# $\int_{-\infty}^{\infty} \frac{\sin^3 x}{x^3} \, dx$ using contour integration.

$$\int_{-\infty}^{\infty}\frac{\sin^3{x}}{x^3}dx$$ using contour integration. Hint: Use this analytic continuation: $$h(z) = -\frac{e^{3iz}-3e^{iz}}{4z^3}-\frac{1}{2z^3}$$, and then take the imaginary part.

EDIT

1. I used the complex exponential definition of $$\sin$$ to show that $$\frac{\sin^3{x}}{x^3} = \operatorname{Im}\left(\frac{e^{3ix}-3e^{ix}}{-4x^3}\right)$$. Therefore, I am really only itnerested in the first term of $$h(x)$$ for my result.

2. I used a contour that is shown in red - two concentric half-circles with radii $$R$$ and $$r$$. The one with radius $$R$$ I let expand to infinity, while the one with radius $$r$$ I let shrink to zero.

1. I argue that on $$\gamma_R$$, the integral vanishes because both the exponential terms and the $$1/z^3$$ terms tend to zero as $$R$$ increases, and, the integral over $$\gamma_r$$ vanishes as well since the exponential terms tend to 1 ar $$r$$ goes to zero. Finally, the total integral is zero by Cauchy.

$$\int_{-\infty}^{\infty} \frac{\sin^3{x}}{x^3} \, dx = \operatorname{Im} \int_{-\infty}^{\infty}\left(\frac{e^{3ix}-3e^{ix}}{-4x^3}\right) \, dz$$

I integrate by usage of the hinted analytic continuiation:

\begin{align*} &\int_{\gamma} \left(-\frac{e^{3iz}-3e^{iz}}{4z^3}-\frac{1}{2z^3}\right) \, dz \\ &\quad= \operatorname{v.p.} \int_{-\infty}^{\infty} \left(\frac{e^{3ix}-3e^{ix}}{-4x^3}-\frac{1}{2x^3}\right) \, dz \\ &\hspace{3em} + \int_{\gamma_R} \left(-\frac{e^{3iz}-e^{iz}}{4z^3}-\frac{1}{2z^3}\right) \, dz + \int_{\gamma_r} \left(-\frac{e^{3iz}-3e^{iz}}{4z^3}-\frac{1}{2z^3}\right)dz \\ &\quad= 0 \end{align*}

1. I only want the integral of the first term of $$h(x)$$, I transferred the integrals of $$1/x^3$$ to the other side and integrated them to get zero again.

$$\operatorname{v.p.} \int_{-\infty}^{\infty} \left(\frac{e^{3ix}-3e^{ix}}{-4x^3}\right) \, dz = \operatorname{v.p.} \int_{-\infty}^{\infty}\frac{1}{2x^3}dx = 0$$

The correct answer should be $$3\pi/4$$, which is given in the solutions, and I confirmed with wolfram alpha. However, I have been staring at my method for hours, and I can't find what doesn't check out.

In case something is unclear, or if I made any typos, I also provide a picture of my working.

• I find it a bit pretentious to greet a newbie on the site with two downvotes on a question where they claim to be stuck for two hours with it Jun 3, 2022 at 12:40
• Hi I have tidied up your question a bit Jun 3, 2022 at 12:43
• Thank you so much, although I admit that I may have been a little negligent with some of my details, so the criticism was probably not unwarranted Jun 3, 2022 at 12:45
• I have a question: In this equation $$\int_{-\infty}^{\infty} \frac{\sin^3{x}}{x^3} \, dx = \operatorname{Im} \int_{-\infty}^{\infty}\left(\frac{e^{3ix}-e^{ix}}{-4x^3}\right) \, dz$$ , why you didnt write -1/2z^3 term? also what does v.p. mean? Jun 3, 2022 at 12:46
• If you expand $\sin^3{x}/x^3 = (\frac{e^{ix}-e^{-ix}}{2ix})^3 = \frac{e^{3ix} - e^{-3ix} + 3e^{ix} - 3e^{-ix}}{-8ix^3}$ and then you use $Im(z) = \frac{z - z*}{2i}$ the last term does not appear. I suspected the term is in the hint to make the integration easier in some way. Jun 3, 2022 at 12:51

Consider the function

$$g(z) = - \frac{e^{3iz} - \bbox[color:red;padding:3px;border:1px red dotted;]{3}e^{iz}}{4z^3} - \frac{1}{2z^3}.$$

Then we have $$\operatorname{Im}(g(x)) = \frac{\sin^3 x}{x^3}$$ for real $$x$$. Now using OP's contour and applying the Cauchy integration theorem,

\begin{align*} 0 &= \int_{\gamma} g(z) \, \mathrm{d}z \\ &= \int_{[-R,-r]\cup[r,R]} g(z) \, \mathrm{d}z + \int_{\gamma_r} g(z) \, \mathrm{d}z + \int_{\gamma_R} g(z) \, \mathrm{d}z. \end{align*}

1. As $$R \to \infty$$, as OP argued, we can show that $$\int_{\gamma_R} g(z) \, \mathrm{d}z \to 0$$.

2. Using the power series expansion of the exponential function, it is not hard to check that $$g(z)$$ has a simple pole at $$z = 0$$ with

$$\mathop{\mathrm{Res}}_{z=0} g(z) = \frac{3}{4}.$$

Since the change of argument along the path $$\gamma_r$$ is $$-\pi$$, by using the well-known lemma (or simply substituting $$z = re^{i\theta}$$ where $$\theta$$ varies from $$\pi$$ to $$0$$ and letting $$r \to 0^+$$), it follows that

$$\lim_{r \to 0^+} \int_{\gamma_r} g(z) \, \mathrm{d}z = -\pi i \mathop{\mathrm{Res}}_{z=0} g(z) = -\frac{3\pi i}{4}.$$

Therefore it follows that

$$\int_{-\infty}^{\infty} \frac{\sin^3 x}{x^3} \, \mathrm{d}x = \lim_{\substack{r \to 0^+ \\ R \to \infty}} \operatorname{Im}\left( \int_{[-R,-r]\cup[r,R]} g(z) \, \mathrm{d}z \right) = \operatorname{Im}\left(\frac{3\pi i}{4}\right) = \frac{3\pi}{4}.$$

In light of this solution, we can realize what went wrong in OP's solution:

• With the function $$h(z)$$ given in the hint, we have $$\operatorname{Im}(h(x)) \neq \frac{\sin^3}{x^3}$$.

• OP made an incorrect claim that the contribution from $$\int_{\gamma_r} g(z) \, \mathrm{d}z$$ vanishes as $$r \to 0^+$$, where in fact, it may contribute to the final answer.

• Thank you so much! the missing factor of three is a typo, so I apologize if it caused trouble. But I completely missed the fact that the $\gamma_r$ integral does not vanish because of the power series expansion. Jun 3, 2022 at 13:33
• You can edit the question you know @NX37B Jun 3, 2022 at 14:03

Here is an alternate approach that is fairly simple, as well. \begin{align} \int_{-\infty}^\infty\frac{\sin^3(x)}{x^3}\,\mathrm{d}x &=\int_{-i-\infty}^{-i+\infty}\frac{\sin^3(z)}{z^3}\,\mathrm{d}z\tag1\\ &=\frac i8\int_{-i-\infty}^{-i+\infty}\frac{e^{3iz}-3e^{iz}+3e^{-iz}-e^{-3iz}}{z^3}\,\mathrm{d}z\tag2\\ &=\frac i8\int_{\gamma_+}\frac{e^{3iz}-3e^{iz}}{z^3}\,\mathrm{d}z+\frac i8\int_{\gamma_-}\frac{3e^{-iz}-e^{-3iz}}{z^3}\,\mathrm{d}z\tag3\\ &=-\frac{2\pi}8\frac{(3i)^2-3i^2}2\tag4\\[3pt] &=\frac{3\pi}4\tag5 \end{align} Explanation:
$$\text{(1)}$$: the integral along $$[-\infty,\infty]$$ is the same as along $$[-i-\infty,-i+\infty]$$
$$\phantom{\text{(1): }}$$since the integrals along $$[R,R-i]$$ and $$[-R-i,-R]$$ both vanish as $$R\to\infty$$
$$(2)$$: $$\sin(x)=\frac{e^{ix}-e^{-ix}}{2i}$$
$$(3)$$: $$\gamma_+=[-R-i,R-i]\cup Re^{+i[0,\pi]}-i$$ and
$$\phantom{\text{(3): }}\gamma_-=[-R-i,R-i]\cup Re^{-i[0,\pi]}-i$$
$$\phantom{\text{(3): }}$$where the integrals along the circular arcs vanishes as $$R\to\infty$$
$$(4)$$: the integral is $$2\pi i$$ times the sum of the residues inside $$\gamma_+$$
$$\phantom{\text{(4): }}$$there are no singularities inside $$\gamma_-$$
$$\phantom{\text{(4): }}$$the residue of $$\frac{e^{i\lambda z}}{z^3}=-\frac{\lambda^2}2$$
$$(5)$$: simplify