# Integral of x^{2} e^{-x^2} dx

I ran into an integral while working on response of a signal processing filter, it looks like:

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

While trying integration by parts u = $$x^{2}$$ we get du = 2xdx but can't proceed with dv = $$e^{-x^{2}}$$ because then
v = $$\int e^{-x^{2}}$$
can't be integrated unless we use the limits.

Can anyone suggest an approach for this?

Thanks,
Niks

• Bruno Cavalcante and Ranvir

## Answers and Replies

HallsofIvy
Science Advisor
Homework Helper
What exactly are you trying to do? As you point out, your v is not any elementary function, and that tells you that neither is
$$\int x^2e^{-x^2} dx$$
You might be able to do that in terms of the "error function", Erf(x), which is defined to be
$$\int e^{-x^2} dx$$

If you just want to calculate the definite integral, I don't see why you wouldn't want to include the limits when integrating by parts?

HallsofIvy
Science Advisor
Homework Helper
His point, about the limits of integration, was that it is well known that
$$\int_{-\infty}^{\infty} e^{x^2}dx= 2\sqrt{\pi}[/itex] while the anti-derivative is not an elementary function. in other words, he could, theoretically, do it as a definite integral with the right limits but not as an indefinite integral. Was that adressed to me? Anyway, Maple tells me that: [tex] \int_{-\infty}^{\infty} x^{n} e^{-x^{2}}\rm{d} x=\frac{1}{2}\Gamma\left(\frac{n}{2}+\frac{1}{2}\right)\left(1+(-1)^n\right)$$ ,
which should be possible to prove by induction.

PS. HallsofIvy: You have forgotten the minus-sign in your integrand.

nicksauce
Science Advisor
Homework Helper
A cute way to solve this is to recall that

$$\int_{-\infty}^{\infty}e^{-ax^2}=\frac{\sqrt{\pi}}{\sqrt{a}}$$

Then use Feynman's favorite trick and differentiate both sides with respect to a, and evaluate at a = 1.

• Curieuse and Ranvir
His point, about the limits of integration, was that it is well known that
$$\int_{-\infty}^{\infty} e^{x^2}dx= 2\sqrt{\pi}[/itex] while the anti-derivative is not an elementary function. in other words, he could, theoretically, do it as a definite integral with the right limits but not as an indefinite integral. For one, it's $e^{-x^2}$ and for two, it's $\sqrt{\pi}$. As you point out, your v is not any elementary function, I think I should use this as a guideline for future problems. Both u and v should be elementary functions otherwise integration by parts becomes too messy(perhaps impossible). while the anti-derivative is not an elementary function. in other words, he could, theoretically, do it as a definite integral with the right limits but not as an indefinite integral. Yes, that was what I had in mind. That's why I got stuck there. Anyway, Maple tells me that: [tex] \int_{-\infty}^{\infty} x^{n} e^{-x^{2}}\rm{d} x=\frac{1}{2}\Gamma\left(\frac{n}{2}+\frac{1}{2}\r ight)\left(1+(-1)^n\right)$$
which should be possible to prove by induction.
Thanks!! That will help me move forward.

Thanks to everyone who replied, I learnt a lot from this thread.

-Niks

Using lots of substitutions and integration by parts I get this:

$$\int x^{2}e^{-{x^2}}dx=xe^{x^{2}}\left[1-\sum_{n=1}^{\infty}\frac{\prod_{k=2}^{n}\left(2k-3\right)}{2^{n}x^{2n}}\right]$$

I would go over the derivation but LaTex is killing me.

Quite frankly, I think using Leibniz' rule, as suggested by Nicksauce, would by far be the simplest method in this case.

Quite frankly, I think using Leibniz' rule, as suggested by Nicksauce, would by far be the simplest method in this case.

Quite true.

But, if doing by parts, then the proper selection of u an dv is

$$u=x$$

$$dv= x*e^{-x^2}dx$$

and then things won't be so messy - however, it will involve the definite integral $\int^{\infty}_{-\infty}{ e^{-{x^2}}dx}$ which we know equals $\sqrt{\pi}$.

• Curieuse
umm mathematica gives me $$\frac{\sqrt{\pi}}{2}$$

and for the indefinite :

$$\frac{1}{4} \sqrt{\pi } \text{erf}(x)-\frac{1}{2} e^{-x^2} x$$

quadraphonics
Well, perhaps the very simplest approach is to recognize that the integral is $\sqrt{\pi}$ times the variance of a Gaussian random variable with mean 0 and standard deviation $\frac{1}{\sqrt{2}}$. That's certainly all I'd bother doing in the signal processing context the OP mentioned.