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.


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.

  • 2
    $\begingroup$ What do you know about the Gamma function? $\endgroup$ – carmichael561 Mar 1 '16 at 4:26
  • $\begingroup$ @carmichael561 I just editted my question with some of my thoughts. $\endgroup$ – illysial Mar 1 '16 at 4:40
  • $\begingroup$ 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. $\endgroup$ – carmichael561 Mar 1 '16 at 4:55
  • $\begingroup$ @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. $\endgroup$ – illysial Mar 1 '16 at 4:58

Have you by any chance seen$\dots$

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

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.

  • $\begingroup$ I hope this isn't unhelpful. $\endgroup$ – GiantTortoise1729 Mar 1 '16 at 5:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.