integration - Evaluating the contour integral $int_{0}^{infty}frac{sin^{3}(x)}{x^{3}}mathrm dx$

I am trying to show

$$\int_{0}^{\infty}\frac{\sin^{3}(x)}{x^{3}}\mathrm dx = \frac{3\pi}{8}.$$

I believe the contour I should use is a semicircle in the upper half plane with a slight bump at the origin, so I miss the singularity.

Lastly I have the hint to consider
$$\int_{0}^{\infty}\frac{e^{3iz}-3e^{iz}+2}{z^{3}}\mathrm dz$$

around the contour I mentioned. Thanks for any help or hints!

Answer

Your coutour will work perfectly, so I wonder why you're hesitating to proceed with your calculation. Anyway, here is a solution:

Since $\sin^3 z = (\sin 3z - 3\sin z)/4$, we have

$$\int_{0}^{\infty} \frac{\sin^3 x}{x^3} \, dx = \frac{1}{8} \Im \lim_{\epsilon \to 0} \int_{\mathbb{R} \backslash (-\epsilon, \epsilon)} \frac{3 e^{iz} - e^{3iz} -2}{z^3} \, dz,$$

where the term $-2$ is introduced in order to cancel out the pole of order 3, without affecting the value of the integral. Consider the counterclockwise-oriented upper semicircle $C$ of radius $R$, centered at the origin, with semicircular indent of radius $\epsilon$. Let $\Gamma_{R}^{+}$ and $\gamma_{\epsilon}^{-}$ denote semicircular arcs of $C$ of radius $R$ and $\epsilon$, respectively. If we put

$$f(z) = \frac{3 e^{iz} - e^{3iz} - 2}{z^3},$$

then we find that

• On $\Gamma_R^+$, we have $|f(z)| \leq 6R^{-3}$ and thus
  $$\int_{\Gamma_{R}^{+}} f(z) \, dz \to 0 \quad \text{as } R \to \infty.$$




• Notice that
  $$f(z) = \frac{3}{z} + O(1) \quad \text{near } z = 0.$$
  So by the direct computation,
  $$\int_{\gamma_{\epsilon}^{-}} f(z) \, dz = -\int_{0}^{\pi} f(\epsilon e^{i\theta}) i\epsilon e^{i\theta} \, d\theta = -\int_{0}^{\pi} (3i + O(\epsilon)) \, d\theta \to -3\pi i \quad \text{as } \epsilon \to 0.$$
  (This is exactly the same as $-\pi i$ times the residue of $f$ at $z = 0$. The emergence of residue can be attributed to the fact that $f$ has only simple pole at $z = 0$.)

Since $f(z)$ has no pole on the region enclosed by $C$, we have
$$\lim_{\epsilon \to 0} \int_{\mathbb{R} \backslash (-\epsilon, \epsilon)} \frac{3 e^{iz} - e^{3iz} - 2}{z^3} \, dz = 3\pi i.$$
This proves the desired identity.