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

Question

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.

Answer 1

Have you by any chance seen$\dots$

$$\sin \left({\pi z}\right) = \pi z \prod_{n \mathop \ne 0} \left({1 - \frac z n}\right) \exp \left({\frac z n}\right)$$ $$\frac 1 {\Gamma \left({z}\right)} = z e^{\gamma z} \prod_{n \mathop = 1}^\infty \left({1 + \frac z n}\right) \exp \left({-\frac z n}\right)$$

With these the proof of your identity is quick. Of course, to prove these you could use contour integration, which isn't exactly what you wanted.