0

How can I evaluate the value of $$ \frac{\Gamma(\frac{1}{2}-x)}{\Gamma(x)} $$

Is there a simple way to write this term, (Simple in the sense that there is no denominator term)?

I have tried to use the duplication and the reflection formula of the Gamma function, but it doesn't help much. Maybe they need to be applied multiple times. Any ideas\suggestions are appreciated.

Bill Dubuque
  • 272,048
Kashif
  • 710

2 Answers2

5

Hint

Use $$\frac{\Gamma \left(\frac{1}{2}-x\right)}{\Gamma (1-2 x)}=\sqrt{\pi }\frac{ 2^{2 x}}{\Gamma (1-x)}$$ and then the duplication or reflection formula.

Claude Leibovici
  • 260,315
  • 3
    And, one can remove the Gamma function in the denominator with Eulers reflection formula: $$\frac{1}{\Gamma(1-x)}=\frac{\Gamma(x)\sin(\pi x)}{\pi}$$ – K.defaoite Jul 19 '21 at 13:10
  • 2
    @K.defaoite. If my memory is good (not sure at my age), there are a few other ways to remove the denominator. Cheers :-) – Claude Leibovici Jul 19 '21 at 13:14
1

For getting a representation involving zeta functions use

$$\frac{\Gamma\left(\frac{\xi}{2} \right)}{\Gamma\left(\frac{1-\xi}{2} \right)}=\frac{\zeta(1-\xi)}{\zeta(\xi)}\pi^{\xi -1/2}$$

for some $\xi \in \mathbb{C}$ with $Re(\xi) >0$

A S D
  • 356