Strange contour integral: $\int_0^{\infty} \frac{e^{\cos(x)}\sin(\sin(x))}{x} dx$ using residue calculus

Question

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?

Answer 1

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.

Answer 2

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.

Answer 3

(A few years late, but I just saw this a few hours ago.)

Rewrite the integral as $$ \begin{align} I &:= \int_{0}^{\infty}\frac{e^{\cos x}\sin\left(\sin x\right)}{x}dx \\ &= \frac{1}{2}\int_{-\infty}^{\infty}\frac{e^{\cos x}\sin\left(\sin x\right)}{x}dx \\ &= \frac{1}{2}\int_{-\infty}^{0}\Im\left(\frac{\exp\left(e^{ix}\right)}{x}\right)dx+\frac{1}{2}\int_{0}^{\infty}\Im\left(\frac{\exp\left(e^{ix}\right)}{x}\right)dx \\ &= \frac{1}{2}\lim_{R\to\infty}\lim_{r\to0^+}\Im\int_{-R}^{-r}\left(\frac{\exp\left(e^{ix}\right)}{x}\right)dx+\frac{1}{2}\lim_{R\to\infty}\lim_{r\to0^+}\Im\int_{r}^{R}\left(\frac{\exp\left(e^{ix}\right)}{x}\right)dx\, \end{align} $$

where $0 < r \ll R$.

Let $\displaystyle f(z) = \frac{\exp\left(e^{iz}\right)}{z}$. There is a simple pole at $z=0$ because the principal part of the Laurent expansion of $f(z)$ is $\displaystyle \frac{e}{z}$. We can take the OP's idea and construct this contour.

(I forgot to label the small circular arc $\gamma$.) By Cauchy's Integral Theorem, we write $\displaystyle \oint_C f(z)dz$ as

$$ 0 = \int_{-R}^{-r}f(z)dz + \int_{\gamma}f(z)dz + \int_r^R f(z)dz + \int_{\Gamma}f(z)dz\,. $$

We can avoid divergence issues by equating $\Im$ first, applying the limits, then multiplying by $\displaystyle \frac{1}{2}$ like

$$ 0 = I + \frac{1}{2}\lim_{r\to0^+} \Im \int_{\gamma} f(z)dz + \frac{1}{2}\lim_{R\to\infty} \Im \int_{\Gamma}f(z)dz\,. $$

We solve the integral over $\gamma$ like this.

$$ \lim_{r\to0^+} \Im \int_{\gamma} \frac{\exp\left(e^{iz}\right)}{z}dz = -\Im i \pi\mathop{\mathrm{Res}}_{z=0} \frac{\exp\left(e^{iz}\right)}{z} = -\pi \Re \lim_{z\to0}(z-0)\cdot\frac{\exp\left(e^{iz}\right)}{z} = -\pi e \,. $$

To solve the other contour integral, we can first rewrite it as

$$ \begin{align} \Im \int_{\Gamma} \frac{\exp\left(e^{iz}\right)}{z}dz &= \Im \int_{\Gamma} \frac{1}{z} \sum_{n=0}^{\infty} \frac{e^{niz}}{n!}dz \\ &= \Im \sum_{n=0}^{\infty}\frac{1}{n!}\int_{\Gamma}^{ }\frac{e^{niz}}{z}dz \tag{1}\\ &= \Im \sum_{n=0}^{\infty}\frac{1}{n!}\int_{0}^{\pi}\frac{\exp\left(niRe^{it}\right)}{Re^{it}}iRe^{it}dt \\ &= \Re \sum_{n=0}^{\infty}\frac{1}{n!}\int_{0}^{\pi}\exp\left(niRe^{it}\right)dt \\ &= \Re \frac{1}{0!}\int_{0}^{\pi}\exp\left(0\cdot iRe^{it}\right)dt + \Re \sum_{n=1}^{\infty}\frac{1}{n!}\int_{0}^{\pi}\exp\left(niRe^{it}\right)dt \,. \\ \end{align} $$

For $(1)$, since $\exp(e^{iz})$ is entire, its radius of convergence is $\infty$. Note that $\Gamma$ is inside the circle of convergence of the infinite series, which justifies the $\sum$ and $\int$ swap.

The first integral from that last equality equals $\pi$. We bound the second integral like $$ \begin{align} 0 &\leq \left|\Re\sum_{n=1}^{\infty}\frac{1}{n!}\int_{0}^{\pi}\exp\left(niRe^{it}\right)dt\right| \\ &\leq \left|\sum_{n=1}^{\infty}\frac{1}{n!}\int_{0}^{\pi}\exp\left(niRe^{it}\right)dt\right| \\ &\leq \sum_{n=1}^{\infty}\frac{1}{n!}\left|\int_{0}^{\pi}\exp\left(niRe^{it}\right)dt\right| \\ &\leq \sum_{n=1}^{\infty}\frac{1}{n!}\int_{0}^{\pi}\left|\exp\left(niRe^{it}\right)\right|dt \\ &= \sum_{n=1}^{\infty}\frac{1}{n!}\int_{0}^{\pi}e^{-Rn\sin t}dt \\ &< \sum_{n=1}^{\infty}\frac{1}{n!} \cdot \frac{\pi}{Rn}\,. \tag{2} \\ \end{align} $$

For $(2)$, we use Jordan's Inequality to find an upper bound tight enough for that integral.

We take $R \to \infty$ and use the Squeeze Theorem to show that the second integral converges to $0$. We eventually conclude that

$$ \lim_{R\to\infty}\Im \int_{\Gamma} f(z)dz = \pi\,. $$

Finally, $$ \begin{align} 0 &= I - \frac{1}{2}\pi e + \frac{1}{2}\pi \\ \implies I &= \frac{\pi}{2} (e-1) \\ \end{align} $$

and we're finished!