Show that $\int_0^\infty\frac{\arctan(\mathrm{πx})-\arctan(\mathrm x)}xdx=\;\frac{\mathrm\pi}2\ln(\pi)$ 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..
 A: As pointed out this can be solved through differentiation under the integral sign (Feynman's trick), or simply invoking Frullani's theorem. Since I doubt they are topics in 12th grade Calculus, I will try to outline a single-variable approach as elementary as possible. By integration by parts
$$ \int_{0}^{+\infty}\frac{\arctan(\pi x)-\arctan(x)}{x}\,dx = \int_{0}^{+\infty}\left(\frac{1}{1+x^2}-\frac{\pi}{1+\pi^2 x^2}\right)\log(x)\,dx $$
and by the substitution $x=e^t$ the RHS turns into
$$ \int_{-\infty}^{+\infty}\left(\frac{1}{1+e^{2t}}-\frac{\pi}{1+\pi^2 e^{2t}}\right) te^t\,dt=\pi\int_{0}^{+\infty}\left(\frac{1}{\pi^2+ e^{2t}}-\frac{1}{1+\pi^2 e^{2t}}\right)te^t\,dt $$
since $\int_{-\infty}^{+\infty}g(t)\,dt = \int_{0}^{+\infty}\left(g(t)+g(-t)\right)\,dt$ as soon as $g\in L^1(\mathbb{R})$.
Now we may break the integral into two convergent integrals (we cannot do this at the very beginning since $\frac{\arctan x}{x}$ behaves like $\frac{C}{x}\not\in L^1(1,+\infty)$ for large values of $x$)
$$\mathcal{J}_1=\int_{0}^{+\infty}\frac{t e^t}{\pi^2+e^{2t}}\,dt = \int_{1}^{+\infty}\frac{\log(u)}{\pi^2+u^2}\,du$$
$$\mathcal{J}_2=\int_{0}^{+\infty}\frac{t e^t}{1+\pi^2 e^{2t}}\,dt = \int_{1}^{+\infty}\frac{\log(u)}{1+\pi^2 u^2}\,du=\int_{0}^{1}\frac{-\log(u)}{\pi^2+u^2}\,du$$
then re-combine them into
$$\mathcal{J}_1-\mathcal{J}_2 = \int_{0}^{+\infty}\frac{\log(u)}{\pi^2+u^2}\,du\stackrel{u\mapsto \pi v}{=}\frac{1}{\pi}\int_{0}^{+\infty}\frac{\log\pi+\log v}{1+v^2}\,dv. $$
Last trick: $\int_{0}^{+\infty}\frac{\log v}{1+v^2}\,dv$ equals zero by $v\mapsto e^t$. This leads to:
$$ \mathcal{J}_1-\mathcal{J}_2 = \frac{\log \pi}{\pi}\int_{0}^{+\infty}\frac{dv}{1+v^2}=\frac{\log\pi}{2} $$
and to the claim
$$ \int_{0}^{+\infty}\frac{\arctan(\pi x)-\arctan(x)}{x}\,dx = \color{red}{\frac{\pi\log\pi}{2}}.$$
A: 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 $\frac{\pi}{2}\log(\pi)$.
A: 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, $I(a) = \frac{\pi}{2}\ln(a)$.
