I am given the integral $$\int_0^\infty\frac{\sin^4(x)}{x^2}dx$$ And I must compute it. I know that the answer is $\frac{\pi}{4}$, but I don't really know how to begin solving this.

I am thinking of taking $f(x)=\frac{\sin^4(x)}{x^2}$ and then trying to find its antiderivative and evaluating it at the required limits, but I'm unsure of how to find the antiderivative of such a function.


Integration by parts leads to: $$ \int_{0}^{+\infty}\frac{\sin^4 x}{x^2}\,dx = \int_{0}^{+\infty}\frac{4\cos x\sin^3 x}{x}\,dx=\int_{0}^{+\infty}\frac{\sin(2x)-\frac{1}{2}\sin(4x)}{x}\,dx $$ so we just need to exploit: $$ \forall \alpha > 0,\quad \int_{0}^{+\infty}\frac{\sin(\alpha x)}{x}\,dx=\frac{\pi}{2} $$ to prove the claim.

  • $\begingroup$ Sorry...can you explain how this integration by parts is happening? I'm not seeing this in the form of the traditional $\int udv=uv-\int vdu$ $\endgroup$ – Taylor May 4 '15 at 17:34
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    $\begingroup$ $v=-1/x$ and $u=\sin^4 x$ is the answer to your question. $\endgroup$ – mickep May 4 '15 at 17:40

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