# Computing the integral $\int_0 ^{\infty} t^{3/2}e^{-t}dt$

I am trying to understand the equality $$\displaystyle\int_0 ^{\infty} t^{3/2}e^{-t}\mathrm dt = {\sqrt{\pi}\over2}$$

I tried to use integration by parts but it doesn't seem to work.

Your answer was wrong. The gamma function has the following definition. $$\Gamma \left ( x \right )=\int_{0}^{\infty }t^{x-1}e^{-t}\, \mathrm{d}t~~,~~\Re x>0$$ then use $$\Gamma \left ( 1+x \right )=x\Gamma \left ( x \right )$$ we have $$\int_{0}^{\infty }t^{\frac{3}{2}}e^{-t}\, \mathrm{d}t=\Gamma \left ( \frac{5}{2} \right )=\frac{3}{2}\Gamma \left ( \frac{3}{2} \right )=\frac{3}{2}\cdot \frac{1}{2}\Gamma \left ( \frac{1}{2} \right )=\frac{3\sqrt{\pi }}{4}$$
• And now it all boils down to computing $\Gamma(1/2)$ which can be done : $$\Gamma(1/2) = \int_0^\infty e^{-t} t^{-1/2} dt = \int_0^\infty e^{-x^2} \frac{1}{x}\ 2x\ dx = \sqrt{\pi}$$ where we did a change of variable $t = x^2$ and there's plenty of proof of the gaussian integral $\int_0^\infty e^{-x^2} dx = \sqrt{\pi}/2$. Commented Jan 18, 2017 at 11:01