# Gamma function

• January 14th 2011, 05:48 PM
Random Variable
Gamma function
Can you say that $\displaystyle \Gamma \Big(-\frac{1}{4} \Big) = \int_{0}^{\infty} x^{-5/4} e^{-x} \ dx$ even though the integral diverges and by definition $\displaystyle \Gamma \Big(-\frac{1}{4} \Big) \ne \infty$ ?

Specifically what I want to know is does $\displaystyle \Gamma \Big(\frac{5}{4} \Big) \int_{0}^{\infty} x^{-5/4} e^{-x} \ dx$ diverge, or does it equal $\displaystyle \Gamma \Big(\frac{5}{4} \Big) \Gamma \Big(\frac{-1}{4} \Big) = \frac{\pi}{sin \frac{5 \pi}{4}} = - \sqrt{2} \pi$ ?
• January 14th 2011, 06:12 PM
chisigma
The definition...

$\displaystyle \Gamma(z)= \int_{0}^{\infty} t^{z-1}\ e^{-t}\ dt$ (1)

... is valid only for $\mathcal{R} (z) >0$. For the value of z with non positive real part You have to compute the so called 'analytic extension' of $\Gamma (*)$, that pratically means to use the 'Euler's formula'...

$\displaystyle \Gamma(z)\ \Gamma(1-z)= \frac{\pi}{\sin \pi\ z}$ (2)

Kind regards

$\chi$ $\sigma$
• January 14th 2011, 06:19 PM
Random Variable
So then $\displaystyle \Gamma \Big(\frac{5}{4} \Big) \int_{0}^{\infty} x^{-5/4} e^{-x} \ dx$ diverges?
• January 14th 2011, 06:19 PM
dwsmith
Quote:

Originally Posted by Random Variable
$\displaystyle \Gamma \Big(\frac{5}{4} \Big) \Gamma \Big(\frac{-1}{4} \Big) = \frac{\pi}{sin \frac{5 \pi}{4}} = - \sqrt{2} \pi$ ?

$\displaystyle\Gamma\left(-\frac{1}{2}\right)=-\sqrt{2}\pi$

I know

$\displaystyle\Gamma\left(-\frac{1}{2}\right)=-2\Gamma\left(\frac{1}{2}\right)$

I am not sure if that implies

$\displaystyle\Gamma\left(-\frac{1}{4}\right)=-4\Gamma\left(\frac{1}{4}\right)\mbox{?}$
• January 14th 2011, 06:31 PM
chisigma
Quote:

Originally Posted by Random Variable
So then $\displaystyle \Gamma \Big(\frac{5}{4} \Big) \int_{0}^{\infty} x^{-5/4} e^{-x} \ dx$ diverges?

... the integral $\displaystyle \int_{0}^{\infty} t^{-\frac{5}{4}}\ e^{-t}\ dt$ diverges...

Kind regards

$\chi$ $\sigma$
• January 14th 2011, 06:33 PM
dwsmith
The gamma integral doesn't converge $\forall x<0$.