I´m trying to find out if this integral is convergent (and for what values of $a$) or not:
\begin{equation*} \int^{+\infty}_0 \frac{\arctan x}{x^a} \sin x \ dx,\quad a \in \mathbb R \end{equation*}
The problem is almost the same if $x$ is near 0 or infinity, so I'll show here only one part (the second one should be similar, as I've said).
I've tried this:
If x is near 0:
I wanted to use Abel's test, so in my opinion, the function $\arctan{x}$ is monotonic (and differentiable) on the interval $(0,1]$. Then the "remaining function" is $\frac{\sin x}{x^a}$ and
$\int^1_0 \frac{\sin x}{x^a} ~ dx$ is convergent if and only if $a<2$.
So then, $\int^1_0 \frac{\arctan x}{x^a} \sin x ~ dx$ is convergent for $a<2$ according the Abel's test, however ABEL'S TEST ISN'T THE SENTENCE IN FORM OF EQUIVALENCE, so this doesn't have to be the final result.
It only says to me that if $a<2$, this integral converges, but it doesn't says nothing about the situation when $a \geq 2$.
Am I right? And if I am, how can I solve the rest of problem?
Thanks for any help!
Note that $\lim_{x\to\infty}\arctan(x)=\pi/2$ and $\lim_{x\to 0^+}\arctan(x)/x=1$; in other words, for large $x$ $\arctan(x)$ is roughly constant, and $\arctan(x)\sim x$ for $x$ near zero. We split the integral into $\int_0^1$ and $\int_1^{\infty}$, say $I_1$ and $I_2$.
By Dirichlet's Test, a necessary and sufficient condition for $I_2$ to converge is $a>0$, since $\sin(x)\arctan(x)$ is uniformly bounded by $\pi/2$ and then $x^{-a}$ is positive and decreasing.
For $I_1$, near $x=0$ one could use a Taylor series: $$ x^{-a}\arctan(x) \sin(x) = x^{-a}\left(x-\frac{x^3}{3}+\cdots\right)\left(x-\frac{x^3}{6}+\cdots\right)= O(x^{2-a}) $$This is improperly integrable iff $2-a>-1$, i.e. $3<a$.
Combining these conditions yields $0<a<3$.