Evaluate $\Gamma(-\frac{1}{2})$ directly from the definition of $\Gamma()$

Question

Can you evaluate $\space\Gamma(-\frac{1}{2})$ directly from the definition $\Gamma(s) = \int_0^\infty e^{-t}t^{s-1}dt, $

not using $\space \Gamma(\frac{1}{2}) = \sqrt{\pi}$ and $\Gamma(s+1) = s\Gamma(s)$?

I failed to do the integral just like the way to find $\space \Gamma(\frac{1}{2})$ using polar coordiates, but it looks like it goes to infinity.

Answer 1

If we put $s=-\frac12$ in the definition, we get $$\int_0^{\infty}\frac{e^{-t}}{t^{3/2}}dt>\int_0^1\frac{e^{-t}}{t^{3/2}}dt>\int_0^1\frac{e^{-1}}{t^{3/2}}dt=\lim_{\delta\to 0^+}\frac{2}{e}(\frac{1}{\sqrt\delta}-1)=\infty.$$ So, the improper integral is divergent.