How to determine whether or not these integrals converge?

Question

Below are the two integrals under consideration. I know that for both integrals the 'problem areas' are at $x = 0$ and $x = \infty$. For integrals that are are more simple, I usually check that the power of $x$ in the numerator should be less than $1$ for the interval $x = 0$ to $x = 1$ and greater than $1$ for the interval $x = 1$ to $x = ∞$, but I am unsure what exactly to do here. Thank you

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

2. $$\int_0^\infty x\sin\left(\frac{1}{x^{3/2}}\right) \,dx$$

Answer 1

The first one converges, since at $0$ the limit is $1,$ while at infinity $\arctan(x)$ goes to $\pi/2,$ while the denominator decays quadratically.

The second one diverges, since $\sin(x^{-3/2})$ behaves like $x^{-3/2}$ at infinity, so the integrand decays like $1/\sqrt{x},$ not fast enough.

Answer 2

A couple of good hints:

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

$$\lim_{x\to0}\frac{\arctan(x)}{x(1+x^2)^{1/2}}<\infty$$

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$$x\sin\frac1{x^{3/2}}>\frac1{x^{1/2}}-\frac1{6x^{7/2}}$$