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Problem of the Week #82 - October 21st, 2013

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Chris L T521

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Jan 26, 2012
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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:
First express the integral as an iterated integral. Then reverse the order of integration and evaluate.


Remember to read the POTW submission guidelines to find out how to submit your answers!
 
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Chris L T521

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Jan 26, 2012
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This week's problem was correctly answered by MarkFL. You can find his solution below.

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.


Here's my solution as well (an alternate to Mark's):

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

\[\begin{aligned}\int_1^{\pi}\int_0^{\infty} \frac{1}{1+(xy)^2}\,dx\,dy &= \int_1^{\pi}\left.\left[\frac{\arctan(xy)}{y}\right]\right|_0^{\infty}\,dy \\ &= \int_1^{\pi} \frac{1}{y}\left[ \lim_{b\to\infty} \arctan(bx) - \arctan(0)\right]\,dy \\ &= \frac{\pi}{2}\int_1^{\pi}\frac{1}{y}\,dy \\ &= \frac{\pi}{2}\left.\left[\ln|y|\right]\right|_1^{\pi}\\ &= \frac{\pi}{2}\left[\ln \pi - \ln 1\right] \\ &= \frac{\pi}{2}\ln\pi\end{aligned}\]
 
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