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Use contour Integration to evaluate $\int_0^{\infty} \frac{\sin ^3 t}{t} dt$.

First note that $\sin^3 (t)= \Im (\frac{3e^it-e^{3it}}{t})$.

Here I have something like a half donut in the upper half plane.

$\gamma_1$ is the semi-circle with radius $R$ going counterclockwise, where $\theta\in [0,\pi]$

$\gamma_2$ is the line from $-R$ to $-\epsilon$.

$\gamma_3$ is the semi-circle with radius $\epsilon$ going counterclockwise, where $\theta\in [0,\pi]$

$\gamma_4$ is from $\epsilon$ to $R$.

For $\gamma_3$, we have $|\int_{\gamma_3} \frac{\sin^3(t)}{t} dt| \leq \frac{\sin^3(\epsilon)}{\epsilon} \pi \epsilon \rightarrow 0$ as $\epsilon \rightarrow 0$.

However, for $\gamma_1$, we can't do this anymore, so we have to evaluate this directly.

So for $\gamma_1$, we have $\int_{\gamma_1} \frac{\sin^3(t)}{t} dt$ But it seems very complicated to evaluate. So I wonder if there's any simpler way of evaluating this, or am I missing something in my calculation?

Thanks in advance!

insipidintegrator
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Toasted_Brain
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  • Since $0$ isn't a true pole (but rather a removable singularity) of $\frac{\sin^3t}{t}$ - which is why the integral on $\gamma_3$ has $\lim_{\epsilon_to0^+}$ equal to $0$ - we can in fact use an ordinary semicircular contour, if we spot e.g.$$\sin^3t=\Im\frac{3e^{it}-e^{3it}-2}{4}$$for $t\in\Bbb R$. – J.G. Aug 11 '22 at 06:30

3 Answers3

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Is there any simpler way of evaluating this?

I’ll provide a solution with the Lobachevsky’s Dirichlet Integral Formula: $$\int_0^\infty \frac{\sin x}{x} f(x)dx=\int_0^\tfracπ2 f(x)dx $$ so here we take $f(x)=\sin^2x$. So $$\int_0^\infty \frac{\sin^3 x}{x} dx=\int_0^\tfracπ2\sin^2 xdx$$$$=\int_0^\tfracπ2\frac{1-\cos 2x}{2}dx$$$$=\fracπ4.$$

insipidintegrator
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It's a good idea to use $\sin^3 t=\Im(3e^{it}-e^{3it})/4$. Then $$0=\left[\int_{\gamma_1}+\int_{\gamma_2}-\int_{\gamma_3}+\int_{\gamma_4}\right]\frac{3e^{iz}-e^{3iz}}{z}\,dz,$$

$$\lim_{\substack{R\to\infty\\\epsilon\to 0}}\left[\int_{\gamma_2}+\int_{\gamma_4}\right]\frac{3e^{iz}-e^{3iz}}{z}\,dz=8i\int_0^\infty\frac{\sin^3 t}{t}\,dt,$$

$$\lim_{R\to\infty}\int_{\gamma_1}\frac{3e^{iz}-e^{3iz}}{z}\,dz=0,\tag{1}\label{jordan}$$

$$\lim_{\epsilon\to 0}\int_{\gamma_3}\frac{3e^{iz}-e^{3iz}}{z}\,dz=\pi i\operatorname*{Res}_{z=0}\frac{3e^{iz}-e^{3iz}}{z}=2\pi i\tag{2}\label{halfres}$$

(where \eqref{jordan} follows from Jordan's lemma, and \eqref{halfres} holds by "half-residue" trick).

metamorphy
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We have $\int_{0}^{+\infty}\frac{\sin x}{x}\,dx = \frac{\pi}{2}$, so via $x\mapsto nz$ we also have $\int_{0}^{+\infty}\frac{\sin(nz)}{z}\,dz = \frac{\pi}{2}$ for any $n\in\mathbb{N}^+$. Since

$$ \sin^3(z) = \left(\frac{e^{iz}-e^{-iz}}{2i}\right)^3 = -\frac{e^{3iz}-3e^{iz}+3e^{-iz}-e^{-3iz}}{8i}=\frac{3\sin z-\sin(3z)}{4}$$ it follows that

$$ \int_{0}^{+\infty}\frac{\sin^3(z)}{z}\,dz = \frac{3-1}{4}\cdot\frac{\pi}{2}=\color{blue}{\frac{\pi}{4}}.$$ More generally, for any $m\in\mathbb{N}$ $$ \int_{0}^{+\infty}\frac{\sin^{2m+1}(z)}{z}\,dz = \frac{\pi}{2}\cdot\frac{\binom{2m}{m}}{4^m} $$

Jack D'Aurizio
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