3
$\begingroup$

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.

$\endgroup$
2

1 Answer 1

2
$\begingroup$

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 $$\int_{0}^{\infty }t^{\frac{3}{2}}e^{-t}\, \mathrm{d}t=\Gamma \left ( \frac{5}{2} \right )=\frac{3}{2}\Gamma \left ( \frac{3}{2} \right )=\frac{3}{2}\cdot \frac{1}{2}\Gamma \left ( \frac{1}{2} \right )=\frac{3\sqrt{\pi }}{4}$$

$\endgroup$
5
  • 1
    $\begingroup$ 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 $\int_0^\infty e^{-x^2} dx = \sqrt{\pi}/2$. $\endgroup$
    – Zubzub
    Commented Jan 18, 2017 at 11:01
  • 1
    $\begingroup$ @Zubzub Faut le savoir ! $\endgroup$ Commented Jan 18, 2017 at 12:04
  • $\begingroup$ How did you leech such an amount of reputation out of this standard integrals? i'm impressed $\endgroup$
    – tired
    Commented Jan 18, 2017 at 16:06
  • $\begingroup$ @tired You know bro, there always be someone who don't know or unfamiliar with gamma function. $\endgroup$ Commented Jan 19, 2017 at 3:32
  • $\begingroup$ as it turned out, my question wasn't to dump $\endgroup$
    – tired
    Commented Jan 26, 2017 at 18:10

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .