Evaluate $\int_{-\infty }^{\infty } \frac{\sin^2x}{x^2} \, dx$

Question

How to evaluate:

$$\int_{-\infty }^{\infty } \frac{\sin^2x}{x^2} \, dx$$

This is a problem in my complex variables course. This integral is intended to be solved by contour, but I’m struggling with it. Thanks for anyone can give me some hints!

Answer 1

Well, we can write:

$$\mathscr{I}_{\space\text{n}}:=\int_{-\infty}^\infty\frac{\sin^2\left(\text{n}\cdot x\right)}{x^2}\space\text{d}x=2\cdot\int_0^\infty\frac{\sin^2\left(\text{n}\cdot x\right)}{x^2}\space\text{d}x\tag1$$

Because it is an even function.

Using Laplace transform, we can write:

$$\mathcal{I}_{\space\text{n}}\left(\text{s}\right):=\mathscr{L}_x\left[\frac{1}{x}\cdot\frac{\sin^2\left(\text{n}\cdot x\right)}{x}\right]_{\left(\text{s}\right)}=\int_\text{s}^\infty\mathscr{L}_x\left[\frac{\sin^2\left(\text{n}\cdot x\right)}{x}\right]_{\left(\sigma\right)}\space\text{d}\sigma=$$ $$\int_\text{s}^\infty\int_\sigma^\infty\mathscr{L}_x\left[\sin^2\left(\text{n}\cdot x\right)\right]_{\left(\theta\right)}\space\text{d}\theta\space\text{d}\sigma=\int_\text{s}^\infty\int_\sigma^\infty\mathscr{L}_x\left[\frac{1-\cos\left(2x\right)}{2}\right]_{\left(\theta\right)}\space\text{d}\theta\space\text{d}\sigma=$$ $$\frac{1}{2}\cdot\int_\text{s}^\infty\int_\sigma^\infty\left(\mathscr{L}_x\left[1\right]_{\left(\theta\right)}-\mathscr{L}_x\left[\cos\left(2x\right)\right]_{\left(\theta\right)}\right)\space\text{d}\theta\space\text{d}\sigma=$$ $$\frac{1}{2}\cdot\int_\text{s}^\infty\int_\sigma^\infty\left(\frac{1}{\theta}-\frac{\theta}{4+\theta^2}\right)\space\text{d}\theta\space\text{d}\sigma\tag2$$

Where I used the frequency-domain integration property of the Laplace transform.

Now, when $\Im\left(\sigma\right)\ne0\space\vee\space\Re\left(\sigma\right)>0$:

$$\mathcal{I}_{\space\text{n}}\left(\text{s}\right)=\frac{1}{2}\cdot\int_\text{s}^\infty\frac{1}{2}\cdot\left(\ln\left|4+\sigma^2\right|-2\ln\left|\sigma\right|\right)\space\text{d}\sigma=\frac{1}{4}\cdot\int_\text{s}^\infty\left(\ln\left|4+\sigma^2\right|-2\ln\left|\sigma\right|\right)\space\text{d}\sigma\tag3$$