# How to compute the following integral using complex analysis?

I'm trying to compute the following integral using complex analysis: $$$$\int_0^{2\pi}\sin(\exp(e^{i \theta}))d\theta$$$$ 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 $$$$\int_{|z|=1}\frac{\sin(\exp(z))}{iz}dz = \operatorname{Res}(f,0) = \lim_{|z|\to0}-i\sin(\exp(z)) = -i\sin(1)$$$$ It doesn't seem right, though. Can anyone please help me out?

$$\int_{|z|=1} \frac{\sin(\exp(z))}{iz}\mathrm{d}z = 2\pi i\text{ Res}\left(\frac{\sin(\exp(z))}{iz}, 0\right) = 2\pi \lim_{z\to 0} \sin(\exp(z)) = 2\pi \sin(1)$$

• Oh, so I just forgot to add the $2\pi i$. Wonderful! Thank you so much! Commented Aug 22, 2021 at 21:02

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 $$0: $$\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 $$\mathbb{C}$$).

$$\frac{1}{2\pi}\int_0^{2\pi} \sin(\exp(e^{it}))dt=f(0)=\sin(1).$$

• This one is quite interesting, thanks for answering! Commented Aug 23, 2021 at 14:17

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

$$\int_{0}^{2\pi}\sin(\exp(e^{i\theta}))\,d\theta = 2\pi\sum_{n\geq 0}\frac{(-1)^n}{(2n+1)!}=\color{red}{2\pi\sin(1)}. \tag{2}$$

• Wow, even with power series. This was really beautiful. Thanks for replying! Commented Aug 23, 2021 at 14:19