I would like to calculate the following definite integral
$$\int_{-\pi}^{\pi}e^{-x\,u\,i}e^{R\,e^{\,u\,i}}du$$
where $x,R>0$ in terms of elemental or special functions.
A first attempt
$$\int_{-\pi}^{\pi}e^{-x\,u\,i}e^{R\,e^{\,u\,i}}du=-i\,\int_Cz^{-x-1}\,e^{R\,z}\,dz$$
with $C$ the unit circle.
It's correct? In this case, how continue?
Any help or any other way to solve it will be welcomed
On
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\on}[1]{\operatorname{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} &\bbox[5px,#ffd]{\left. \int_{-\pi}^{\pi} \expo{-\ic xu}\expo{R\expo{\ic u}}\dd u \,\right\vert_{\ x, R\ >\ 0}} = \left.\oint_{\verts{z}\ =\ 1} z^{-x}\,\expo{Rz}\,{\dd z \over \ic z}\,\right\vert_{\ z\ =\ \exp\pars{\ic u}} \\[5mm] = &\ -\ic\oint_{\verts{z}\ =\ 1} z^{-x - 1}\,\expo{Rz}\,\dd z \\[5mm] \stackrel{\mrm{as}\ \epsilon\ \to\ 0^{+}}{\sim}\,\,\,&\,\, \ic\int_{-1}^{-\epsilon}\pars{-\xi}^{-x - 1}\, \expo{\ic\pi\pars{-x - 1}}\,\expo{R\xi}\,\dd \xi \\[2mm] + &\ \ic\int_{\pi}^{-\pi}\bracks{\epsilon^{-x - 1}\, \expo{\ic\pars{-x - 1}\theta}} \bracks{\epsilon\expo{\ic\theta}\ic}\dd\theta \\[2mm] + &\ \ic\int_{-\epsilon}^{-1}\pars{-\xi}^{-x - 1}\, \expo{-\ic\pi\pars{-x - 1}}\,\expo{R\xi}\,\dd \xi \\[5mm] = &\ -\ic\expo{-\ic\pi x}\int_{\epsilon}^{1}\xi^{-x - 1}\, \,\expo{-R\xi}\,\dd \xi + {2\epsilon^{-x}\sin\pars{\pi x} \over x} \\[2mm] &\ \,\,\, + \ic\expo{\ic\pi x}\int_{\epsilon}^{1}\xi^{-x - 1}\, \,\expo{-R\xi}\,\dd \xi \\[5mm] = &\ -2\sin\pars{\pi x}\int_{\epsilon}^{1}\xi^{-x - 1}\, \,\expo{-R\xi}\,\dd \xi + {2\epsilon^{-x}\sin\pars{\pi x} \over x} \\[5mm] \stackrel{\mrm{as}\ \epsilon\ \to\ 0^{+}}{\sim}\,\,\,\,\, &\ 2\expo{-R}\,{\sin\pars{\pi x} \over x} + 2R\,{\sin\pars{\pi x} \over x} \int_{\epsilon}^{1}\xi^{-x}\expo{-R\xi}\,\dd\xi \end{align} The last integral converges, when $\ds{\epsilon \to 0^{+}}$, whenever $\ds{\Re\pars{x} < 1}$. In such a case, \begin{align} &\bbox[5px,#ffd]{\left. \int_{-\pi}^{\pi} \expo{-\ic xu}\expo{R\expo{\ic u}}\dd u \,\right\vert_{\substack{R\ >\ 0 \\[1.5mm] \Re\pars{x}\ <\ 1}}} \\[5mm] = &\ 2\expo{-R}\,{\sin\pars{\pi x} \over x} + 2R^{x}\,{\sin\pars{\pi x} \over x} \int_{0}^{R}\xi^{-x}\expo{-\xi}\,\dd\xi \\[5mm] = &\ 2\expo{-R}\,{\sin\pars{\pi x} \over x} + 2R^{x}\,{\sin\pars{\pi x} \over x} \bracks{\int_{0}^{\infty}\xi^{-x}\expo{-\xi}\,\dd\xi - \int_{R}^{\infty}\xi^{-x}\expo{-\xi}\,\dd\xi} \\[5mm] = &\ \bbx{2\expo{-R}\,{\sin\pars{\pi x} \over x} + 2R^{x}\,{\sin\pars{\pi x} \over x} \bracks{\Gamma\pars{-x + 1} - \Gamma\pars{-x + 1,R}}} \\ & \end{align} $\ds{\Gamma\pars{\cdots}\ \mbox{and}\ \Gamma\pars{\cdots,\cdots}}$ are the Gamma and Incomplete Gamma Functions, respectively.
Your attempt is correct if $x$ is an integer (and in that case you get $2\pi R^x/x!$ following, say, this approach). Otherwise, you have a multivalued integrand, so the contour integral isn't well-defined. But one can do $$\int_{-\pi}^{\pi}e^{-xui}e^{Re^{ui}}du=\sum_{n=0}^\infty\frac{R^n}{n!}\int_{-\pi}^{\pi}e^{i(n-x)u}~du=2\sin x\pi\sum_{n=0}^\infty\frac{(-R)^n}{n!(n-x)}$$ where (analytic continuation of) the lower incomplete gamma function appears.