How to evaluate $\int_{-\infty}^\infty \sin^3(x)/x^3 dx$ without finding an analytic continuation, still using complex analysis.

Question

There is a similar post regarding the integral $\int_{-\infty}^\infty \sin^3(x)/x^3 dx$ on Stack. The reason I have some trouble with this post is because it gives an analytic continuation of the function $\sin^3(x)/x^3=\left(\frac{e^{iz}-e^{-iz}}{2iz}\right)^3$, namely $h(z)=-\frac{e^{3iz}-3e^{iz}}{4z^3}-\frac{1}{2z^3}$. Firstly, I'd like to know how such an analytic continuation is obtained. Regarding the sinc function, I know that sine has a zero of order 1, and the denominator also does, so at $z=0$, there is a removable singularity and thus an analytic continuation. I also know that $\lim_{z\to 0}\text{sinc}(z)=1$ (meaning 1 is an analytic continuation to $\mathbb{C}$?)

Regarding the actual integral itself, I have considered the usual indented semicircle, of radius $R$, with an $\epsilon$ semicircle about the essential singularity 0. Without using that analytic continuation $h(z)$, this gives, letting $f(z)=\frac{e^{3iz}-3e^{iz}+3e^{-iz}-e^{-3iz}}{z^3}$ $$0=\frac{i}{8}\left[\int_{[-R,-\epsilon]\cup [\epsilon, R]} f(z)dz+\oint_{\gamma_R}f(z)dz+\oint_{\gamma_\epsilon}f(z)dz\right]$$ where $\gamma_R$ is the contour of the large semicircle of radius $R$, and $\gamma_\epsilon$ is the contour around the singularity $0$ of radius $\epsilon$. Using Jordan's Lemma, I have managed to show that the integral over $\gamma_R$ goes to 0 as $R\to\infty$. The trouble comes in with the integral over $\gamma_\epsilon$ as $\epsilon\to 0$. I have come across a lemma here on Stack that says if $f$ is some meromorphic function (which, by definiton means that if $z_0$ is an isolated singularity of $f$ there exists some $r>0$ such that $1/f$ extends to a holomorphic function on $B_r(z_0)$) that has a simple pole at a point $z_0$, then $\lim_{\epsilon\to z_0}\oint_{\gamma_\epsilon} f(z)d{z} = i(\gamma_\epsilon(b)-\gamma_\epsilon(b))\text{Res}(z_0,f(z))$. This is in fact the reason I asked about the analytic continuation; since the singularity of sinc is essential, this lemma does not apply unless we extend it to some other function that has a pole, at least as I understand it. I also know the answer should be $3\pi/4$.

If anyone could help, I would appreciate it. The link to the post is: $ \int_{-\infty}^{\infty} \frac{\sin^3 x}{x^3} \, dx$ using contour integration.

Answer 1

Here is a method which doesn't use analytic continuation (or contours). Begin with the known integral $$ \int_0^{\infty} \frac{\sin x}{x} \, dx = \frac{\pi}{2} $$ Integrating by parts once and then twice gives \begin{align*} \frac{\pi}{2} = \left. \frac{1-\cos x}{x} \right|_0^{\infty} - \int_0^{\infty} (1-\cos x) \frac{-1}{x^2} \, dx &= \int_0^{\infty} \frac{1-\cos x}{x^2} \, dx \\ &= \left. \frac{x-\sin x}{x^2} \right|_0^{\infty} - \int_0^{\infty} (x-\sin x) \frac{-2}{x^3} \, dx \\ &= 2\int_0^{\infty} \frac{x-\sin x}{x^3} \, dx \end{align*} so $$ \int_0^{\infty} \frac{x-\sin x}{x^3} \, dx = \frac{\pi}{4} $$ Substituting $3x$ for $x$ gives $$ \int_0^{\infty} \frac{3x-\sin(3x)}{x^3} \, dx = \frac{9\pi}{4} $$ Now use the identity $\sin(3x) = 3\sin x-4\sin^3x$ to write $$ \frac{9\pi}{4} = \int_0^{\infty} \frac{3x-3\sin x +4\sin^3x}{x^3} \, dx = 3\int_0^{\infty} \frac{x-\sin x}{x^3} \, dx +4\int_0^{\infty} \frac{\sin^3x}{x^3} \, dx $$ and rearrange. This method can be used to compute all of the integrals $\int_0^{\infty} \frac{\sin^nx}{x^n} \, dx, \; n \in \mathbb{N}$ recursively.

Answer 2

It would be a good application of Laplace techniques for an even function. \begin{align} \int_{-\infty}^{\infty}\sin^3\left(x\right)\frac{1}{x^3}\mathrm{d}x&=\int_0^{\infty}\mathcal{L}\left[\sin^3\left(t\right)\right](x)\mathcal{L}^{-1}\left[\frac{2}{s^3}\right](x)\mathrm{d}x\\ &=\int_0^{\infty}\frac14\left(\frac{3}{x^2+1}-\frac{3}{x^2+9}\right)x^2\mathrm{d}x\\ &=\int_0^{\infty}\frac{27}{4}\frac{1}{x^2+9}-\frac34\frac{1}{x^2+1}\mathrm{d}x\\ &=\frac34\pi. \end{align}

$\mathcal{L}[\sin^3t]=\frac14\mathcal{L}[3\sin(t)-\sin(3t)]$ from trigonometry formulas.

Answer 3

Here is another method which just uses IBP, not using analytic continuation (or contours). In fact, \begin{align} \int_{-\infty}^{\infty}\sin^3\left(x\right)\frac{1}{x^3}\;\mathrm{d}x&=2\int_{0}^{\infty}\sin^3\left(x\right)\frac{1}{x^3}\;\mathrm{d}x\\ &=-\int_0^{\infty}\sin^3(x)\;\mathrm{d}(x^{-2})\\ &\overset{IBP}=-3\int_0^{\infty}\sin^2(x)\cos(x)\;\mathrm{d}(x^{-1})\\ &\overset{IBP}=\frac34\int_0^{\infty}\frac{-\sin(x)+3\sin(3x)}{x}\;\mathrm{d}x\\ &=\frac34(-\frac\pi2+\frac{3\pi}{2})\\ &=\frac34\pi. \end{align}