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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! |
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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 $\int_{\Gamma_{R}^{+}} f(z) \, dz \to 0$ as $R \to \infty$ as $R \to \infty$, and $\int_{\gamma_{\epsilon}^{-}} f(z) \, dz \to -\pi i \, \mathrm{Res}_{z = 0} f(z)$ as $\epsilon \to 0$. Since $f(z)$ has no pole on the region enclosed by $C$, we have $$\lim_{\epsilon \to 0} \int_{\mathbb{R} \backslash (-\epsilon, \epsilon)} \frac{3 e^{iz} - e^{3iz} - 2}{z^3} \, dz = \pi i \, \mathrm{Res}_{z = 0} f(z) = 3\pi i.$$ This proves the desired identity. |
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$\int \frac{\sin^3x}{x^3}\,\mathrm dx$ $=\frac{1}{4}\int\frac{3\sin x-\sin3x}{x^3}\,\mathrm dx$ $=\frac{1}{4}\left[(3\sin x-\sin3x)(\frac{-1}{2x^2})+\frac12\int \frac{3\cos x-3\cos3x}{x^2}\right]\,\mathrm dx$ $=\frac14\left[(3\sin x-\sin3x)(\frac{-2}{x^2})+\frac32\left[(\cos x-\cos3x)(\frac{-1}{x})+\int \frac{-\sin x+3\sin3x}{x}\right]\right]\,\mathrm dx$ $=\frac14\left[(3\sin x-\sin3x)(\frac{-2}{x^2})+\frac32(\cos x-\cos3x)(\frac{-1}{x})+\frac32\int \frac{-\sin x+3\sin3x}{x}\right]\,\mathrm dx$ For limits $x=0$ and $x \to \infty$ the first two terms disappear[each vanishes for both the limits]. The third term, on considering the integral $\int_0^\infty\frac{\sin x}{x}\, \mathrm dx=\pi/2$ evaluates to: $\frac{3}{8}[-\pi/2+3\pi/2] $ $=\frac{3\pi}{8}$ |
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