$\int_{0}^{\infty}0.5 (0.5x)^{d-1} e^{-0.25x^{2}}dx=0.5\Gamma(0.5d)$

Question

I wanted to solve this:$\int_{0}^{\infty}(e^{-r^2}r^{d-1}) dr$

Set $x=2r$ and I got

$\int_{0}^{\infty}0.5 (0.5x)^{d-1} e^{-0.25x^{2}}dx$

According to Wolfram Alpha, this is equal to $0.5\Gamma(0.5d)$.

But I have no idea how to prove that. What should I do?

Answer 1

Integrate by substituting $u = r^2$: \begin{align*} \int_{0}^{\infty}(e^{-r^2}r^{d-1})\,dr &= \int_{0}^{\infty}(e^{-u}u^{(d-1)/2})\cdot \frac{1}{2\sqrt{u}} du\\ &= \frac{1}{2}\int_{0}^{\infty}(e^{-u}u^{d/2 - 1}) du\\ &= \frac{1}{2}\Gamma\left( \frac{d}{2}\right) \end{align*} leaving us with the result from Wolfram Alpha.