How to prove convergence of the integral $\int_{0}^{\infty} x^{-a}\sin(x)dx$ for all $0<a<2$.

Sketch of a proof:

Set $f_a(x)=x^{-a}\sin x$. Split the integral from $0$ to $\pi/2$ and then from $\pi/2$ say $(2m+1)\pi/2$.

If $0<a<1$ there is nothing to worry about $\int^{\pi/2}_0f(x)\,dx$ since then $|f_a(x)|\leq x^{-a}$ and $x^{-a}\in L_1(0,\pi/2)$. If $1\leq a$, then $|f_a(x)|=\Big|\frac{\sin x}{x}\Big|\Big|\frac{1}{x^{a-1}}\Big|\leq \Big|\frac{1}{x^{a-1}}\Big|\in L_1(0,\pi/2)$.

Thus, the only concern is $\lim_{b\rightarrow\infty}\int^b_{\pi/2} x^{-a}\sin x$. For $1<a<2$, there is nothing to worry about since $x^{-a}\in L_1(\pi/2,\infty)$. For $0<a\leq1$ you may proceed as follows.

It will be enough to consider $b$ growing along the sequence $b_n=(2n+1)\pi/2$.

$$ \int^{(2n+1)\pi/2}_{\pi/2} x^{-a}\sin x\,dx=\sum^{n-1}_{k=1}(-1)^k\int^{(2k+3)\pi/2}_{(2k+1)\pi/2}|f_a(x)|\,dx$$ Then it is not difficult to see that the sum on the right is in fact an alternating sum with terms decreasing to $0$.

Some details should be filled in, but nothing to complicated.

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For $0<a<1$, the limit of the existence of limit $\lim_{b\rightarrow\infty}\int^b_{\pi/2}f_a(x)\,dx$ DOES NOT imply that $f_a\in L_1(\pi/2,\infty)$ (in the sense of Lebesgue) of course, in fact $f_a\notin L_1([\pi/2,\infty)$, as once can see by noticing that $\int^{(2m+3)\pi/2}_{(2m+1)\pi/2}|f_a|\geq c\frac{1}{m^a}$ for some constant.