Strange contour integral 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?
 A: 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.
A: 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.
