integration - Show that $int_0^inftyfrac{arctan(mathrm{πx})-arctan(mathrm x)}xdx=;frac{mathrmpi}2ln(pi)$ using 12th grade calculus.

I have to show that $\int_0^\infty\frac{\arctan(\mathrm{πx})-\arctan(\mathrm x)}xdx=\;\frac{\mathrm\pi}2\ln(\pi)$ using 12th grade calculus wich means single variable calculus.

What I've tried:

First I tried to make a single arctan from those 2: $\arctan(\mathrm{πx})-\arctan(\mathrm x)=\arctan(\frac{(\mathrm\pi-1)\mathrm x}{1+\mathrm{πx}^2})$ as you can see it's still not very pleasant... and it looks like I don't get anywhere with this..

Answer

Let $I(a) = \int_0^\infty \frac{\arctan(a x)-\arctan(x)}{x}dx$. Then $I'(a) = \int_0^\infty \frac{1}{x}\frac{x}{1+(ax)^2}dx = \frac{1}{a}\frac{\pi}{2}$. So, $I(a) = \frac{\pi}{2}\ln(a)+C$. Letting $a=1$, we see that $0=I(1)=\frac{\pi}{2}\ln(1)+C \implies C=0$. So, $I(a) = \frac{\pi}{2}\ln(a)$.