Prove $\int_{0}^{\infty}{e}^{a\cos\left(bx\right)}\sin\left(a\sin\left(bx\right)\right)\frac{\mathrm{d}x}{x}=\frac{\pi}{2}\left({e}^{a}-1\right)$ Show that if $a > 0$ and $b > 0$ that
\begin{align}
 \int_{0}^{\infty} {e}^{a \cos \left(b x\right)} \sin \left(a \sin \left(b x\right)\right) \frac{\mathrm{d}x}{x} = \frac{\pi}{2} \left({e}^{a} - 1\right) \\
\end{align}
Attempt:
I attempted to simplify the integral by putting it into a form of complex exponentials.
\begin{align}
 & \int_{0}^{\infty} {e}^{a {e}^{i b x}} \frac{\mathrm{d}x}{x} \\
 = & \int_{0}^{\infty} {e}^{a \cos \left(b x\right) + i a \sin \left(b x\right)} \frac{\mathrm{d}x}{x} \\
 = & \int_{0}^{\infty} {e}^{a \cos \left(b x\right)} \cos \left(\sin \left(b x\right)\right) \frac{\mathrm{d}x}{x} + i \int_{0}^{\infty} {e}^{a \cos \left(b x\right)} \sin \left(a \sin \left(b x\right)\right) \frac{\mathrm{d}x}{x} \\
\end{align}
We then have the result by linearity that
\begin{align}
 & \int_{- \infty}^{\infty} {e}^{a {e}^{i b x}} - {e}^{a {e}^{- i b x}} \frac{\mathrm{d}x}{x} \\
 & = \int_{- \infty}^{\infty} {e}^{a \cos \left(b x\right)} \left({e}^{i a \sin \left(b x\right)} - {e}^{i a \sin \left(b x\right)}\right) \frac{\mathrm{d}x}{x} \\
 & = 2 i \int_{- \infty}^{\infty} {e}^{a \cos \left(b x\right)} \sin \left(a \sin \left(b x\right)\right) \frac{\mathrm{d}x}{x} \\
 & = 4 i I \\
\end{align}
where
\begin{align}
 I = \int_{0}^{\infty} {e}^{a \cos \left(b x\right)} \sin \left(a \sin \left(b x\right)\right) \frac{\mathrm{d}x}{x} \\
\end{align}
I am confused because when I take the residue at $0$, I find
\begin{align}
 & {\text{Res}}_{z = 0} \frac{{e}^{a {e}^{i b z}} - {e}^{a {e}^{- i b z}}}{z} \\
 & = {e}^{a} - {e}^{a} = 0 \\
\end{align}
 A: Let $bx\to x$ to simplify the integral to\begin{align}
& \int_{0}^{\infty} {e}^{a \cos x}\  \sin (a \sin x)\ \frac{dx}{x} \\
= &\  \Im \int_{0}^{\infty} {e}^{a e^{ix}}\ \frac{dx}{x}
= \Im \int_{0}^{\infty} \sum_{k=0}^\infty\frac{(a e^{ix})^k}{k!}\frac{dx}x\\
= &\sum_{k=1}^\infty\frac{a^k}{k!} \int_{0}^{\infty} \frac{\sin(kx)}x\ dx
= \sum_{k=1}^\infty\frac{a^k}{k!}\cdot \frac\pi2  =  \frac{\pi}{2} \left({e}^{a} - 1\right)
\end{align}
A: $$\int_0^\infty e^{a e^{ib x}}\frac{dx}{x}=\int_0^\infty e^{a \cos(b x)+ia\sin(bx)}\frac{dx}{x}$$
$$=\int_0^\infty e^{a \cos (b x)}\Big(\cos(a\sin(bx))+i \sin (a \sin (b x))\Big) \frac{dx}{x}$$
Due to the fact that
$$\int_{-\infty}^\infty e^{a \cos (b x)}\cos(a\sin(bx)) \frac{dx}{x}=0\,\,(\text{the integrand is odd})$$
and that
$$\int_{-\infty}^0 e^{a \cos (b x)}\sin (a \sin (b x)) \frac{dx}{x}=\int_0^\infty e^{a \cos (b x)}\sin (a \sin (b x)) \frac{dx}{x}\,\,(\text{the integrand is even})$$
we can present the integrand in the form
$$I=\int_0^\infty e^{a \cos (b x)}\sin (a \sin (b x)) \frac{dx}{x}=-\frac{i}{2}\int_{-\infty}^\infty e^{a e^{ib x}}\frac{dx}{x}=-\frac{i}{2}\int_{-\infty}^\infty e^{a e^{i x}}\frac{dx}{x}$$
where the integral is evaluated in the principal value sense. We go in the complex plane, adding a small arch (of the radius $r\to0$, around $x=0$) and a big arch (of the radius  $R\to\infty$), closing the contour counter-clockwise, from $R$ to$-R$.
There are no singular points inside the contour; therefore
$$-\frac{i}{2}\oint e^{a e^{i z}}\frac{dz}{z}=0=I+I_r+I_R\qquad(1)$$
where $I_r, I_R$ are the integrals along the small and big arches correspondingly:
$$I_r=-\pi i\underset{z=0}{\operatorname{Res}}\bigg(-\frac{i}{2} e^{a e^{i z}}\frac{1}{z}\bigg)=-\frac{\pi}{2}e^a\qquad(2)$$
(we integrate along the small arch clockwise, in the negative direction), and
$$I_R=-\frac{i}{2}\int_0^\pi e^{ae^{iRe^{i\phi}}}id\phi=\frac{1}{2}\int_0^\pi e^{ae^{-R\sin\phi}e^{iR\cos\phi}}d\phi\to\frac{1}{2}\int_0^\pi d\phi=\frac{\pi}{2}\,\text{at}\,R\to\infty\quad(3)$$
Putting (2) and (3) into (1)
$$I=-I_r-I_R=\frac{\pi}{2}\Big(e^a-1\Big)$$
