Improper integral $\int_0^{+\infty} \frac{\arctan x}{\sqrt{x}} \sin(2x) dx$

Question

Check absolute and conditional convergence of following improper integral $$\int_0^{+\infty} \frac{\arctan x}{\sqrt{x}} \sin(2x) dx$$

Can someone please check if my solution is alright.

Here is what I have done, first my intentions were to find all singularities, here that is only $+ \infty$ because in zero limit exists (it is equal to zero).

Using Dirichlet's test, integral $$\int_0^{+\infty} \frac{\sin(2x)}{\sqrt{x}}$$ converges because $$\sin(2x)$$ has bounded primitive function and $$\frac{1}{\sqrt{x}}$$ converges to $0$ in $+\infty$.

Now, using Abel's test, we can conclude that former integral converges because, $$\int_0^{+\infty} \frac{\sin(2x)}{\sqrt{x}} $$ converges and $$\arctan x$$ is bounded function.

So, in conclusion, I would say that this integral converges both absolutely and conditionaly. Is this okay? Is there any other way to check convergence here?

Answer 1

We have $${\arctan x\over \sqrt{x}}={{\pi \over 2}\over \sqrt{x}} -{{\pi \over 2}-\arctan x\over \sqrt{x}}$$ Both functions on RHS are nonnegative, decreasing, with limit $0$ at $\infty.$ On multiplying by $\sin 2x$ we can apply the Dirichlet test for the interval say $[\pi,\infty).$

Answer 2

There are no integrability issues on a right neighbourhood of the origin since there $\frac{\arctan x}{\sqrt{x}}$ behaves like $\sqrt{x}$.
By taking $\frac{1-\cos(2x)}{2}=\sin^2(x)$ as a primitive for $\sin(2x)$, integration by parts leads to

$$ \int_{0}^{M}\frac{\arctan(x)}{\sqrt{x}}\sin(2x) = \left[\frac{\arctan(x)}{\sqrt{x}}\sin^2(x)\right]_{0}^{M}-\int_{0}^{M}\sin^2(x)\cdot\frac{d}{dx}\left(\frac{\arctan(x)}{\sqrt{x}}\right)\,dx. $$ As $M$ tends to $+\infty$ the first term in the RHS converges to zero since $\arctan(x)$ is bounded. The derivative of $\frac{\arctan(x)}{\sqrt{x}}$ in absolute value behaves like $\frac{1}{\sqrt{x}}$ in a right neighbourhood of the origin and like $\frac{1}{x\sqrt{x}}$ in a left neighbourhood of $+\infty$, $\sin^2(x)$ is non-negative and bounded, hence the integral is convergent.