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I'm trying to simply prove the Basel Series

$$ \sum_{k=1}^{\infty} \frac{1}{k^2} = 1 + \frac{1}{4} + \frac{1}{9} + \frac{1}{16} + \dots = \frac{\pi^2}{6} $$

by deconstructing it into a sum of Sub-Series.

I was able to show using WolframAlpha that

$$ \frac{1}{(2n - 1)^2} = \int_{0}^{1} \int_{0}^{1} x^{2n-2} y^{2n-2} \, dx \, dy $$

Thus, $$ \sum_{n=1}^{\infty} \frac{1}{(2n - 1)^2} = \sum_{n=1}^{\infty} \int_{0}^{1} \int_{0}^{1} x^{2n-2} y^{2n-2} \, dx \, dy = \int_{0}^{1} \int_{0}^{1} \frac{1}{1 - x^2 y^2} \, dx \, dy $$

But, to complete the Basel proof and avoid any circular arguments, I need to show that

$$ \int_{0}^{1} \int_{0}^{1} \frac{1}{1 - x^2 y^2} \, dx \, dy = \frac{\pi^2}{8} \tag{1} $$

Proof

Every natural number $n$ can be written in one and only one way as $2^mq$ where $q=(2n-1)$ is an odd number.

Define the Sub-Series:

$$ S_m = \sum_{n=1}^{\infty} \frac{1}{\left(2^m (2n - 1)\right)^2} = \frac{1}{4^m} \sum_{n=1}^{\infty} \frac{1}{(2n - 1)^2} $$

Notice the sum of all $S_m$ reconstructs the Basel series:

$$ \sum_{k=1}^{\infty} \frac{1}{k^2} = \sum_{m=0}^{\infty} S_m = \sum_{m=0}^{\infty} \frac{1}{4^m} \sum_{n=1}^{\infty} \frac{1}{(2n - 1)^2} = \frac{4}{3} \sum_{n=1}^{\infty} \frac{1}{(2n - 1)^2} $$

The result follows from $(1)$:

$$ \sum_{k=1}^{\infty} \frac{1}{k^2} = \frac{4}{3} \sum_{n=1}^{\infty} \frac{1}{(2n - 1)^2} = \frac{4}{3} \cdot \frac{\pi^2}{8} = \frac{\pi^2}{6} \quad \blacksquare $$

vengy
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    I saw an article once that evaluates the integral of $\frac{1}{1-x^2 y^2}$ with the substitution $x = \frac{\sin u}{\cos v}, y = \frac{\sin v}{\cos u}$. (I have no idea how they came up with that substitution, though, and it seems miraculous the way things work out.) – Daniel Schepler Nov 04 '24 at 22:16
  • I'm curious if that integral introduces a circular argument by using the known Basel series result? Thanks. – vengy Nov 04 '24 at 22:25
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    The non-circular strategy is to evaluate the integral without using the Basel result, then write it as a multiple of $\zeta(2)$ to solve the Basel problem. @DanielSchepler's comment gives the approach you want. You'll get $\int_0^{\pi/2}\int_0^{\pi/2-v}\mathrm{d}u\mathrm{d}v$. – J.G. Nov 04 '24 at 22:47
  • I think the article was a survey of ways to prove $\zeta(2) = \frac{\pi^2}{6}$. If I remember correctly, one of the other alternatives was the common proof involving finding the Fourier series for (the periodic extension of) $f(x) = x$ then applying Parseval's identity. Yet another was to write a polynomial with $\cot(k\pi/(2n+1)), k=1, \ldots, n$ as roots, then use a Vieta type identity to find $\sum_{k=1}^\infty \cot^2(k\pi/(2n+1))$, then use the bounds $\cot \theta < \theta < \csc \theta$ for $\theta\in (0, \pi/2)$ to give a Squeeze Theorem argument. – Daniel Schepler Nov 04 '24 at 23:27
  • @DanielSchepler I hope you can dig up a link to that article, although there will be an MSE question of similar content. That squared cotangent one is my favourite actually, if we exclude simultaneous treatments of all the even zeta constants. – J.G. Nov 04 '24 at 23:32
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    See Proof 2 here. @DanielSchepler Are you thinking of the linked article? – cqfd Nov 05 '24 at 00:15
  • You might also like looking at links in the comments and the answer to https://math.stackexchange.com/questions/1884418/how-to-evaluate-int-01-int-01-frac11-xy-dy-dx-to-prove-sum-n/1885211#1885211 – peter a g Nov 06 '24 at 02:17

2 Answers2

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\begin{align}\int_{0}^{1} \int_{0}^{1} &\frac{1}{1 - x^2 y^2} dy \, dx =\int_0^1 \frac{\tanh^{-1}x}{x}dx\\ =& \int_0^1\int_0^1 \frac{2t}{(1+x)^2-4xt^2}dt\ dx = \int_0^1 \frac{\sin^{-1}t}{\sqrt{1-t^2}}dt= \frac{\pi^2}{8} \end{align}

Quanto
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  • Nice! Just to clarify your proof, it's non-circular to deduce $$\int_{0}^{1} \int_{0}^{1} \frac{1}{1 - x^2 y^2} , dx , dy = \frac{\pi^2}{8} \tag{1}$$ I'm always fearful of Basel $\frac{\pi^2}{6}$ insidiously inserting itself into proofs. Thanks. – vengy Nov 05 '24 at 00:38
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    @vengy There is no circularity. The double integral is evaluated elementarily.. – Quanto Nov 05 '24 at 00:42
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    Completed the Basel Proof using your solution. – vengy Nov 05 '24 at 02:32
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$$ \begin{aligned} \int_0^1 \int_0^1 \frac{1}{1-x^2 y^2} d x d y & =\frac{1}{2} \int_0^1 \int_0^1\left(\frac{1}{1+x y}+\frac{1}{1-x y}\right) d x d y \\ & =\frac{1}{2} \int_0^1\left[\frac{\ln (1+x y)}{y}-\frac{\ln (1-x y)}{y}\right]_0^1 \\ & = \frac{1}{2} \int_0^1 \frac{\ln \left(\frac{1+y}{1-y}\right)}{y} d y \\& =- \int_0^1 \frac{\ln t}{1-t^2} d t, \quad \textrm{ where }t=\frac{1-y}{1+y}. \\ & =- \left(-\frac{\pi^2}{8} \right)\\ & =\frac{\pi^2}{8} \end{aligned} $$

where the second last answer comes from my post or post without the use of the formula $\sum_{n=1}^{\infty} \frac{1}{n^2}=\frac{\pi^2}{6}$.

Lai
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