Solve Complex Integral Homework

[IHateMayonnaise]

Homework Statement

I need to solve:

\int_{-\infty}^{\infty}xe^{(a-x)^2}dx

Homework Equations

The Attempt at a Solution

My first intuition would be to rewrite this as:

\oint_cze^{(a-z)^2}dz

and then use Cauchy's Residue theorem to calculate the integral. There is one singularity at x_o=0 when x->\infty. To calculate the residue,Res(z_o)=(z-z_o)f(z) |_{z=z_o}

where in this case

f(z)=ze^{(a-x)^2}

So, we have

Res(0)=(z-0)ze^{(a-z)^2}|_{z=0}

=0

which is clearly not right (mathematica gives a\sqrt{\pi}. What am I doing wrong? Any hints? Thanks!

EDIT: if you take the derivative of the residue twice and then taking the limit you get 2e^{-a^2}, and multiplying this by 2\pi i still doesn't give the answer!

 

[hunt_mat]

No, the solution is much simplier. Make the substitution u=a-x, then note that:
<br /> \frac{d}{dx}\left(\frac{1}{2}e^{x^{2}}\right) =xe^{x^{2}}<br />

 

[hunt_mat]

Just thought, did you mean to write:
<br /> \int_{-\infty}^{\infty}xe^{-(a-x)^{2}}<br />
as the integral you wrote down was unbounded

 

[IHateMayonnaise]

Just thought, did you mean to write:
<br /> \int_{-\infty}^{\infty}xe^{-(a-x)^{2}}<br />
as the integral you wrote down was unbounded

Yes isn't that what I had? I am supposed to integrate it over the complex plane, this is what the problem asks and I won't get credit if I do it any other way.

 

[hunt_mat]

Right, what are the poles? What contour are you going to integrate it over?

 

[IHateMayonnaise]

Right, what are the poles? What contour are you going to integrate it over?

There's only one pole right? At x-&gt;\infty?

Typically I would integrate over the upper hemisphere.

 

[hunt_mat]

The problem for the residues is that the integrand is well defined for all values of x (and hence z), So I think that you're going to have to do a Laurent series expansion about z=infinty and work the residues from there.

 

[IHateMayonnaise]

The problem for the residues is that the integrand is well defined for all values of x (and hence z), So I think that you're going to have to do a Laurent series expansion about z=infinty and work the residues from there.

I am not seeing how to do this.. my math methods books don't seem to be of much help. Can you recommend some literature specific to this?

thanks for your help by the way!

 

[hunt_mat]

Hmm, The problem here is the integral screams not to be done via complesx analysis.

 

[hunt_mat]

Try looking at the Laurent expansion about infinity, I.e. expand in a series in z^{-1} and find the coefficent of z^{-1} by the usual method and that should be your residue.

 

[Dickfore]

There's only one pole right? At x-&gt;\infty?

Wrong.