \frac{\ln(1-x)}{x} \ dx = \frac{\pi^{2}}{6} \) ?

The only solution I've found is to write it as \(\displaystyle \int^{1}_{0} \int^{1}_{0} \frac{1}{1+xy} \ dy \ dx \) and then make a crazy change of variables that rotates the coordinate system by \(\displaystyle \frac{\pi}{4}\). But that gets really messy.
\(\displaystyle J=-\int_0^1\frac{\ln(1-x)}{x}dx\). Let \(\displaystyle -\ln(1-x)=z\implies 1-e^{-z}=x\implies e^{-z}dz=dx\)

So, our integral becomes \(\displaystyle J=\int_0^{\infty}\frac{z e^{-z}}{1-e^{-z}}dz\) multiplying top and bottom gives \(\displaystyle J=\int_0^{\infty}\frac{z}{e^z-1}dz=\cdots\) ah crap, you need the zeta function.

Why not just do Fourier series or use contour integration?