I'm trying to compute the following integral using complex analysis: \begin{equation} \int_0^{2\pi}\sin(\exp(e^{i \theta}))d\theta \end{equation} I know that there has to be an easy way out, but I can't see it.
I've tried the following: by changing of variable $z = e^{i\theta}$, we get to \begin{equation} \int_{|z|=1}\frac{\sin(\exp(z))}{iz}dz = \operatorname{Res}(f,0) = \lim_{|z|\to0}-i\sin(\exp(z)) = -i\sin(1) \end{equation} It doesn't seem right, though. Can anyone please help me out?
$$\int_{|z|=1} \frac{\sin(\exp(z))}{iz}\mathrm{d}z = 2\pi i\text{ Res}\left(\frac{\sin(\exp(z))}{iz}, 0\right) = 2\pi \lim_{z\to 0} \sin(\exp(z)) = 2\pi \sin(1)$$
In general for a holomorphic function $f$ on a disc say ($D(a,R)$) (and even for a much more class of functions), we have for every $0<r<R$: $$ \frac{1}{2\pi}\int_0^{2\pi} f(a+re^{it})dt=f(a). $$ This is called the mean propriety, and the proof is direct using Cauchy formula, in fact \begin{eqnarray} f(a)&=&\frac{1}{2i\pi}\int_{|z-a|=r}\frac{f(z)}{z -a}dz\\ &=&\frac{1}{2i\pi} \int_0^{2\pi} \frac{f(a+re^{it})}{re^{it}}ire^{it}dt \qquad (z=a+re^{it}) \\ &=& \frac{1}{2\pi}\int_0^{2\pi} f(a+re^{it})dt. \end{eqnarray}
In your case, taking $f(z)=\sin(e^z)$ this function is entire (holomorphic on $\mathbb{C}$).
$$ \frac{1}{2\pi}\int_0^{2\pi} \sin(\exp(e^{it}))dt=f(0)=\sin(1). $$
You can just use power series, since both $\sin$ and $\exp$ are entire functions. $$ \sin(\exp(e^{i\theta})) = \sum_{n\geq 0}\frac{(-1)^n}{(2n+1)!}\exp((2n+1)e^{i\theta})=\sum_{n\geq 0}\frac{(-1)^n}{(2n+1)!}\sum_{m\geq 0}\frac{(2n+1)^m e^{m i\theta}}{m!}\tag{1} $$ but $\int_{0}^{2\pi}e^{mi\theta}\,d\theta = 0$ unless $m=0$, so
$$ \int_{0}^{2\pi}\sin(\exp(e^{i\theta}))\,d\theta = 2\pi\sum_{n\geq 0}\frac{(-1)^n}{(2n+1)!}=\color{red}{2\pi\sin(1)}. \tag{2}$$
