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I would like to prove Euler's reflection formula $$\Gamma(z)\Gamma(1-z) = \frac{\pi}{\sin(\pi z)}$$ using Wielandt's theorem:

Let $f$ be a function that is bounded on the strip $1 \le \mathrm{Re}(z) < 2$ and holomorphic on an open set containing it. If $f(z+1) = z f(z)$ for all $z$ with $z,z+1 \in D$, then $f(z) = f(1)\Gamma(z)$.

It is not hard to see that $f(z) = \frac{\pi}{\Gamma(1-z)\sin(\pi z)}$ satisfies the functional equation $f(z+1) = z f(z)$. I'm having trouble showing that it is bounded in the given strip. Is there an easy proof?

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  • $\begingroup$ I guess using the product representations would kind of defeat the purpose, wouldn't it? But that's the easiest thing that comes to mind. $\endgroup$ – Daniel Fischer Nov 23 '13 at 22:45

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