Let, $s=\sigma+it$. For $\sigma >0$ prove that
$$\int_0^{\infty}x^{\frac s2-1}e^{-n^2\pi x}\,dx=\frac{\Gamma(s/2)}{n^s\pi^{s/2}}$$
How can I prove this ? I think it can be prove by complex integration with suitable contour which I've to choose with a suitable complex function. But I'm unable to do that.
$$ \int_{0}^{\infty}x^{-1+s/2}e^{-n^2\pi x}dx \\ = (n^2\pi)^{-s/2}\int_{0}^{\infty}(n^2\pi x)^{-1+s/2}e^{-(n^2\pi x)}d(n^2\pi x) \\ = (n^2\pi)^{-s/2}\int_{0}^{\infty}u^{-1+s/2}e^{-u}du \\ = (n^2\pi)^{-s/2}\Gamma(s/2). $$