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Integral of x^{2} e^{x^2} dxby niks
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#1
May208, 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 


#2
May208, 06:19 AM

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PF Gold
P: 39,363

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] 


#3
May208, 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?



#4
May208, 09:40 AM

Math
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PF Gold
P: 39,363

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 antiderivative 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. 


#5
May208, 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 minussign in your integrand. 


#6
May208, 10:47 AM

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P: 1,275

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. 


#7
May208, 09:13 PM

P: 85




#8
May208, 09:18 PM

P: 2

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


#9
May208, 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(2k3\right)}{2^{n}x^{2n}}\right][/tex] I would go over the derivation but LaTex is killing me. 


#10
May308, 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.



#11
May308, 07:02 AM

P: 182

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]. 


#12
May308, 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] 


#13
May308, 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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