Finding values where an improper integral converges

Question

I’n an exercise I’ve been asked to find the values for $a$ such that:

$\int^{\infty}_0 \frac{x^a \sin(x)}{1+x^2} dx $

converges

I think I have to split the integral into two parts at some positive number, $\epsilon $, because the behaviour of the function (call it $f(x)$) is concerning at $0$ and $\infty$.

I can’t evaluate the integral.

Beyond that I don’t know what to do.

Using $1+x^2\leq x^2 $ and $-1 \leq \sin x \leq 1$, I found simpler integrals that are greater than the given integral such as $\int ^{\infty}_\epsilon x^{a-2} dx$ and found where they converge. But this doesn’t tell me whether the given integral converges when the greater integral diverges to infinity.

I’ve also tried using the substitutions $x=\sec u$ and $u=\arctan x$ but both make the integral more complicated. Then I tried $u=x^{a+1}$ which also failed.

I thought of using integration by parts or the maclaurin expansion of sin or that $\sin x \approx x$ near $0$ but didn’t know how’d they help.

Answer 1

This is not a proof.

As I wrote in comments, I think that the integral converges if $|a|<2.

Searching my old cookbook, I found two of them $$I_1=\int_0^\infty \frac{x^{3/2} \sin (x)}{x^2+1}\,dx$$ $$I_1=\frac{e \,\pi }{2 \sqrt{2}}\text{erf}(1)-\frac{\pi }{2 \sqrt{2} e} \text{erfi}(1)+\sqrt{\frac{\pi }{2}}-\frac{\pi \sinh (1)}{\sqrt{2}}$$

$$I_2=\int_0^\infty \frac{x^{-3/2} \sin (x)}{x^2+1}\,dx$$ $$I_2=\sqrt{2\pi}-\frac{\pi }{2 \sqrt{2} e} \left(e^2 \text{erf}(1)+\text{erfi}(1)-2 e \sinh (1)\right)$$

Answer 2

This is a mixture of type one and type two improper integral, so we have to split it into two parts.
That is $\int_{0}^{\infty}\frac{x^asin(x)}{1+x^2}=\int_{0}^{1}\frac{x^asin(x)}{1+x^2}+\int_{1}^{\infty}\frac{x^asin(x)}{1+x^2}$
We first look at the type two improper integral.
I will use the limit comparison test here.
If we set $f(x)=\frac{x^asin(x)}{1+x^2}$, then i will set $g(x)=\frac{1}{x^{-1-a}}$.
So $lim_{x\rightarrow 0}\frac{f(x)}{g(x)}=lim_{x\rightarrow 0}\frac{sin(x)}{x(1+x^2)}dx=lim_{x\rightarrow 0}\frac{sin(x)}{x}*lim_{x\rightarrow 0}\frac{1}{1+x^2}=1*1=1$
So we got the property of converge of $\int_{0}^{1}\frac{x^asin(x)}{1+x^2}$ is equal to which of the improper integral $\int_{0}^{1}\frac{1}{x^{-1-a}}dx$,
So we got the limitation of the domain of a is a>-2 at least.
Now we have to look at the type one improper integral.
I want to do some work in advance first, i want to change this question into that for an improper integral $\int_{1}^{\infty}x^msin(x)dx$ about when will it converge, because i can predict that through the limitation comparison i will meet this problem eventually.
For this integral to converge, the oscillatory behavior of sinx must have to be balances by $x^m$ as x approaches infinity.
This part will be really difficult.
I will try to learn it first, and try to solve this problem here.
But up to now we have a>-2 to be a necessary condition.