We consider two strictly positive real numbers $a$ and $b$, and we set $\rho = \frac{b-a}{b+a}$. We call $\Psi$ the application from $\mathbf{R}$ to $\mathbf{R}$ defined by:
$$\forall x \in \mathbf{R}, \Psi(x) = \ln(a^2 \cos^2 x + b^2 \sin^2 x)$$
and $\sigma(x) = \sum_{k=1}^{+\infty} \frac{x^k}{k^2}$.
Conclude that
$$\int_0^{\pi} \Psi(x)^2 \mathrm{d}x = 4\pi\left(\ln\left(\frac{a+b}{2}\right)\right)^2 + 2\pi\sigma(\rho^2)$$