# Thread: Gamma function of -1/2 = ?

1. ## 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
$\Gamma(-1/2)= \int_{0 }^{ \infty } {u }^{-2 }{e }^{ {-u }^{2 } }du$
I cannot find anyway to solve it.
Any help would be really appreciated. Thanks a lot.

2. 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

3. 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?