Fourier coefficients of $e^{xe^{it}}$

Question

I am given an arbitrary $x \in \mathbb R$ and the $2 \pi$-periodic function $$f(t) = e^{xe^{it}}.$$ The Fourier coefficients are for any $n$ given by \begin{equation*} 2 \pi c_n ( f) = \int_{-\pi}^{\pi}f(t)e^{-int}dt \end{equation*} This is not very useful as it results in a very ugly integral. Since the function is differentiable on the interval we can use the relation $c(f') = i n c(f)$. The derivative of the function is $ix e^{xe^{it}+it}$ which gives the coefficients: \begin{equation*} 2 \pi c_n ( f') = \int_{-\pi}^{\pi} ix e^{xe^{it}+it} e^{-int}dt \end{equation*} An obvious substitution is $z = xe^{it}$ which gives $x^n \int e^z z^{-n} dz$ but this seems to mess up the integration bounds as $-x = x e^{i \pi} = x e^{- i \pi}$. I therefore tried to find some symmetry in the integrand, but to no avail. Also is this method possible? Do we need to ensure that $f'$ has the same period as $f$? Any hints will be appreciated.

Answer 1

Compare $f(t)=\displaystyle\sum_{n\in\mathbb Z}c_n(f)\mathrm e^{\mathrm int}$ and $\displaystyle\exp(x\mathrm e^{\mathrm it})=\sum_{n\geqslant0}\frac{(x\mathrm e^{\mathrm it})^n}{n!}=\sum_{n\geqslant0}\frac{x^n}{n!}\mathrm e^{\mathrm int}.$