# Thread: Improper Integral

1. ## Improper Integral

I'm having trouble determining if this integral is convergent or not:

$\int_0^{\infty}\frac{xArctan(x)}{(1+x^2)^2}dx=\lim _{t->\infty}\int_0^{t}\frac{xArctan(x)}{(1+x^2)^2}dx$

If I graph the integrand, it appears that

$\frac{xArctan(x)}{(1+x^2)^2}\leq \frac{Arctan(x)}{1+x^2}$ for $x>0$

So I attempted the comparison theorem:

$\lim_{t->\infty}\int_0^{\infty}\frac{Arctan(x)}{1+x^2}dx$

$u=Arctan(x)\Rightarrow du=\frac{dx}{1+x^2}$

$\lim_{t->\infty}\int_0^{Arctan(t)}udu$

$=\int_0^{\frac{\pi}{2}}udu$

This is divergent right? So this was useless??

2. wait

or perhaps...

$\lim_{t->\infty}\int_0^{Arctan(t)}udu$

$=\lim_{t->\infty}\frac{1}{2}(Arctan(t))^2-0$

$=\frac{\pi^2}{8}$

3. split it into two sets, one for $[0,1]$ and the other one for $[1,\infty),$ so for each $x\ge1$ is $\frac{x\arctan x}{\left( 1+x^{2} \right)^{2}}<\frac{\arctan x}{x^{3}}<\frac{\pi }{2x^{3}},$ and the integral converges.

4. I understand everything except the point of writing:

$\int_0^{\infty}\frac{xArctan(x)}{(1+x^2)^2}dx=\int _0^1\frac{xArctan(x)}{(1+x^2)^2}dx+\int_1^{\infty} \frac{xArctan(x)}{(1+x^2)^2}dx$

Why split the interval?

5. i could've picked any number there, but to make my life easier i took a "one."

besides, one doesn't care about the first piece, since is a continuous function being integrated over a compact set.