# Gamma function of -1/2 = ?

• Jul 14th 2010, 06:59 PM
ttna90
Gamma function of -1/2 = ?
I've been struggling to solve this the whole day today :(. Basically, I followed the same process as demonstrated in this post, but when I got to this equation
Quote:

$\Gamma(-1/2)= \int_{0 }^{ \infty } {u }^{-2 }{e }^{ {-u }^{2 } }du$
I cannot find anyway to solve it.
Any help would be really appreciated. Thanks a lot.
• Jul 14th 2010, 07:14 PM
mr fantastic
Quote:

Originally Posted by ttna90
I've been struggling to solve this the whole day today :(. Basically, I followed the same process as demonstrated in this post, but when I got to this equation

I cannot find anyway to solve it.
Any help would be really appreciated. Thanks a lot.

$\displaystyle {\int_{0}^{+\infty} u^{-2} e^{-u^2} \, du = \frac{1}{2} \int_{0}^{+\infty} t^{-3/2} e^{-t} \, dt}$ making an obvious substitution

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

See Gamma function - Wikipedia, the free encyclopedia
• Jul 15th 2010, 04:45 AM
Ackbeet
I would probably use the following two well-known properties of the gamma function:

$\Gamma(\frac{1}{2})=\sqrt{\pi},$ and
$\Gamma(z+1)=z\Gamma(z).$

Taken together, can you tell me what $\Gamma(-\frac{1}{2})$ is?