in my calculus class we are currently dealing with improper integrals and I was tackled with the following:

$$ \int_{0}^{\infty} x^4 e^{-x^2}dx = $$


$$ \int_{0}^{\infty} x^5 e^{-x^2}dx = $$

I have no idea on either one as I am new to these sort of tricks so I am writing here so hopefully someone can show me how to work these. I thank all helpers.


3 Answers 3


Using two integration by part,

$$ \int_0^{+\infty} x^4 e^{-x^2}~\mathrm{d}x = \dfrac{-1}{2}\left[x^3e^{-x^2} \right]^{+\infty}_0 +\dfrac{3}{2}\int_0^{+\infty} x^2 e^{-x^2}~\mathrm{d}x$$


$$\int_0^{+\infty} x^2 e^{-x^2}~\mathrm{d}x= \dfrac{-1}{2}\left[xe^{-x^2} \right]^{+\infty}_0 +\dfrac{1}{2}\int_0^{+\infty} e^{-x^2}~\mathrm{d}x$$


$$\int_0^{+\infty} e^{-x^2}~\mathrm{d}x=\dfrac{\sqrt{\pi}}{2}$$

as a well-known integral.


The second integral is elementary.

One may write, by a change of variable and integrations by parts, $$ \begin{align} \int_{0}^{\infty} x^5 e^{-x^2}dx &=\frac12\int_{0}^{\infty} u^2 e^{-u}du \quad (u=x^2) \\\\&=\left.\frac12\:u^2(-e^{-u})\right|_0^\infty+\int_{0}^{\infty} u\: e^{-u}du \\\\&=0+\left.u(-e^{-u})\right|_0^\infty+\int_{0}^{\infty} e^{-u}du \\\\&=1. \end{align} $$ The first integral reduces to a classic gaussian integral, $$ \begin{align} \int_{0}^{\infty} x^4 e^{-x^2}dx&=\left.\frac12\:x^3(-e^{-x^2})\right|_0^\infty+\frac32\int_{0}^{\infty} x^2\: e^{-x^2}dx \\\\&=0+\left.\frac34\:x(-e^{-x^2})\right|_0^\infty+\frac34\int_{0}^{\infty} e^{-x^2}dx \\\\&=\frac{3 \sqrt{\pi }}{8}. \end{align} $$

  • $\begingroup$ @EricTowers Typo corrected, thanks. $\endgroup$ Mar 27, 2016 at 20:28

Taking into account the fact that you already received answers to your specific integrals (and that they nicely explain the method to be used in order to compute them), applying the same methods you could arrive to $$I_n=\int_{0}^{\infty} x^n e^{-x^2}dx=\frac{1}{2}\Gamma \left(\frac{n+1}{2}\right)$$ provided $\Re(n)>-1$ where appears the gamma function.

For even values of $n$ you have rational fractions; for odd values of $n$, the same multiplied by $\sqrt \pi$.


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