I want to prove Euler's 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*}


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.

  • $\begingroup$ It seems like you are looking for a version of Leibniz's rule (for differentiation under the integral sign) for complex-analytic functions. This is standard, and should be in just about any textbook in complex analysis. By the way, you can use Stirling's bound on the gamma function to show that $|f(s)|$ is bounded in any fixed strip. $\endgroup$ – Matt Young Apr 13 '16 at 18:33
  • $\begingroup$ Of course, $f$ is analytic (it is constant!). The answer really depends on what you want to use about $\sin$ and $\Gamma$. $\endgroup$ – Alexandre Eremenko Apr 13 '16 at 18:38
  • $\begingroup$ @AlexandreEremenko I was asking about the analyticity of $f_m$ and I still have to show that $f$ is constant. That's what the proof is all about. $\endgroup$ – fje Apr 13 '16 at 18:48
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    $\begingroup$ @AlexandreEremenko : it is obvious that he starts from $\Gamma(s) = \int_0^\infty x^{s-1} e^{-x} dx, \Gamma(s+1) = s \Gamma(s)$. that $f(s)$ is entire is easy to prove, but that $\Gamma(s)$ has no zero is much less. $\endgroup$ – reuns Apr 13 '16 at 19:03
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    $\begingroup$ The actual argument $\sigma$ should not appear in the formula for the Fourier coefficient, see the answer below. $\endgroup$ – Sebastian Goette Apr 13 '16 at 19:53

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.

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  • $\begingroup$ Oh, thank you! I didn't realise that my $f_m$ was just wrong. $\endgroup$ – fje Apr 13 '16 at 20:06

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