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

[niks]

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

[HallsofIvy]

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

[Big-T]

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]

His point, about the limits of integration, was that it is well known that
[tex]\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.

[Big-T]

Was that adressed to me?

Anyway, Maple tells me that:

\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.

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

[nicksauce]

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.

[gamesguru]

His point, about the limits of integration, was that it is well known that
[tex]\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}.

[niks]

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:


\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

[thebetapirate]

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.

[Big-T]

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

[TheoMcCloskey]

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}.

[ice109]

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.