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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.

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    Search for the Gamma function. And you might want to try substitution with logarithm: $t = \log{x}$. https://en.wikipedia.org/wiki/Gamma_function – 2017-01-18
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    See also [this question](http://math.stackexchange.com/q/650546). Found [using Approach0](https://approach0.xyz/search/?q=%24%5Cint_0%20%5E%7B%5Cinfty%7D%20t%5E%7B3%2F2%7De%5E%7B-t%7Ddt%24&p=1). – 2017-01-30

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

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