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, 2016 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 $\Gamma$. $\endgroup$ Apr 13, 2016 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, 2016 at 18:48
  • 3
    $\begingroup$ @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, 2016 at 19:03
  • 1
    $\begingroup$ The actual argument $\sigma$ should not appear in the formula for the Fourier coefficient, see the answer below. $\endgroup$ Apr 13, 2016 at 19:53

1 Answer 1


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.

  • $\begingroup$ Oh, thank you! I didn't realise that my $f_m$ was just wrong. $\endgroup$
    – fje
    Apr 13, 2016 at 20:06

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .