I am interested in calculating $$\int_0^{\infty} \frac{e^{\cos(x)}\sin(\sin(x))}{x} dx$$ using the residue calculus. I thought the thing to do was to consider $$f(z) = \frac{e^{e^{iz}}}{z}$$ since $\textrm{Im}(f(x)) = \frac{e^{\cos(x)}\sin(\sin(x))}{x}$ and integrate on an indented semicircular contour with inner radius $\epsilon$, outer radius $R$ and let $\epsilon \to 0, R \to \infty$. I calculated $$\lim_{\epsilon \to 0}\int_0^{\pi} f(\epsilon e^{i\theta})i\epsilon e^{i\theta} d\theta$$ using that $f(z) = \frac{e}{z} + O(z)$ near $0$, but the problem is that the integral along the large semicircle does not vanish as $R \to \infty$. Actually calculating this integral seems hopeless. Any hints?

  • $\begingroup$ Are you only interested in contour based approaches? $\endgroup$
    – user679268
    Jul 3, 2019 at 4:36
  • $\begingroup$ @KevinNivek Yes $\endgroup$ Jul 3, 2019 at 5:23
  • $\begingroup$ I think the integral is diverge. Isn't $e^{\color{red}{-}\cos}$ ? $\endgroup$
    – Nosrati
    Jul 3, 2019 at 5:39
  • $\begingroup$ @Nosrati $\cos$ is a bounded periodic function, so that doesn't make a difference. $\sin[\sin(x)]/x$ is providing the convergence. $\endgroup$ Jul 3, 2019 at 5:52

2 Answers 2


As shown by @eyeballfrog, $$\int_0^{\infty} \frac{e^{\cos(x)}\sin(\sin(x))}{x} dx =\int_0^\pi \frac{\sin[\sin(u)]}{\sin(u)}\sin^2\left(\frac{u}{2}\right)e^{-\cos(u)}du$$

Further manipulating: $$\begin{align} \int_0^\pi \frac{\sin[\sin(u)]}{\sin(u)}\sin^2\left(\frac{u}{2}\right)e^{-\cos(u)}du &=\int_0^\pi \frac{\sin[\sin(u)]}{\sin(u)}\left(\frac{1-\cos u}{2}\right)e^{-\cos(u)}du \\ &=\frac12\int_{-\pi}^\pi \frac{\sin[\sin(u)]}{\sin(u)}\left(\frac{1-\cos u}{2}\right)e^{-\cos(u)}du \\ &=\frac14\Im\int_{-\pi}^\pi \frac{1-\cos u}{\sin(u)}e^{-\cos(u)+i\sin u}du \\ &=\frac14\Im\int_{-\pi}^\pi \frac{1-\cos u}{\sin(u)}\exp(-e^{-iu})du \\ &=-\frac14\Im\int_{-\pi}^\pi \frac{1-\cos u}{\sin(u)}\exp(-e^{iu})du \\ \end{align} $$

Let $z=e^{iu}$, then $$\begin{align} \int_0^\pi \frac{\sin[\sin(u)]}{\sin(u)}\sin^2\left(\frac{u}{2}\right)e^{-\cos(u)}du &=-\frac14\Im\int_{-\pi}^\pi \frac{2-2\cos u}{2\sin(u)}\exp(-e^{iu})du \\ &=-\frac14\Im\oint_{|z|=1} \frac{2-z-z^{-1}}{(z-z^{-1})/i}\exp(-z)\frac{dz}{iz} \\ &=\frac14\Im\oint_{|z|=1} \frac{z^2-2z+1}{z^2-1}\frac{e^{-z}}z dz\\ &=\frac14\Im\underbrace{\oint_{|z|=1} \frac{z-1}{z+1}\frac{e^{-z}}z dz}_{I}\\ \end{align} $$

Note that the contour integral has to be understood in the Cauchy principal value sense, since a pole lies on the path of integration.

Consider the contour $C$, a unit circle with a semicircle indent to the right at $z=-1$.

By residue theorem, $$\oint_C \frac{z-1}{z+1}\frac{e^{-z}}z dz=2\pi i\operatorname*{Res}_{z=0}\frac{z-1}{z+1}\frac{e^{-z}}z$$ $$\implies I+\int_{\text{indent}}\frac{z-1}{z+1}\frac{e^{-z}}z dz=-2\pi i$$

Since the indent is a semicircle and goes clockwisely, it is not difficult to prove that $$\int_{\text{indent}}\frac{z-1}{z+1}\frac{e^{-z}}z dz=-\frac12\cdot 2\pi i\operatorname*{Res}_{z=-1}\frac{z-1}{z+1}\frac{e^{-z}}z=-2\pi i\cdot e$$

Hence, $$I=2\pi i (e-1)$$

As a result, $$\int_0^{\infty} \frac{e^{\cos(x)}\sin(\sin(x))}{x} dx=\frac\pi 2(e-1)$$ which has been confirmed numerically.


Partial answer

This transform gives you a finite integral that is much more amenable to numeric methods:

\begin{multline} \int_0^\infty \frac{e^{\cos(x)}\sin[\sin(x)]}{x}dx \\ = \sum_{n=0}^\infty \left(\int_{2n\pi}^{(2n+1)\pi}\frac{e^{\cos(x)}\sin[\sin(x)]}{x}dx+\int_{(2n+1)\pi}^{2(n+1)\pi}\frac{e^{\cos(x)}\sin[\sin(x)]}{x}dx\right) \\ = \sum_{n=0}^\infty \left(\int_{0}^{\pi}\frac{e^{-\cos(u)}\sin[\sin(u)]}{(2n+1)\pi -u}du-\int_{0}^{\pi}\frac{e^{-\cos(u)}\sin[\sin(u)]}{(2n+1)\pi + u}du\right) \\ = \int_0^\pi e^{-\cos(u)}\sin[\sin(u)]\left[\sum_{n=0}^\infty \frac{2u}{(2n+1)^2\pi^2-u^2}\right]du \\ = \frac{1}{2}\int_0^\pi e^{-\cos(u)}\sin[\sin(u)]\tan\left(\frac{u}{2}\right)du = \int_0^\pi \frac{\sin[\sin(u)]}{\sin(u)}\sin^2\left(\frac{u}{2}\right)e^{-\cos(u)}du \end{multline} This last integral has a bounded integrand within the domain of integration, and thus is good for numeric methods.

We can also get rather concise form of the integral using a trig transform: $$ \int_0^\infty \frac{e^{\cos(x)}\sin[\sin(x)]}{x}dx = \frac{1}{2}\int_0^\pi e^{-\cos(u)}\sin[\sin(u)]\tan\left(\frac{u}{2}\right)du = \int_{-1}^1\frac{e^{-y}\sin(\sqrt{1-y^2})}{2(1+y)}dx $$ but Mathematica still doesn't want to do it analytically. It does have branch points at the ends of the interval, so perhaps this kind of contour integration could go somewhere.

  • 1
    $\begingroup$ Really beautiful transformation! $\endgroup$ Aug 2, 2019 at 7:07

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.