What is the Integral of x^{2} e^{-x^2} dx in Signal Processing?

[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]<br /> 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.[/tex]

 

[Big-T]

Was that adressed to me?

Anyway, Maple tells me that:
$$\int_{-\infty}^{\infty} x^{n} e^{-x^{2}}\rm{d} x=\frac{1}{2}\Gamma\left(\frac{n}{2}+\frac{1}{2}\right)\left(1+(-1)^n\right)$$ ,
which should be possible to prove by induction.

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

 

[nicksauce]

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]<br /> 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.[/tex]

[tex] For one, it's $e^{-x^2}$ and for two, it's $\sqrt{\pi}$.[/tex]

 

[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:
$$<br /> \int_{-\infty}^{\infty} x^{n} e^{-x^{2}}\rm{d} x=\frac{1}{2}\Gamma\left(\frac{n}{2}+\frac{1}{2}\r ight)\left(1+(-1)^n\right) <br />$$
which should be possible to prove by induction.

Thanks! That will help me move forward.

Thanks to everyone who replied, I learned 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.