Story
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I want to prove [Euler's reflection formula](https://en.wikipedia.org/wiki/Reflection_formula) by showing that


\begin{equation\*}
 f(s) = \sin(\pi s) \Gamma(s) \Gamma(1 - s)
\end{equation\*}


is constant, where $s = \sigma + it$. It's easy to see that $f$ is entire and $f(s + 1) = f(s)$, so for fixed $t$ we have $f \in C(\mathbb{R})^\infty$ and $f(\sigma + it)$ is 1-periodic. Therefore $f$ has a rapidly decreasing Fourier series for fixed $s$


\begin{equation\*}
f(s) = \sum \_{n \in \mathbb{Z}} c\_n(t) e^{i 2 \pi n \sigma}.
\end{equation\*}


Let's have a look at the $m$-th term. By definition we have
\begin{equation\*}
 f\_m(s) = \left(\int \_0 ^1 f(s + x) e^{- i 2 \pi m x} dx\right) e^{i 2 \pi m \sigma} = c\_m(t) e^{i 2 \pi m \sigma}.
\end{equation\*}


Problem
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Now to my problem and question: **Is $f\_m$ analytic and why**? I can't find an argument and I need one to justify the use of the Cauchy-Riemann equations later on.



Proof of Euler's reflection formula via rapidly decreasing Fourier series
The function $\sigma\mapsto f(\sigma+it)$ is periodic with period $1$, for any value
of $t$. Therefore
$$f(\sigma+it)=\sum\_{n\in Z} c\_n(t)e^{2\pi i n\sigma}.$$
It follows that 
$$c\_n(t)=\int\_0^1 f(x+it) e^{-2\pi i n x}\,dx.$$
The $n$-th term in the Fourier expansion is, with $s=\sigma+it$
\begin{multline\*}
f\_m(s)=f\_m(\sigma+it)=c\_n(t)e^{2\pi i n\sigma}=e^{2\pi i n\sigma}
\int\_0^1f(x+it) e^{-2\pi i n x}\,dx\\
=\int\_0^1 f(x+it) e^{-2\pi i n (x-\sigma)}\,dx
= \int\_{-\sigma}^{1-\sigma} f(x+\sigma+it) e^{-2\pi i n x}\,dx
\end{multline\*}


By the periodicity the limits of the integral can be changed to $0$ and $1$, 
therefore
$$f\_m(s)=\int\_0^1f(s+x)e^{-2\pi i n x}\,dx.$$
That is not what you write in your question. In this form the analyticity 
of $f\_m$ is easy. For example consider Theorem 5.4 in the book


E. M. Stein and R. Shakarchi, Complex Analysis, Princeton University Press, 
2003. 


