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


by niks
Tags: integral
niks
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#1
May2-08, 05:46 AM
P: 2
I ran into an integral while working on response of a signal processing filter, it looks like:

[tex]\int_{-\infty}^{\infty} x^{2} e^{-x^{2}} dx [/tex]

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

Can anyone suggest an approach for this?

Thanks,
Niks
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HallsofIvy
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#2
May2-08, 06:19 AM
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PF Gold
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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
[tex]\int x^2e^{-x^2} dx[/tex]
You might be able to do that in terms of the "error function", Erf(x), which is defined to be
[tex]\int e^{-x^2} dx[/tex]
Big-T
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#3
May2-08, 07:30 AM
P: 64
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
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#4
May2-08, 09:40 AM
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PF Gold
P: 38,898

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


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
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#5
May2-08, 10:43 AM
P: 64
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)
[/tex] ,
which should be possible to prove by induction.

PS. HallsofIvy: You have forgotten the minus-sign in your integrand.
nicksauce
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#6
May2-08, 10:47 AM
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P: 1,276
A cute way to solve this is to recall that

[tex]\int_{-\infty}^{\infty}e^{-ax^2}=\frac{\sqrt{\pi}}{\sqrt{a}}[/tex]

Then use Feynman's favorite trick and differentiate both sides with respect to a, and evaluate at a = 1.
gamesguru
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#7
May2-08, 09:13 PM
P: 85
Quote Quote by HallsofIvy View Post
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 [itex]e^{-x^2}[/itex] and for two, it's [itex] \sqrt{\pi}[/itex].
niks
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#8
May2-08, 09:18 PM
P: 2
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)

[/tex]
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
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#9
May2-08, 09:51 PM
P: 9
Using lots of substitutions and integration by parts I get this:

[tex]\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][/tex]

I would go over the derivation but LaTex is killing me.
Big-T
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#10
May3-08, 03:10 AM
P: 64
Quite frankly, I think using Leibniz' rule, as suggested by Nicksauce, would by far be the simplest method in this case.
TheoMcCloskey
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#11
May3-08, 07:02 AM
P: 181
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

[tex]u=x[/tex]

[tex]dv= x*e^{-x^2}dx[/tex]

and then things won't be so messy - however, it will involve the definite integral [itex]\int^{\infty}_{-\infty}{ e^{-{x^2}}dx}[/itex] which we know equals [itex]\sqrt{\pi}[/itex].
ice109
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#12
May3-08, 04:48 PM
P: 1,705
umm mathematica gives me [tex]\frac{\sqrt{\pi}}{2}[/tex]

and for the indefinite :

[tex]
\frac{1}{4} \sqrt{\pi } \text{erf}(x)-\frac{1}{2} e^{-x^2} x
[/tex]
quadraphonics
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#13
May3-08, 05:19 PM
P: 270
Well, perhaps the very simplest approach is to recognize that the integral is [itex]\sqrt{\pi}[/itex] times the variance of a Gaussian random variable with mean 0 and standard deviation [itex]\frac{1}{\sqrt{2}}[/itex]. That's certainly all I'd bother doing in the signal processing context the OP mentioned.


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