TITLE: Show that $\int_0^\infty\frac{\arctan(\mathrm{πx})-\arctan(\mathrm x)}xdx=\;\frac{\mathrm\pi}2\ln(\pi)$ using 12th grade calculus.
QUESTION [5 upvotes]: 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..

REPLY [2 votes]: While not really 12th grade level, Glaisher's theorem should be of interest to the OP, given the interest in learning about integration methods. Glaisher's theorem is a special case of Ramanujan's master theorem, it says:
If $f(x)$ is an even function with series expansion around $x = 0$:
$$f(x)= \sum_{k=0}^{\infty}(-1)^kc_k x^{2k}$$
and the integral over the real line converges, then we have:
$$\int_{0}^{\infty}f(x) dx = \frac{\pi}{2} c_{-\frac{1}{2}}$$
$c_n$ is a priori only defined for integer $n$, but when an analytic expression for $c_n$ is known then one should be able to substitute $n = -\frac{1}{2}$. Here expressions involving factorials should be replaced by gamma functions, often one deals with simple rational functions in which case putting $n = -\frac{1}{2}$ poses no problems. In some cases one needs to take the limit $n \rightarrow -\frac{1}{2}$.
In this case, we have:
$$c_n = \frac{\pi^{2n+1} - 1}{2 n + 1}$$
for $n = -\frac{1}{2}$ this becomes ill defined, but as mentioned above, we then need to take the limit for $n \rightarrow -\frac{1}{2}$. This yields:
$$\lim_{n \rightarrow -\frac{1}{2}}c_n = \log(\pi)$$
The integral thus equals $\frac{\pi}{2}\log(\pi)$.