How can I integrate $\exp(−\sin(θ))\cos(θ+\cos(θ))$?

Question

The question refers to the contour integration of $\exp(−\sin(θ))\cos(θ+\cos(θ))$

I know I have to use residue theorem somehow, but I can't realise

Here's what I tried to do, but it doesn't match to the statement

Answer 1

If you mean to find $$\int_0^{2\pi} \exp(-\sin\theta)\cos(\theta+\cos\theta)d\theta$$a famous method for solving such contour integral is to assume $z=\exp(i\theta)$ with the contour being $|z|=1$. In this case, $dz=izd\theta$, $\sin\theta={1\over 2i}(z-z^{-1})$, $\cos\theta={1\over 2}(z+z^{-1})$ and this integral is equivalent to $${1\over i}\int \exp({i\over 2}(z-z^{-1}))\left[{1\over 2}(z+z^{-1})\cos \left({1\over 2}(z+z^{-1})\right)-{1\over 2i}(z-z^{-1})\sin \left({1\over 2}(z+z^{-1})\right)\right]{dz\over z}$$which has a singularity in $z=0$. The rest of the solution is left to you by using the Laurant series of the integrand around $z=0$.

(Hint: you only need to find the constant term in $$\exp({i\over 2}(z-z^{-1}))\left[{1\over 2}(z+z^{-1})\cos \left({1\over 2}(z+z^{-1})\right)-{1\over 2i}(z-z^{-1})\sin \left({1\over 2}(z+z^{-1})\right)\right]$$ )

Answer 2

$e^{iz}$ is an entire function, so by Cauchy's Theorem $$\oint_{|z|=1} e^{iz}dz =0$$

On the other hand, using the parametrization of the unit circle $\gamma(t)=e^{it}=\cos t+i\sin t$, $t\in[0,2\pi]$, we have that $$0=\oint_{|z|=1} e^{iz}dz = \int_0^{2\pi} e^{i{\gamma(t)}}\gamma'(t)dt = \int_0^{2\pi} e^{-\sin t+i\cos t} i e^{it}dt = i \int_0^{2\pi} e^{-\sin t+(t+\cos t)i}dt =$$ $$=i\int_0^{2\pi} e^{-\sin t}\left(\cos(t+\cos t) + i \sin(t+\cos t)\right)dt $$ Hence, $$\int_0^{2\pi}e^{-\sin t}\cos(t+\cos t)dt=0=\int_0^{2\pi}e^{-\sin t}\sin(t+\cos t)dt$$