Evaluate the integral $\int_{0}^{+\infty}\frac{\arctan \pi x-\arctan x}{x}dx$ Compute improper integral : $\displaystyle I=\int\limits_{0}^{+\infty}\dfrac{\arctan \pi x-\arctan x}{x}dx$.

We have
$$\int_a^b \dfrac{\arctan \pi x-\arctan x}{x}dx=\int_{\pi a}^{\pi b}\dfrac{\arctan x}{x}dx-\int_{ a}^{ b}\dfrac{\arctan x}{x}dx\\=\int_{ b}^{\pi b}\dfrac{\arctan x}{x}dx-\int_{ a}^{ \pi a}\dfrac{\arctan x}{x}dx$$
and since the function $\arctan$ is increasing so
$$\arctan( b)\log\pi=\arctan( b)\int_b^{\pi b}\frac{dx}{x}\leq\int_{ b}^{\pi b}\dfrac{\arctan x}{x}dx\\ \leq\arctan(\pi b)\int_b^{\pi b}\frac{dx}{x}=\arctan(\pi b)\log\pi$$
so if $b\to\infty$ we have
$$\int_{ b}^{\pi b}\dfrac{\arctan x}{x}dx\to\frac{1}{2}\pi\log\pi$$
and by a similar method we prove that
$$\int_{ a}^{ \pi a}\dfrac{\arctan x}{x}dx\to 0,\quad a\to0$$
hence we conclude
$$\displaystyle I=\int\limits_{0}^{+\infty}\dfrac{\arctan \pi x-\arctan x}{x}dx=\frac{1}{2}\pi\log\pi$$
