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}

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}
