Integral calculating using complex analysis

Question

There is an integral given: $$\int \limits_0^{2 \pi}e^{\cos t} \cos(nt - \sin t) \mbox{d}t$$ Of course the integrand has no antiderivative.
Firstly I thought of calculating residues, but our function has no poles.
I wonder how that integral can be solved.

Answer 1

Presumably $n$ is an integer. The integral is the real part of $$\int_0^{2\pi}e^{\cos(t)}e^{int-i\sin(t)}\,dt=\int_0^{2\pi}e^{int}e^{e^{-it}}\,dt=\int_0^{2\pi}e^{int}\sum_k\frac1{k!}e^{-ikt}\,dt= \begin{cases}\frac{2\pi}{n!},&(n\ge0), \\0,&(n<0).\end{cases}$$

Answer 2

$$ \begin{aligned} I & =\int_0^{2 \pi} e^{\cos t} \cos (\sin t-n t) d t+i \int_0^{2 \pi} e^{\cos t} \sin (\sin t-n t) d t \\ & =\int_0^{2 \pi} e^{\cos t} e^{(\sin t-n t) i} d t \\ & =\int_0^{2 \pi} \frac{e^{e^{t i}}}{e^{n t i}} d t \end{aligned} $$ Integrating along the unit circle $\kappa(0,1)$ by putting $z=e^{ti}$ as $$ \begin{aligned} I & =\frac{1}{i} \int_{\kappa(0,1)} \frac{e^z}{z^{n+1}} d z \\ & =\frac{1}{i}\cdot 2 \pi i \operatorname{Res}\left(\frac{e^z}{z^{n+1}} ,z=0\right) \\ & =\frac{2 \pi}{n !} \lim _{z \rightarrow 0}\left(z^{n+1} \cdot \frac{e^z}{z^{n+1}}\right)^{(n)} \\ & =\frac{2 \pi}{n !} \end{aligned} $$ Comparing the real and imaginary parts on both sides yields $$ \begin{aligned} & \int_0^{2 \pi} e^{\cos t} \cos (\sin t-n t) d t=\operatorname{Re}(I)=\frac{2 \pi}{n !} \\ & \therefore \quad \boxed{\int_0^{2 \pi} e^{\cos t} \cos (nt-\sin t) d t=\frac{2 \pi}{n !} }\\ \text { As a bonus, } \\ & \int_0^{2 \pi} e^{\cos t} \sin (n t-\sin t) d t=0 \\ & \end{aligned} $$