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The integral is:

$$ I =\int_{0}^{\infty} \frac{\sin^{5}(\alpha x)}{x^{3}}dx; $$

I think that the Dirichlet Integral can helps me: $$\int\limits_0^{+\infty} \frac{\sin \alpha x}{x}\,dx = \frac{\pi}{2}\mathrm{sgn}\,\alpha $$ But I have no idea how to get rid of the $x^{3}$ in the denominator.

Any hint or advice would be much appreciated!

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    the resuld should be $$\frac{5}{32}\pi \alpha^2 \sgn (\alpha)$$ – 2017-02-27
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    For now, I provide hints: consider adding $e^{-xt}$ in the integral, differentiate w.r.t. $t$ three times, expand $\sin^5(x)$ into $a\sin(5x)+b\sin(3x)+c\sin(x)$, solve, and integrate w.r.t. $t$ three times. Then let $t\to\infty$ and $\alpha=0$ etc. to solve for the constants of integration. – 2017-02-27

1 Answers 1

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Outline: You can write $$ \sin^5{\theta} = \frac{1}{16}\sin{5\theta} -\frac{5}{16}\sin{3\theta}+\frac{5}{8}\sin{\theta} $$ (this is basically the reverse of de Moivre's formula) to deal with the power of sine in the numerator.

For the power of $x$ in the denominator, you can with sufficient care integrate by parts twice to get down to things of the form $\frac{\sin{k\alpha x}}{x}$.