log and poisson-like integral Here is a fun looking one some may enjoy. 
Show that:
$$\int_{0}^{1}\log\left(\frac{x^{2}+2x\cos(a)+1}{x^{2}-2x\cos(a)+1}\right)\cdot \frac{1}{x}dx=\frac{\pi^{2}}{2}-\pi a$$

Starting from
$$
{\rm Log}(1+x e^{ia})=\sum_{n=1}^\infty\frac{(-1)^{n-1}e^{ina}}{n}x^n
$$
we see that
$$
\int_0^1{\rm Log}(1+x e^{ia})\cdot \frac{1}{x}dx=\sum_{n=1}^\infty\frac{(-1)^{n-1}e^{ina}}{n^2}
$$
Taking real parts we get
$$
\int_0^1 \log|1+x e^{ia}|\cdot \frac{1}{x}dx=\sum_{n=1}^\infty\frac{(-1)^{n-1}\cos(na)}{n^2}
$$
applying this to $a+\pi$ instead of $a$ we obtain also
$$
\int_0^1 \log|1-x e^{ia}|\cdot \frac{1}{x}dx=-\sum_{n=1}^\infty\frac{\cos(na)}{n^2}
$$
Subtracting these two formulas:
$$
\int_0^1 \log\frac{|1+x e^{ia}|}{|1-x e^{ia}|}\cdot \frac{1}{x}dx=
\sum_{n=1}^\infty\frac{((-1)^{n-1}+1)\cos(na)}{n^2}
=\sum_{n=0}^\infty\frac{2\cos((2n+1)a)}{(2n+1)^2}
$$
or
$$
\int_0^1 \log\left(\frac{1+2x\cos(a)+x^2}{ 1-2x \cos(a)+x^2 }\right)\cdot \frac{1}{x}dx
=\sum_{n=0}^\infty\frac{4\cos((2n+1)a)}{(2n+1)^2}\tag{1}
$$
On the other hand if $f$ is the $2\pi$-periodic even function that coincides with $a\mapsto \frac{\pi^2}{2}-\pi a$ on $[0,\pi]$ then it is straightforward to check that the Fourier series expansion of $f$ coincides with the right side of $(1)$. So, we have shown that
$$
\int_0^1 \log\left(\frac{1+2x\cos(a)+x^2}{ 1-2x \cos(a)+x^2 }\right)\cdot \frac{1}{x}dx
=\frac{\pi^2}{2}-\pi |a|
$$
for $a\in[-\pi,\pi]$.$\qquad\square$
