$\int_{-\infty}^{\infty}\frac{\sin^3{x}}{x^3}dx$ 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.


  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.

enter image description here

  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.

  • 12
    $\begingroup$ 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 $\endgroup$ Jun 3, 2022 at 12:40
  • 2
    $\begingroup$ Hi I have tidied up your question a bit $\endgroup$ Jun 3, 2022 at 12:43
  • $\begingroup$ 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 $\endgroup$
    – NX37B
    Jun 3, 2022 at 12:45
  • $\begingroup$ 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? $\endgroup$ Jun 3, 2022 at 12:46
  • $\begingroup$ 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. $\endgroup$
    – NX37B
    Jun 3, 2022 at 12:51

2 Answers 2


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.

  • $\begingroup$ 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. $\endgroup$
    – NX37B
    Jun 3, 2022 at 13:33
  • $\begingroup$ You can edit the question you know @NX37B $\endgroup$ 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


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .