# Existence of improper integral

Prove that $$\int_{0}^{\infty} \frac{(\arctan x)^2}{x^2} dx$$ converges.

This is my attempt: The above integral is equal to $$\int_{1}^{\infty} \frac{(\arctan x)^2}{x^2} dx + \int_{0}^{1} \frac{(\arctan x)^2}{x^2} dx.$$

The first integral exists since it's smaller than $$\frac{\pi^2}{4} \int_{1}^{\infty} \frac{1}{x^2} dx < \infty.$$

For the existence of the second integral, I first defined $f(x) = \frac{(\arctan x)^2}{x^2}$ for $x>0$ and $f(0) = 1$. $f$ is continuous and thus realises its maximum on the interval $[0,1]$, call this number $M$. The second integral then is smaller than $M$. In fact, isn't $f$ decreasing such that $M = 1$ ?

Is this approach correct? In particular, is my proof of the second integral correct?

• Its value is $\pi\ln2$. – Lucian Jul 23 '14 at 13:26

Yes correct. Notice that you can also (and it's more simple) use the asymptotic equivalence of the function at $0$ and at $+\infty$. In fact, we have
$$\frac{\arctan^2x}{x^2}\sim_\infty\frac{\pi^2}{4x^2}\in L^1([1,+\infty))$$ and $$\frac{\arctan^2x}{x^2}\sim_01\in L^1((0,1])$$ so the given integral is convergent.
• Just you should know the Taylor series at $0$ of the usual functions and since $\lim_{x\to\infty}\arctan x=\frac\pi2$ then we have $$\arctan x\sim_\infty\frac\pi2$$ – user63181 Jul 23 '14 at 12:52