How Do You Prove This Integral Equals \(\frac{\pi}{2}\ln \pi\)?

[Chris L T521]

Thanks again to those who participated in last week's POTW! Here's this week's problem!

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Problem: Show that $\displaystyle \int_0^{\infty} \frac{\arctan(\pi x) - \arctan x}{x}\,dx = \frac{\pi}{2}\ln \pi$.

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Hint: [sp]First express the integral as an iterated integral. Then reverse the order of integration and evaluate.[/sp]

 

[Chris L T521]

This week's problem was correctly answered by MarkFL. You can find his solution below.

[sp]We are given to demonstrate:

\(\displaystyle I=\int_0^{\infty} \frac{\tan^{-1}(\pi x) - \tan^{-1}(x)}{x}\,dx = \frac{\pi}{2}\ln(\pi)\)

Expressing the integral as an iterated integral, we have:

\(\displaystyle I=\int_0^{\infty}\int_x^{\pi x}\frac{1}{y^2+1}\,dy\,\frac{1}{x}\,dx\)

The region of integration is:

\(\displaystyle 0\le x\le\infty\)

\(\displaystyle x\le y\le\pi x\)

Which can also be described by:

\(\displaystyle \frac{y}{\pi}\le x\le y\)

\(\displaystyle 0\le y\le\infty\)

Hence, changing the order of integration gives us:

\(\displaystyle I=\int_0^{\infty}\frac{1}{y^2+1}\int_{\frac{y}{\pi}}^{y}\frac{1}{x}\,dx\,dy\)

Applying the FTOC to the inner integral, we find:

\(\displaystyle I=\int_0^{\infty}\frac{1}{y^2+1}\left[\ln|x| \right]_{\frac{y}{\pi}}^{y}\,dy\)

\(\displaystyle I=\int_0^{\infty}\frac{1}{y^2+1}\left(\ln(y)-\ln\left(\frac{y}{\pi} \right) \right)\,dy\)

\(\displaystyle I=\ln(\pi)\int_0^{\infty}\frac{1}{y^2+1}\,dy\)

Applying the FTOC to the outer integral gives us:

\(\displaystyle I=\ln(\pi)\left[\tan^{-1}(y) \right]_0^{\infty}\)

Since \(\displaystyle \tan^{-1}(0)=0\) and \(\displaystyle \lim_{t\to\infty}\tan^{-1}(y)=\frac{\pi}{2}\) there results:

\(\displaystyle I=\frac{\pi}{2}\ln(\pi)\)

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]