Evaluate Integral: Arctan(πx) - Arctan(x) from 0 to 2

[shwin]

How does one evaluate \int (arctan(pi*x) - arctan(x))dx from 0 to 2 by rewriting the integrand as an integral?

 

[HallsofIvy]

The point is that you have F(b)- F(a) which can be written as
\int_a^b f(x) dx[/itex] where f is the derivative of F. The derivative of arctan(x) is <br /> \frac{x^2+ 1}[/itex] so this could be&lt;br /&gt; \int_0^2 \int_x^{\pi x} \frac{1}{t^2+ 1} dt dx[/itex]&amp;lt;br /&amp;gt; &amp;lt;br /&amp;gt; Now see what happens if you reverse the order of integration:&amp;lt;br /&amp;gt; For every x, t goes from x to \pi x so boundaries on the region on which you are integrating are t= x, t= \pi x, and x= 2 (x= 0 just gives the intersection of the two straight lines). &amp;lt;br /&amp;gt; &amp;lt;br /&amp;gt; If you reverse the order of integration, you will need to take t going from 0 to 2\pi. The limits of integration for x are a bit more complicated. The left end will be at t= \pi x or x= t/\pi. For t between 0 and 2, the right end will be t= x or x= t, but for t between 2 and 2\pi the right end is at x= 2. The integral is&amp;lt;br /&amp;gt; \int_{t=0}^2 \int_{x= t/\pi}^t \frac{1}{1+ t^2} dx dt+ \int_{t= 2}^{2\pi} \int_{x= t/\pi}^2 \frac{1}{1+t^2} dx dt[/itex]