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?
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.