Show that the standard integral: $\int_{0}^{\infty} x^4\mathrm{e}^{-\alpha x^2}\mathrm dx =\frac{3}{8}{\left(\frac{\pi}{\alpha^5}\right)}^\frac{1}{2}$

Question

In my physics course this standard formula is used a lot without proof so it would be interesting to see a neat proof for it. From a previous thread by me I know the proof for

$\int x\mathrm{e}^{-\alpha x^2}\mathrm dx =\dfrac{-1}{2\alpha} \mathrm e^{-\alpha x^2}$ + Constant

which was kindly shown to me by a community member by means of the substitution $-\alpha x^2=u \Rightarrow -2\alpha x \mathrm dx=\mathrm du$, then $x\mathrm dx=\frac {-1}{2\alpha}\mathrm du$ such that

$\displaystyle\int xe^{-\alpha x^2}\,dx=\frac {-1}{2\alpha}\displaystyle\int e^u \mathrm du= \frac {-1}{2\alpha}\displaystyle e^{-\alpha x^2}+$C. Where C , $\alpha$$=$constants

The previous thread was simply a paradigm to set the scene for this question. I tried a similar approach to prove the formula in question by letting $u=-\alpha x^2$, but this does not get me anywhere, I have also tried doing it by parts but still without success.

Is there a simple way to prove $\int_{0}^{\infty} x^4\mathrm{e}^{-\alpha x^2}\mathrm dx =\dfrac{3}{8}{\left(\dfrac{\pi}{\alpha^5}\right)}^\frac{1}{2}$?

Note: This question is not strictly a duplicate of the thread asking to prove that $\int_{-\infty}^{\infty} x^4\mathrm{e}^{-\alpha x^2}\mathrm dx =\dfrac{3}{4}{\left(\dfrac{\pi}{\alpha^5}\right)}^\frac{1}{2}$ as the limits are different in that post. But I acknowledge that it is almost the same thing.

Any proof will do.

Thank you.

Kindest Regards,

Blaze.

Answer 1

One way is to use 'Feynman's trick', and differentiate under the integral. Note $$\int_0^\infty x^4e^{-\alpha x^2}=\int_0^\infty \frac{\partial^2}{\partial \alpha^2}e^{-\alpha x^2}=\frac{\partial^2}{\partial \alpha^2}\int_0^\infty e^{-\alpha x^2}$$

Calculate the inner definite integral first, then differentiate it twice with respect to $\alpha$.

Answer 2

Get rid of the $x^4$ by integrating by parts twice. Then look for one of the many ways (some probably explained on this site) to evaluate the integral over a Gaussian. My favorite is squaring it and transforming the resulting double integral to polar coordinates – which, by the way, you can also do directly with your integral without first integrating by parts, if you prefer.

Answer 3

Here, you'll want to integrate by part to find the Gauss integral :

$$I = \int_0^{\infty} x^4 e^{-\alpha x^2} dx = \int_0^{\infty} x^3 \times x e^{-\alpha x^2} dx$$

Now, you let $u = x^3$ and $v' = x e^{-\alpha x^2} $, and this gives you

$$I = \left [ -x^3 \frac{e^{-\alpha x^2}}{2\alpha} \right]_0^{\infty}+\int_0^{\infty} \frac{3}{2\alpha} x^2 e^{-\alpha x^2}dx = \frac{3}{2\alpha}\int_0^{\infty} x^2 e^{-\alpha x^2}dx $$

Now, you let $u = x$ and $v' = x e^{-\alpha x^2} $, and this gives you

$$I = - \frac{3}{2\alpha}\left[ x \frac{e^{-\alpha x^2}}{2\alpha} \right]_0^{\infty} + \frac{3}{2\alpha} \int_0^{\infty} \frac{e^{-\alpha x^2}}{2\alpha} dx$$

$$= \frac{3}{4 \alpha^2} \int_0^{\infty}e^{-\alpha x^2}$$

Now you do the change of variable $t = \sqrt\alpha\times x$,

$$I= \frac{3}{4 \alpha^2} \frac{1}{\sqrt{\alpha}} \int_0^{\infty}e^{-t^2} dt$$

And this is equal to

$$I = \frac{3\sqrt{\pi}}{8 \alpha^\frac{5}{2}}$$

Answer 4

Another way to look at it (that probably isn't simpler) : Let $u=x^2$, then the integral becomes:

$$\frac{1}{2} \int_0^\infty u^{3/2}e^{-\alpha u}du$$ which is the Laplace transform of $u^{3/2}$, times ${1}/{2}$. We know that the Laplace transform of $u^n$ is $n!*\alpha^{-n-1}$ or generally $\Gamma(n+1)*\alpha^{-n-1}$. Here, $n=3/2$ so $$\Gamma(3/2 +1)=\Gamma(2+1/2)=3\sqrt{\pi}/4$$ $$\alpha^{-n-1}=\frac{1}{\alpha^{5/2}}$$ Finally: $$\frac{1}{2} \int_0^\infty u^{3/2}e^{-\alpha u}du=\frac{3}{8}\sqrt{\frac{\pi}{\alpha^5}}$$