MHB Show Integral of $f(z)$ with Exponential Form

[polygamma]

Show that $ \displaystyle \text{Im} \ \exp \left( ae^{be^{i c \ x}} \right) = \exp \left( ae^{b\cos(cx)\cos (b\sin cx)} \right) \sin \left( a\sin (b\sin c x) e^{b\cos cx} \right) $.EDIT: Then by integrating $ \displaystyle f(z) = \frac{z \exp ( ae^{be^{ic \ x}} )}{z^{2}+d^{2}}$ around a contour that consists of the real axis and the upper half of the circle $|z|=R$, show that

$$ \int_{0}^{\infty}\frac{ x\exp \left( ae^{b\cos(cx)\cos (b\sin cx)}\right) \sin \left( a\sin (b\sin cx) e^{b\cos cx} \right)}{x^{2}+d^{2}}\ dx = \frac{\pi}{2}\left( \exp \left( ae^{be^{-c \ d}} \right)-e^{a}\right)$$This requires showing that $ \displaystyle \lim_{R \to \infty} \int_{C_{R}} f(z) \ dz = i \pi e^{a}$ where $C_{R}$ is the upper half of the circle $|z|=R$.

 

[polygamma]

Spoiler

$$ \int_{C_{R}}f(z)\ dz =\int_{C_{R}}\frac{z}{z^{2}+d^{2}}\ dz+a\int_{C_{R}}\frac{ze^{be^{icx}}}{z^{2}+d^{2}}\ dz+\frac{a^{2}}{2!}\int_{C_{R}}\frac{ze^{2be^{icx}}}{z^{2}+d^{2}}\ dz+\ldots $$

$$ =\int_{C_{R}}\frac{z}{z^{2}+d^{2}}\ dz+a\left(\int_{C_{R}}\frac{z}{z^{2}+d^{2}}\ dz+b\int_{C_{R}}\frac{ze^{icz}}{z^{2}+d^{2}}+\frac{b^{2}}{2!}\int_{C_{R}}\frac{z e^{2icz}}{z^{2}+d^{2}}\ dz+\ldots\right) $$

$$ +\ \frac{a^{2}}{2!}\left(\int_{C_{R}}\frac{z}{z^{2}+d^{2}}\ dz+2b\int_{C_{R}}\frac{ze^{2icz}}{z^{2}+d^{2}}+ \frac{4b^{2}}{2!} \int_{C_{R}} \frac{z e^{4icz}}{z^{2}+d^{2}}\ dz+\ldots\right)+\ldots $$Then using Jordan's lemma and the fact that $ \displaystyle \lim_{|z| \to \infty} z \frac{z}{z^{2}+d^{2}} = 1$,

$$ \lim_{R\to\infty}\int_{C_{R}}f(z)\ dz =\pi i+a (\pi i+0+0+\ldots)+\frac{a^{2}}{2!}(\pi i+0+0+\ldots)+\ldots $$

$$ = i\pi\left( 1+a+\frac{a^{2}}{2!}+\ldots\right) = i\pi e^{a} $$