# Tricky integral

I need a hint. Wolfram alpha says it equals pi/2, but I don't know how to get that.

\int_{-\infty}^{\infty} \frac{\sin^4(x)}{x^2}\, dx

Did you make a typo?

I don't think so. Here's the Wolfram alpha solution.

Are you familiar with contour integration?

Yes. I thought about trying something like that, but the usual substitution z = e^{ix} to write the sin part as an exponentials doesn't work with the x^2 on the bottom.

What were you thinking about with the contour integration?

My other thought was to try to relate it back to \int sin(x)/x which is something I can do.

I haven't worked it out but if you write out the sin^4 and then use contour integration it will probably work:
$$\sin^4\theta = \frac{3 - 4 \cos 2\theta + \cos 4\theta}{8}\$$

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You're right, that would probably work. I'll give it a go. Thanks for your help!

Alternatively, I figured out how to reduce it:

$$\int_{-\infty}^{\infty} \frac{\sin^4{x}}{x^2}\, dx = \int_{-\infty}^{\infty} \frac{4 \sin^3{x} \cos{x}}{x}\, dx$$
(IBP)
$$= \int_{-\infty}^{\infty} \frac{\sin{2x}(1 - \cos{2x})}{x}\, dx$$
(trig identities)
$$= \int_{-\infty}^{\infty} \frac{\sin{2x}}{x}\, dx - \int_{-\infty}^{\infty} \frac{\sin{4x}}{2x}\, dx$$
$$= \int_{-\infty}^{\infty} \frac{\sin{x}}{x}\, dx - \frac{1}{2} \int_{-\infty}^{\infty} \frac{\sin{x}}{x}\, dx$$
(change of variables)
$$= \pi - \frac{\pi}{2}$$
(sinc integral)

Yes, well found!