<s>[INST] Compute $\int_{0}^{\infty} \frac{\sin^4x}{x^4}dx$ Compute
$$
\int_{0}^{\infty} \frac{\sin^4x}{x^4}dx\;\;.
$$
Well, I'm not sure how to approach this. I thought about using Plancheral, and using $f(x) = \frac{\sin^2x}{x^2}$, and finding the fourier transform of $f(x)$. Any other better approach?
 [/INST] I think you sould use integration by parts to do it more easily in the following way:
$$   \int \frac{\sin^4(x)}{x^4}dx =  -\frac{1}{3} \frac{\sin^4(x)}{x^3} + \frac{4}{3} \int \frac{\cos(x) \sin^3(x) }{x^3} dx $$
$$ =
  -\frac{1}{3} \frac{\sin^4(x)}{x^3} -\frac{2 \cos(x) \sin^3(x)}{3 x^2} + \frac{2}{3} \int \frac{3 \cos^2(x) \sin^2(x) - \sin^4(x)}{x^2} dx$$
$$= -\frac{1}{3} \frac{\sin^4(x)}{x^3} -\frac{2 \cos(x) \sin^3(x)}{3 x^2} + \frac{2}{3} \int \left(\frac{\sin^2(2x)}{x^2}  - \frac{\sin^2(x)}{x^2} \right) dx$$
Last line I have used $\sin^4(x) = \sin^2(x)(1-cos^2(x)) $ and simplify it.
In general we know that $\int_0^\infty \frac{\sin^2x}{x^2} = \frac{\pi}{2}$ using complex analysis.
Also put $2x = y$ and find the integral foe the other part.
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