Calculate $\int\limits_{-\infty}^{\infty}\frac{\sin^2(x)\cos(wx)}{x^2}dx$ using complex analysis technique

Question

In a complex analysis test an exercise asks to calculate ($w \in \Bbb{R}$): $$\displaystyle\int_{-\infty}^{\infty}\frac{\sin^2(x)\cos(wx)}{x^2}dx$$ Of course I need to solve it with complex analysis technique. I have used all the tricks I know such as integrate on a semi-circumference of radius $R$ in $\Bbb{C}$ and then let $R \to \infty$ but everything I have tried to do just seemed useless. Thank you for every hint or solution to this problem!

Answer 1

If I had to use complex methods in this, the place I'd do it is in proving$$\int_{\mathbb{R}}\frac{\sin^{2}x}{x^{2}}dx=\pi$$(e.g. by the residue theorem; see here for the gory details) so$$\left|k\right|\pi=\int_{\mathbb{R}}\frac{\sin^{2}kx}{x^{2}}dx=\int_{\mathbb{R}}\frac{1-\cos2kx}{2x^{2}}dx.$$Hence$$\int_{\mathbb{R}}\frac{\sin^{2}x\cos wx}{x^{2}}dx=\int_{\mathbb{R}}\frac{2\cos wx-\cos\left(w-2\right)x-\cos\left(w+2\right)x}{4x^{2}}dx\\=\frac{\pi}{4}\left(\left|w-2\right|+\left|w+2\right|-2\left|w\right|\right).$$As a sanity check, the case $w=0$ needs to give $\pi$, and lo and behold it does.

Answer 2

$$I(w)=\int_{-\infty}^\infty\frac{\sin^2(x)\cos(wx)}{x^2}dx$$ $$I'(w)=-\int_{-\infty}^\infty\frac{\sin^2(x)\sin(wx)}{x}$$ $$I''(w)=-\int_{-\infty}^\infty\sin^2(x)\cos(wx)dx$$ $$I''(w)=-\Re\int_{-\infty}^\infty\sin^2(x)e^{iwx}dx$$ and then try integration by parts. Alternatively: $$I(w)=\Re\int_{-\infty}^\infty\frac{\sin^2(x)e^{iwx}}{x^2}dx$$ $$I(w)=\Re\int_{-\infty}^\infty\sin^2(x)\sum_{n=0}^\infty\frac{(iw)^nx^{n-2}}{n!}dx$$

Answer 3

This is certainly not a complex analytic method, but Fourier inversion formula also can give you the result. It holds that $$ \int_{-1}^1 e^{-i\xi x}dx=\frac{2\sin \xi}{\xi}. $$ Let $f(x) = 1_{[-1,1]}(x)$. This shows $\widehat{f}(\xi)=\frac{2\sin \xi}{\xi}$ and therefore, $\widehat{(f*f)}(\xi)=f(\xi)^2=4\frac{2\sin^2 \xi}{\xi^2}$.
The given integral is $$\begin{eqnarray} \int_{-\infty}^{\infty}\frac{\sin^2(x)\cos(wx)}{x^2}dx&=&\int_{-\infty}^{\infty}\frac{\sin^2(x)e^{iwx}}{x^2}dx\\ &=&2\pi \mathcal{F}^{-1}\left[\frac{\sin^2 x}{x^2}\right](w)\\ &=&\frac{\pi}{2}(f*f)(w). \end{eqnarray}$$ Finally, we have $$\begin{eqnarray} (f*f)(w)&=&\int f(w-y)f(y)dy\\ &=&\int_{|w-y|<1,|y|<1}dy\\ &=&\begin{cases}2-w,\quad w\in[0,2)\\2+w,\quad w\in(-2,0]\\0,\quad|w|>2\end{cases}\\ &=&(2-|w|)1_{|w|<2}. \end{eqnarray}$$ This gives that the given integral is equal to $$ \frac{\pi}{2}(2-|w|)1_{|w|<2}. $$