The Basel problem As I have heard people did not trust Euler when he first discovered the formula (solution of the Basel problem)
$$\zeta(2)=\sum_{k=1}^\infty \frac{1}{k^2}=\frac{\pi^2}{6}$$
However, Euler was Euler and he gave other proofs.
I believe many of you know some nice proofs of this, can you please share it with us?

I saw this proof in an extract of the College Mathematics Journal.
Consider the Integeral  : $I = \int_0^{\pi/2}\ln(2\cos x)dx$
From $2\cos(x) = e^{ix} + e^{-ix}$ , we have:
$$\int_0^{\pi/2}\ln\left(e^{ix} +  e^{-ix}\right)dx = \int_0^{\pi/2}\ln\left(e^{ix}(1 +  e^{-2ix})\right)dx=\int_0^{\pi/2}ixdx + \int_0^{\pi/2}\ln(1 +  e^{-2ix})dx$$
The Taylor series expansion of $\ln(1+x)=x -\frac{x^2}{2} +\frac{x^3}{3}-\cdots$
Thus , $\ln(1+e^{-2ix}) = e^{-2ix}- \frac{e^{-4ix}}{2} + \frac{e^{-6ix}}{3} - \cdots $, then for $I$ :
$$I = \frac{i\pi^2}{8}+\left[-\frac{e^{-2ix}}{2i}+\frac{e^{-4ix}}{2\cdot 4i}-\frac{e^{-6ix}}{3\cdot 6i}-\cdots\right]_0^\frac{\pi}{2}$$
$$I = \frac{i\pi^2}{8}-\frac{1}{2i}\left[\frac{e^{-2ix}}{1^2}-\frac{e^{-4ix}}{2^2}+\frac{e^{-6ix}}{3^2}-\cdots\right]_0^\frac{\pi}{2}$$
By evaluating we get something like this..
$$I = \frac{i\pi^2}{8}-\frac{1}{2i}\left[\frac{-2}{1^2}-\frac{0}{2^2}+\frac{-2}{3^2}-\cdots\right]_0^\frac{\pi}{2}$$
Hence
$$\int_0^{\pi/2}\ln(2\cos x)dx=\frac{i\pi^2}{8}-i\sum_{k=0}^\infty \frac{1}{(2k+1)^2}$$
So now we have a real integral equal to an imaginary number,  thus the value of the integral should be zero.
Thus, $\sum_{k=0}^\infty \frac{1}{(2k+1)^2}=\frac{\pi^2}{8}$
But let $\sum_{k=0}^\infty \frac{1}{k^2}=E$ .We get $\sum_{k=0}^\infty \frac{1}{(2k+1)^2}=\frac{3}{4} E$
And as a result  $$\sum_{k=0}^\infty \frac{1}{k^2} = \frac{\pi^2}{6}$$
