Calculate $\int_{0}^{\infty}\frac{\arctan(x)}{(1+x^2)^{3/2}}dx$ or show that it diverges

Question

Calculate the following improper integral or show that it diverges.

$$\int_{0}^{\infty}\frac{\arctan(x)}{(1+x^2)^{3/2}}dx$$

I'm really lost. Your help would be very appreciated.

Answer 1

Sub $x=\tan{t}$ and the integral is equal to

$$\int_0^{\pi/2} dt \, t \, \cos{t} $$

Which may easily be evaluated by parts:

$$[ t \sin{t}]_0^{\pi/2} - \int_0^{\pi/2} dt \, \sin{t} = \frac{\pi}{2}-1 $$

Answer 2

Hint: For all $x \in (0, \infty)$, we have

$$\frac{\arctan{x}}{(1 + x^2)^{3/2}} \le \frac{\pi/2}{(1 + x^2)^{3/2}}$$

Now what can you say about this integral?

$$\int_0^{\infty} \frac{dx}{(1 + x^2)^{3/2}}$$

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It's clear that $\int_0^1 (1 + x^2)^{-3/2} dx$ is finite, since the function is bounded.

Now on the interval $(1, \infty)$, use the fact that

$$1 + x^2 > x^2 \implies (1+x^2)^{-3/2} < (x^2)^{-3/2} = x^{-3}$$

But what can you say about: $$\int_1^{\infty} x^{-3} dx$$