Rewriting an improper integral confusion

Question

I was struggling with the following integral: $$\displaystyle I=\int\limits_{0}^{+\infty}\dfrac{\arctan \pi x-\arctan x}{x}dx$$

And I found a question about it on this site where the answer said that we can rewrite the above integral as follows:

$$I=\int_0^\infty dx\int_1^\pi dy\ \frac{1}{1+x^2y^2}$$

But I have no clue why this is true, could anybody enlighten me?

Answer 1

You can directly compute $$ \int\frac{1}{1+x^2y^2}\,dy=\frac{\arctan(\pi x)-\arctan x}{x} $$ Just substitute $u=xy$.

Answer 2

$$\int_{0}^{+\infty}\frac{\arctan(ax)-\arctan(x)}{x}\,dx = \int_{0}^{+\infty}\frac{1}{x}\int_{x}^{ax}\frac{1}{1+t^2}\,dt\,dx $$ and by substituting $t=xu$, $dt=x\,du$ in the innermost integral we get: $$ \int_{0}^{+\infty}\int_{1}^{a}\frac{1}{1+x^2 u^2}\,du\,dx \stackrel{\text{Fubini}}{=}\int_{1}^{a}\frac{\pi}{2u}\,du=\color{red}{\frac{\pi}{2}\,\log a}.$$ The same can be deduced from the complex version of Frullani's theorem.

Answer 3

Another way you could do this is by differentiation under the integral $I(\alpha)=\int_{0}^{\infty}\frac{\tan^{-1} \alpha x}{x}\text{d}x\to I'(\alpha)=\int_{0}^{\infty}\frac{1}{1+(\alpha x)^2}\text{d}x=\frac{1}{\alpha}\tan^{-1} (\alpha x)\big\rvert_{0}^{\infty}=\frac{\pi}{2\alpha}.$ integrating it to get back to $I(\alpha)$ we get $\frac{\pi}{2}\ln \alpha+C$ because we have a difference of these 2 functions we know that the constant will cancel out. So using this information we have $\int\frac{\tan^{-1} ax-\tan^{-1} bx}{x}\text{d}x=\frac{\pi}{2}\ln \left (\frac{a}{b}\right )$