Prove ζ(2) = π²/6 using a double integral and geometric series

[fresh_42]

Prove
$$
\zeta(2) = \sum_{n\in \mathbb{N}}\dfrac{1}{n^2} = \dfrac{\pi^2}{6}
$$
by evaluating
$$
\int_0^1\int_0^1\dfrac{1}{1-xy}\,dx\,dy
$$
twice: via the geometric series and via the substitutions $u=\dfrac{y+x}{2}\, , \,v=\dfrac{y-x}{2}$.

 

[etotheipi]

I'm having trouble. I tried$$I = \int_0^1 \int_0^1 \frac{1}{1-xy} dx dy = 2\int_0^{\frac{1}{2}} \int_v^{1-v} \frac{1}{1-(u+v)(u-v)} du dv + 2\int_{-
\frac{1}{2}}^{0} \int_{-v}^{1+v} \frac{1}{1-(u+v)(u-v)} du dv $$because $0 \leq x \leq 1$, $0 \leq y \leq 1$ in the $x$-$y$ space maps to a region $u \leq 1+v$, $u \geq v$, $u \leq 1-v$, $u \geq -v$ in the $u$-$v$ space, and $$\frac{\partial(x,y)}{\partial(u,v)} = \begin{vmatrix}
1 & -1\\
1 & 1\end{vmatrix} = 2
$$That would appear to give$$I = 2\int_0^{\frac{1}{2}} \frac{\text{artanh}(\frac{1-v}{\sqrt{1+v^2}}) - \text{artanh}(\frac{v}{\sqrt{1+v^2}})}{\sqrt{1+v^2}} dv + 2\int_{-\frac{1}{2}}^{0} \frac{\text{artanh}(\frac{1+v}{\sqrt{1+v^2}}) - \text{artanh}(\frac{-v}{\sqrt{1+v^2}})}{\sqrt{1+v^2}} dv$$Wondered if you had any pointers?

 

[fresh_42]

If you draw a picture of the two integration areas (square $(x,y)$ and diamond $(u,v)$), and use the symmetry along the $u$-axis beside the Jacobi determinant for the coordinate change then the integration becomes
$$
\int_{0}^{1}\int_{0}^{1}\dfrac{1}{1-xy}\,dx\,dy=4\int_{0}^{1/2}\left(\int_{0}^{u}\dfrac{dv}{1-u^2+v^2}\right)\,du +4\int_{1/2}^{1}\left(\int_{0}^{1-u}\dfrac{dv}{1-u^2+v^2}\right)\,du
$$
to which you can apply the integration formula for $\int \frac{dx}{a^2+x^2}$ which is basically how far you got except for maybe easier boundaries.

Now you can either substitute $u=\sin \alpha\, , \,u=\cos \beta,$ or more directly differentiate the $\arctan$ term to get integrals of the form $\int g'(u)g(u)\,du$ which can be done by integration by parts.

 

[etotheipi]

OK, I think I got it. To do the first integral, let $u = \sin{\alpha}$, $$4\int_0^{\frac{1}{2}} \frac{1}{\sqrt{1-u^2}} \arctan{\frac{u}{\sqrt{1-u^2}}} du = 4\int_0^\frac{\pi}{6} \arctan{(\tan{\alpha})} d\alpha = 4\left[ \frac{1}{2} \alpha^2 \right]_0^{\frac{\pi}{6}} = \frac{\pi^2}{18}$$For the second integral, we notice that$$\begin{align*}\frac{d}{du} \left(\arctan{\frac{\sqrt{1-u}}{\sqrt{1+u}}} \right)^2 &= -\frac{1}{4} \left( \frac{\sqrt{1+u}}{\sqrt{1-u}} + \frac{\sqrt{1-u}}{\sqrt{1+u}}\right) \cdot 2\arctan{\frac{\sqrt{1-u}}{\sqrt{1+u}}}\\ &= -\frac{1}{2} \frac{1}{\sqrt{1-u^2}} \cdot 2\arctan{\frac{\sqrt{1-u}}{\sqrt{1+u}}}\\

&=-\frac{1}{\sqrt{1-u^2}} \arctan{\frac{\sqrt{1-u}}{\sqrt{1+u}}}\end{align*}$$which means that$$4\int_{\frac{1}{2}}^1 \frac{1}{\sqrt{1-u^2}} \arctan{\frac{\sqrt{1-u}}{\sqrt{1+u}}} du = 4\left[-\left(\arctan{\frac{\sqrt{1-u}}{\sqrt{1+u}}} \right)^2 \right]_\frac{1}{2}^1 =4 \left(\arctan{\frac{1}{\sqrt{3}}} \right)^2 = \frac{\pi^2}{9}$$Hence the sum is$$I = I_1 + I_2 = \frac{\pi^2}{18} + \frac{\pi^2}{9} = \frac{\pi^2}{6}$$Now for the second half. By analogy to the formula for the sum of an infinite geometric series, the integral can be re-written as$$\begin{align*}I &= \int_0^1 \int_0^1 1 + xy + x^2 y^2 + x^3 y^3 + \dots dx dy \\

&= \int_0^1 \left[x + \frac{1}{2}yx^2 + \frac{1}{3}y^2 x^3 + \frac{1}{4}y^3 x^4 + \dots \right]_0^1 dy \\

&= \int_0^1 1 + \frac{1}{2}y + \frac{1}{3}y^2 + \frac{1}{4} y^3 + \dots dy \\

&= \left[y + \frac{1}{4}y^2 + \frac{1}{9}y^3 + \frac{1}{16}y^4 + \dots \right]_0^1 \\

&= \sum_{n \in \mathbb{N}} \frac{1}{n^2}\end{align*}$$Hence,$$\zeta(2) = \sum_{n\in \mathbb{N}}\dfrac{1}{n^2} = \dfrac{\pi^2}{6}$$

 

[fresh_42]

Well done. You used both hints and combined them: one with a trig substitution and one with integration by parts. You really understood the idea of this training ground! Ready for the next lap?

How to get the maximum out of a proof? With corollaries!

Prove that Euler's series $\zeta(2)=\pi^2/6$ is equivalent to
$$
\sum_{k=0}^{\infty} \dfrac{1}{(2k+1)^2} = \dfrac{\pi^2}{8}
$$

 

[etotheipi]

If I understood correctly, I think what you want is to re-write$$\sum_{n \in \mathbb{N}}\dfrac{1}{n^2} = \sum_{k=1}^{\infty} \frac{1}{(2k)^2} + \sum_{k=0}^{\infty} \frac{1}{(2k + 1)^2} = \frac{\pi^2}{6}$$i.e. splitting the sum into even and odd terms. Then subtracting the even sum from the LHS,$$\sum_{k=0}^{\infty} \frac{1}{(2k + 1)^2} = \frac{\pi^2}{6} - \frac{1}{4} \sum_{k=1}^{\infty} \frac{1}{k^2} = \frac{\pi^2}{6} - \frac{\pi^2}{24} = \frac{\pi^2}{8}$$Phew, that wasn't as difficult as last time

 

[fresh_42]

If I understood correctly, I think what you want is to re-write$$\sum_{n \in \mathbb{N}}\dfrac{1}{n^2} = \sum_{k=1}^{\infty} \frac{1}{(2k)^2} + \sum_{k=0}^{\infty} \frac{1}{(2k + 1)^2} = \frac{\pi^2}{6}$$i.e. splitting the sum into even and odd terms. Then subtracting the even sum from the LHS,$$\sum_{k=0}^{\infty} \frac{1}{(2k + 1)^2} = \frac{\pi^2}{6} - \frac{1}{4} \sum_{k=1}^{\infty} \frac{1}{k^2} = \frac{\pi^2}{6} - \frac{\pi^2}{24} = \frac{\pi^2}{8}$$Phew, that wasn't as difficult as last time

No, it wasn't. However, it was merely the information lap to ...
$$
\sum_{k=0}^{\infty} \frac{1}{(2k + 1)^2} = \int_0^1 \int_0^1 \dfrac{1}{1-x^2y^2}\,dx\,dy
$$
To be honest, I have no idea how the authors of this proof came up with their substitution for the RHS to calculate the integral. Maybe the Weierstraß substitution leads to it, or at least close to it. Have a try!

 

[etotheipi]

ahaha, okay. I've got a few things to finish off for tomorrow, but I'll come back and try that one when I get the chance . Thanks for the question!