Referring to a question I posted on MS, I post it here, as I didn't get an answer:

By analogy with the Jacobi–Anger expansion, one expects that $e^{iz\cot(x)}$ has a Fourier expansion of the form : $$e^{iz\cot(\theta)}=\sum_{n=-\infty}^{\infty}\Lambda_{n}(z)e^{in\theta}$$ $\Lambda_{n}(z)$ is given by: $$\Lambda_{n}(z)=\frac{1}{\pi}\int_{-\pi/2}^{\pi/2}e^{iz\cot(\theta)-in\theta}d\theta$$ A simple calculation yields: $$\Lambda_{n}(z)=\frac{2}{n\pi}\sin\left(\frac{\pi n}{2} \right )\sum_{m=0}^{\infty}\frac{z^{m}}{m!}\text{F}_{1}\left(-\frac{n}{2};-m,m;1-\frac{n}{2};1,-1 \right )$$

Where $\text{F}_{1}(\alpha;\beta,\beta^{'};\gamma;x,y)$ is the Appell Hypergeometric Function. Now, I have two questions:

1-For a purely imaginary $z$,$\;\;$ $e^{iz\cot(\theta)}$ has essential singularities at $\theta=\pm n\pi$. How is that reflected in the Fourier expansion?

2-Can we express the infinite sum in terms of other special functions?


I will try to answer your two questions in reverse order.

$2.$ For real $z$ your Fourier coefficient


can be expressed in terms of elementary functions for even $n$ and in terms of the special functions $I_{m}(z)$ (modified Bessel function of the first kind) and $L_{-m}(z)$ (modified Struve function) for odd $n$. The first few Fourier coefficients are


$$\Lambda_1(z)=\Lambda_{-1}(-z)=z L_{-1}(z)+zL_0(z)-|z|I_0(z)-|z|I_1(z)+2/\pi$$


For even $n$ the Fourier coefficient $\Lambda_n(z)\equiv 0$ for $nz>0$ [***]

$1.$ The Fourier coefficients, given above along the real $z$-axis, can be continued analytically into the complex-$z$ plane, but not along the imaginary $z$-axis. The branch cut for imaginary $z$ is a direct consequence of the essential singularity you noticed.


added November 2014 in response to a request for clarification.

since $\Lambda_{n}(z)=\Lambda_{-n}(-z)$, we may take $n>0$. Suppose $n=2m$, $m=1,2,\ldots$ and $z>0$. Then $$ \Lambda_{2m}(z)=\frac{1}{2\pi}{\rm Re}\,\int_{-\pi}^{\pi}e^{im\theta}e^{-iz\cot(\theta/2)}\,d\theta =\frac{1}{2\pi i}{\rm Re}\,\oint_{|\xi|=1^-} \xi^{m-1}e^{-z(1+\xi)/(1-\xi)}\,d\xi=0 $$ For $z<0$ the singularity at $\xi\equiv e^{i\theta}=1$ contributes, so the contour integral does not vanish.


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