This week's problem was correctly answered by MarkFL. You can find his solution below.
[sp]We are given to demonstrate:
$$I=\int_0^{\infty} \frac{\tan^{-1}(\pi x) - \tan^{-1}(x)}{x}\,dx = \frac{\pi}{2}\ln(\pi)$$
Expressing the integral as an iterated integral, we have:
$$I=\int_0^{\infty}\int_x^{\pi x}\frac{1}{y^2+1}\,dy\,\frac{1}{x}\,dx$$
The region of integration is:
$$0\le x\le\infty$$
$$x\le y\le\pi x$$
Which can also be described by:
$$\frac{y}{\pi}\le x\le y$$
$$0\le y\le\infty$$
Hence, changing the order of integration gives us:
$$I=\int_0^{\infty}\frac{1}{y^2+1}\int_{\frac{y}{\pi}}^{y}\frac{1}{x}\,dx\,dy$$
Applying the FTOC to the inner integral, we find:
$$I=\int_0^{\infty}\frac{1}{y^2+1}\left[\ln|x| \right]_{\frac{y}{\pi}}^{y}\,dy$$
$$I=\int_0^{\infty}\frac{1}{y^2+1}\left(\ln(y)-\ln\left(\frac{y}{\pi} \right) \right)\,dy$$
$$I=\ln(\pi)\int_0^{\infty}\frac{1}{y^2+1}\,dy$$
Applying the FTOC to the outer integral gives us:
$$I=\ln(\pi)\left[\tan^{-1}(y) \right]_0^{\infty}$$
Since $$\tan^{-1}(0)=0$$ and $$\lim_{t\to\infty}\tan^{-1}(y)=\frac{\pi}{2}$$ there results:
$$I=\frac{\pi}{2}\ln(\pi)$$
Shown as desired.[/sp]
Here's my solution as well (an alternate to Mark's):
[sp]Note that we can rewrite the given integral as the following iterated integral:
\[\int_0^{\infty}\frac{\arctan(\pi x) - \arctan x}{x}\,dx = \int_0^{\infty}\int_1^{\pi} \frac{1}{1+(xy)^2}\,dy\,dx\]
and since the region we're integrating over is an infinite rectangle where $0\leq x < \infty$, $1\leq y\leq \pi$, reversing the order of integration is very simplistic and leaves us with
\[\begin{aligned}\int_1^{\pi}\int_0^{\infty} \frac{1}{1+(xy)^2}\,dx\,dy &= \int_1^{\pi}\left.\left[\frac{\arctan(xy)}{y}\right]\right|_0^{\infty}\,dy \\ &= \int_1^{\pi} \frac{1}{y}\left[ \lim_{b\to\infty} \arctan(bx) - \arctan(0)\right]\,dy \\ &= \frac{\pi}{2}\int_1^{\pi}\frac{1}{y}\,dy \\ &= \frac{\pi}{2}\left.\left[\ln|y|\right]\right|_1^{\pi}\\ &= \frac{\pi}{2}\left[\ln \pi - \ln 1\right] \\ &= \frac{\pi}{2}\ln\pi\end{aligned}\][/sp]
