Asymptotic behavior of $\int_{0}^r \frac{\sin(x)^2}{x^2} x^{\alpha}dx$.

Question

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.

Answer 1

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 $r\to\infty$.

Answer 2

The solution by @TheSimpliFire fully addresses the problem. It would be interesting however to find a full asymptotic of the integral.

Let's denote $\displaystyle I(\alpha, r)=\int_{0}^r \frac{\sin^2(x)}{x^2} x^{\alpha}\,dx=2^{-\alpha}\Re\int_0^{2r}(1-e^{it})t^{\alpha-1}\,dt$ and suppose for a while that $\alpha>1$.Let's also consider the integral in the complex plane along a closed contour (a quarter-circle): $ s\to 2r\to i2r\to is\to s$. There are no singularities insider the contour; therefore $$2^{-\alpha}\oint_C(1-e^{iz})z^{\alpha-1}\,dz$$ $$=2^{-\alpha}\int_s^{2r}(1-e^{it})t^{\alpha-1}\,dt+I_r+2^{-\alpha}\int_{2ir}^{is}(1-e^{it})t^{\alpha-1}\,dt+I_s=0$$ where we added a small arch of the radius $s\to0$ to close the contour. Integral along this arch $\to 0$ at $s\to 0$ and $\alpha>1$. Hence, we get $$I(\alpha,r)=2^{-\alpha}\Re\left(\int_0^{2ir}(1-e^{it})t^{\alpha-1}dt-I_r\right)$$ $$=\Re\left(2^{-\alpha}\int_0^{2ir}(1-e^{it})t^{\alpha-1}dt-2^{-\alpha}\int_0^{\frac \pi2}(1-e^{2ir e^{i\phi}})\left(2re^{i\phi}\right)^{\alpha-1}i\,d\phi\right)$$ $$=2^{-\alpha}\sin\frac{\pi\alpha}2\int_0^{2r}(1-e^{-t})t^{\alpha-1}dt-2^{-\alpha}\left(2r\right)^{\alpha-1}\int_1^i(1-e^{2irz})z^{\alpha-2}dz$$ where we denoted $\displaystyle z=e^{i\phi}$, and integration with respect to $z$ goes along the arch of the radius $\displaystyle2r$.

Integrating by part and dropping exponentially small terms for $r\gg1$

(for example, $\displaystyle \int_0^rt^{\alpha-1}e^{-t}dt=\Gamma(\alpha)+O(e^{-r}r^\alpha)\,$) $$I(\alpha,r)\sim2^{-\alpha}\frac{\sin\frac{\pi\alpha}2}{\alpha-1}\left(2r\right)^{\alpha-1}-2^{-\alpha}\frac{\sin\frac{\pi\alpha}2\Gamma(\alpha)}{\alpha-1}$$ $$-\,2^{-\alpha}\Re\left(\,\left(2r\right)^{\alpha-1}\frac1{1-\alpha}\big(e^{\frac{\pi i}2(\alpha-1)}-1\big)+\left(2r\right)^{\alpha-2}\frac{e^{2ir}}{2ir}+\left(2r\right)^{\alpha-2}\frac{\alpha-2}{2ir}\int_1^ie^{2irz}z^{\alpha-3}dz\right)$$ Integrating by part several times and making some rearrangement, we see that at $r\gg1$ $$I(\alpha,r)\sim\frac{2^{-\alpha}}{\alpha-1}\left(\left(2r\right)^{\alpha-1}-\Gamma(\alpha)\sin\frac{\pi\alpha}2\right)$$ $$+\,2^{-\alpha}\sin2r\left(-\left(2r\right)^{\alpha-3}+\left(2r\right)^{\alpha-5}(\alpha-2)(\alpha-3)- ...\right)$$ $$+\,2^{-\alpha}\cos2r\left(-\left(2r\right)^{\alpha-4}(\alpha-2)+\left(2r\right)^{\alpha-6}(\alpha-2)(\alpha-3)(\alpha-4)- ...\right)\tag{1}$$ After a slight rearrangement, the asymptotic can also be presented in the form $$I(\alpha,r)\sim\frac{2^{-\alpha}}{\alpha-1}\left(\left(2r\right)^{\alpha-1}-\Gamma(\alpha)\sin\frac{\pi\alpha}2\right)$$ $$+\,\frac{r^{\alpha-1}}{2\Gamma(2-\alpha)}\sum_{k=1}^\infty(-1)^k\left(2r\right)^{-2k}\Gamma(2k-\alpha)\left(\sin2r-\frac{(2k-\alpha)}{2r}\cos2r\right)\tag{2}$$ Employing analytical continuation we see that the asymptotic is valid for $\alpha>-1$. For positive integers $\alpha$, as it follows from (1), we get a limited number of terms (as it should be).

For $\alpha=1$ the main term is ($\alpha=1+\epsilon$) $$I(\alpha,r)\sim\frac{2^{-\alpha}}{\alpha-1}\left(\left(2r\right)^{\alpha-1}-\Gamma(\alpha)\sin\frac{\pi\alpha}2\right)\to\lim_{\epsilon\to0}\frac1{2\epsilon}\left(1+\epsilon\ln2r-(1-\gamma\epsilon)+O(\epsilon^2)\right)$$ $$=\frac12\left(\ln2r+\gamma\right)$$ At $\alpha\to0$ and $r\to\infty$ we get the well-known result $$I(\alpha=0;r\to\infty)=\int_{0}^\infty \frac{\sin^2(x)}{x^2} \,dx=\lim_{\alpha\to 0}\Gamma(\alpha)\sin\frac{\pi\alpha}2=\frac\pi2$$

Answer 3

Using almost the same steps as @Svyatoslav, we can write, if $\alpha > -1$, $$I=\int_{0}^r \frac{\sin(x)^2}{x^2} x^{\alpha}\,dx=\frac{r^{\alpha -1} \sin ^2(r)}{\alpha -1}-\frac{2\, r^{\alpha +1} }{\alpha ^2-1}\, _1F_2\left(\frac{\alpha +1}{2};\frac{3}{2},\frac{\alpha+3 }{2};-r^2\right)$$

Expanded as a series for small $\alpha$ $$I=\text{Si}(2 r)-\frac{\sin ^2(r)}{r}+O(\alpha)$$

If $\alpha=1$, the problem is simple $$I=\frac{1}{2} (\log (2r)-\text{Ci}(2 r)+\gamma )$$

There is another case which is interesting : using the incomplete gamma function and then switching to the exponential integral $$\int \frac{\sin(x)^2}{x^2} x^{\alpha}\,dx=\frac{x^{\alpha -1}}{2 (\alpha -1)}+\frac{1}{4} x^{\alpha -1} (E_{2-\alpha }(-2 i x)+E_{2-\alpha }(2 i x))$$ and, if $|\alpha|<1$ then $$\int_{0}^\infty \frac{\sin(x)^2}{x^2} x^{\alpha}\,dx=-2^{-\alpha }\, \sin \left( \frac{\alpha\pi }{2}\right)\, \Gamma (\alpha -1)$$ which is $$\frac{\pi }{2}-\frac{\pi}{2} \left(\log \left(\frac{2}{e}\right)+\gamma \right)\,\alpha +O\left(\alpha ^2\right)$$