# Show that $\Gamma(z)\Gamma(1-z) \sin \pi z$ is bounded in the complex plane

Attempt

I know that $\Gamma(z)=\int_0^\infty e^{-t}t^{z-1} \ dt$ so $$\lvert \Gamma(z) \rvert \leq \int_0^\infty e^{-t}|t^{z-1}| \ dt=\int_0^{\infty} \frac{e^{-t}}{t} t^{\Re(z)} dt .$$ After this, I'm not really sure what more I can do.

Edit

I am actually trying to show that $$\frac{\pi}{\sin \pi z} =\Gamma(z) \Gamma(1-z)$$ without resorting to contour integration. However I am beginning to think that showing that this function is bounded might be more work than just carrying out the integration.

Sorry for the lack of background in the original question.

• What do you know about the Gamma function? – carmichael561 Mar 1 '16 at 4:26
• @carmichael561 I just editted my question with some of my thoughts. – illysial Mar 1 '16 at 4:40
• If $0<z<1$ then both $\Gamma(z)$ and $\Gamma(1-z)$ have an integral representation, and by combining them and computing a contour integral one can show that $\Gamma(z)\Gamma(1-z)=\frac{\pi}{\sin(\pi z)}$. See Stein's Complex Analysis book for instance. – carmichael561 Mar 1 '16 at 4:55
• @carmichael561 funny enough, that integral is exactly what I was hoping to compute by showing the result in my question. After an hour of fooling around with the Stirling approximation, I guess contour integration cannot be avoided after all. – illysial Mar 1 '16 at 4:58

Have you by any chance seen$\dots$
$$\sin \left({\pi z}\right) = \pi z \prod_{n \mathop \ne 0} \left({1 - \frac z n}\right) \exp \left({\frac z n}\right)$$ $$\frac 1 {\Gamma \left({z}\right)} = z e^{\gamma z} \prod_{n \mathop = 1}^\infty \left({1 + \frac z n}\right) \exp \left({-\frac z n}\right)$$