I would like to compute the integral
$ \int_{0}^{\infty}\frac{1}{\sqrt{2t}}e^{-\frac{1}{2t}}dt $
which wolfram alpha says that it does not converge. However by letting $x=1/2t$ I get $dt=\frac{-1}{2x^2}dx$ and that the integral is transformed to $ \int_{\infty}^{0}x^{0.5}e^{-x}\cdot\frac{-1}{2x^{2}}dx $ or $ \frac{1}{2}\int_{0}^{\infty}x^{-1.5}e^{-x}dx=0.5\Gamma(-0.5) $ which is finite. Something must have gone wrong in the above calculation but I couldn't find out. Any help is appreciated.