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Looking at an proof of $\sum_{n=1}^{\infty}\frac{1}{n^{2}}=\frac{\pi^{2}}{6}$ that argues that $\int_0^1 \int_0^1 \frac{1}{1-xy} \,dy\,dx = \sum_{n=1}^{\infty}\frac{1}{n^{2}}$ but I can't see how this is accomplished since the integral is generalized (it is a limit it the endpoint) and also if using an geometric sum to derive this it is divergent in this endpoint (value $=1$). How is this solved in a rigorous proof?

laissez_faire
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  • See this: http://math.stackexchange.com/questions/1884418/how-to-evaluate-int-01-int-01-frac11-xy-dy-dx-to-prove-sum-n – Robert Z Aug 07 '16 at 18:57
  • Just take the limits as the upper endpoints approaches $1$ when using the geometric series. – Ahmed S. Attaalla Aug 07 '16 at 19:02
  • @Robert Z I can't see how this answers my question. In order to do a variable substitution the functions must be integrateable over the area in question which it is not. Also this question or its answers shines no light over how the realtionship with the series is established. – laissez_faire Aug 07 '16 at 19:04
  • When you integrate with respect to $y$ you get $-\int_{0}^{1} \frac{\ln (1-x)}{x}dx=\int_{0}^{1} \frac{1}{x}\cdot\sum_{k=1}^{\infty} \frac{x^k}{k}\ dx$. – Robert Z Aug 07 '16 at 19:14
  • Take a look here for more detailed proofs: http://empslocal.ex.ac.uk/people/staff/rjchapma/etc/zeta2.pdf – Robert Z Aug 07 '16 at 19:20
  • @laissez_faire "shines no light" $\rightarrow$ "doesn't shed light" – Jean Marie Aug 07 '16 at 19:31

2 Answers2

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Suppose $0<r<1.$ Then $\sum_{n=0}^{\infty}(xy)^n$ converges uniformly to $1/(1-xy)$ on $[0,r]^2.$ Therefore

$$\int_0^r\int_0^r \frac{1}{1-xy}\, dy\, dx = \sum_{n=0}^{\infty}\int_0^r\int_0^r (xy)^n \, dy\, dx.$$

The $n$th integral on the right equals $[r^{n+1}/(n+1)]^2.$ Now you just need to show

$$\lim_{r\to 1^-} \sum_{n=0}^{\infty}[r^{n+1}/(n+1)]^2 = \sum_{n=0}^{\infty}1/(n+1)^2.$$

I'll leave that for now.

zhw.
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  • Just a nitpick about the wording: The first integral on the right has integrand $(xy)^0$, the second has $(xy)^1$, ... , and the $n$th integral has integrand $(xy)^{n-1}$ and is therefore $(r^n/n)^2$. – John Bentin Aug 07 '16 at 21:14
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So there is a post in @Robert Z's link in the comments that proves

$$ \sum_{n=1}^{\infty} \frac{1}{n^2} = \frac{\pi^2}{6} $$

So I won't restate what has already been done, but will just show that

$$ \int_0^{1} \int_0^1 \frac{1}{1 - xy} dx dy = \sum_{n=1}^{\infty} \frac{1}{n^2} $$

Observe that

$$ 1 + x + x^2 + x^3 ... = \frac{1}{1-x}$$ Thus

$$ \int_0^x \frac{1}{1-x'} dx' = x + \frac{1}{2} x^2 + \frac{1}{3}x^3 ... $$

Thus:

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

Moving another step, observe that if we divide by $x$ and integrate AGAIN,

Our terms of the form $$\frac{1}{n} x^n \rightarrow_{divide} \frac{1}{n} x^{n-1} \rightarrow_{integrate} \frac{1}{n} \frac{1}{n} x^n = \frac{1}{n^2} x^n $$

So it then follows

$$ \int_{0}^{x} - \frac{\ln(1-x')}{x'} dx' = x + \frac{1}{4}x^2 + \frac{1}{9} x^3 + ... = \sum_{n=1}^{\infty} \frac{x^n}{n^2} $$

So now suppose we integrate to $x=1$ we then yield:

$$ \int_{0}^{1} - \frac{\ln(1 - x')}{x'} dx' = 1 + \frac{1}{4} + \frac{1}{9} ... = \sum_{n=1}^{\infty} \frac{1}{n^2} $$

So now the final punch. Observe (treating $x$ constant)

$$ \int_0^1 \int_0^1 \frac{1}{1 - xy} dx dy = -\int_0^1 \frac{\ln(1-x)}{x} dx $$

So we now have the desired equivalence.

Sidharth Ghoshal
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