I know that $\int_{\mathbb{R}} \frac{\sin(x)^2}{x^2} dx < \infty$ and more precisely that it equals $\frac{\pi}{2}$ (thanks to Plancherel theorem).
I am interested in the asymptotic behavior of an integral related to this :
$$I_{\alpha}(r) = \int_{0}^r \frac{\sin(x)^2}{x^2} x^{\alpha}dx \sim_{r\to \infty} ?$$
For $0 \leq \alpha < 1$, the integral converge, thus $I_{\alpha}(r) \sim \int_{0}^{\infty}\frac{\sin(x)^2}{x^2} x^{\alpha}dx$.
For $\alpha = 1$, I suspect that $I_{1}(r) \sim cste \times \log(r)$, but I wasn't able to derive a rigorous proof.
For $\alpha > 1$, I suspect that $I_{\alpha}(r) \sim cste \times r^{\alpha -1}$.
I don't know how to proceed because the $\sin$ oscillate and may compensate the divergence. Thus for $\alpha \geq 1$, the divergence may be slower than expected.
I hope that the question is not answered somewhere. I did some reasearch but I didn't find anything.
Thanks.
Best Answer
Using $2\sin^2x=1+\cos2x$ we can rewrite the integral as \begin{align}I_\alpha(r)&=\frac12\int_0^rx^{\alpha-2}\,dx+\frac12\int_0^rx^{\alpha-2}\cos2x\,dx\\&=\frac12\frac{r^{\alpha-1}}{\alpha-1}+\frac12r^{\alpha-1}\int_0^1t^{\alpha-2}\cos2rt\,dt\end{align} for $\alpha>1$. This integral is asymptotically zero due to Riemann-Lebesgue, taking $f(t)=t^{\alpha-2}\chi_{[0,1]}(t)$ which is $L^1[0,\infty)$.
When $\alpha=1$, we can use the asymptotics of the cosine integral. Noting that $\operatorname{Ci}(+\infty)=0$ and and $\operatorname{Ci}(\varepsilon)\sim\gamma+\log\varepsilon$ for small $\varepsilon$, we obtain\begin{align}I_1(r)&=\frac12\lim_{\varepsilon\to0^+}\left[\log\frac r\varepsilon+\operatorname{Ci}(2r)-\operatorname{Ci}(2\varepsilon)\right]\sim\frac12\log r+\frac12(\gamma+\log 2)\end{align} as $r\to\infty$.