Is this integral solvable using integration by parts?

[PICsmith]

This should be a fairly simple integral but I can't get it for some reason. Here's the problem:
A\int_{-\infty}^{\infty} x e^{-\lambda(x-a)^2}dx
Now I know that
\int_{-\infty}^{\infty} e^{-x^2}dx=\sqrt{\pi}
only for those limits.
Okay so I do parts,
u=x
du=dx
dv=e^{-\lambda(x-a)^2}dx
v=?
When you evaluate the integral from dv to get v, you substitue say
s=\sqrt{\lambda}(x-a)
to make it like the second integral i put down, but you can't evaluate it between the limits of -infinity to infinity when doing parts right? And the indefinite integral of this form is not solvable as far as I know.

BTW, This is for my QM class, finding the average/expectation value of x,
<x>

Am I even going about this the right way? I don't know anymore. Please tell me where I screwed up and point me in the right direction.

 

[Curious3141]

Here's a hint :

\frac{d}{dx}e^{-x^2} = -2xe^{-x^2}

Use a substitution to get the integrand to a form like the above. You'll still have a term that needs the error function, but you already know how to handle that.

 

[PICsmith]

Awesome, I got it!
Much thanks Curious :D

 

[Curious3141]

No problem, glad to help.

 

[PICsmith]

doh!

I'm stuck again

Okay, so I found <x> and now I have to find <x^2>, which gives,
A\int_{-\infty}^{\infty} x^2 e^{-\lambda(x-a)^2}dx
I made the same substitution that I made last time, namely,
u=\sqrt{\lambda}(x-a)
which gives me a error function term that just equals \sqrt{\pi}, a term that goes to zero and another term that is,
\int_{-\infty}^{\infty}u^2e^{-u^2}du
and I'm having trouble solving this now.

Is there another substitution that I should have made initially? I couldn't find one. I swear this is the last time I'll need help . Any help is greatly appreciated.

 

[learningphysics]

\int_{-\infty}^{\infty}u^2e^{-u^2}du
and I'm having trouble solving this now.

Is there another substitution that I should have made initially? I couldn't find one. I swear this is the last time I'll need help . Any help is greatly appreciated.

You can solve the above integral using integration by parts.

 

[FulhamFan3]

for an equation of form:

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

do integration by parts with

u=x
dv=xe^{-x^2}dx

 

[PICsmith]

Geez how could I not see that? Thanks for spending your 300th post helping me out learningphysics! And thanks to you too FulhamFan3.

 

[dextercioby]

I hope your first integral was zero... As u had to integrate the odd function on a symmetric interval wrt to the origin...

Daniel.


P.S.The Cauchy principal value...

 

[PICsmith]

I hope your first integral was zero... As u had to integrate the odd function on a symmetric interval wrt to the origin...

I forget to mention that lambda and a are positive real constants (and A = sqrt(lambda/pi)), so the function is symmetric about x=a not the origin, and the first integral turns out to be just <x> = a, and the second one <x^2> = a^2 + 1/(2*lambda) (for anyone who's interested).

 

[dextercioby]

Yes,you're right,it's not symmetric wrt 0...I should have looked better...

Good thing you finally pulled them through.

Daniel.