Need help with an integral (for momentum space wave function)

[arenaninja]

Homework Statement

Given \Psi(x,0)=\frac{A}{x^{2}+a^{2}}, (-\inf<x<\inf)
a) determine A
c) find the momentum space wave function \Phi(p,0), and check that it is normalized

Homework Equations

At t=0, we can find the momentum space wave function by the formula \Phi(p,0)=\frac{1}{\sqrt(2\pi\hbar)}\int_{-\inf}^{inf} e^{-\frac{ipx}{\hbar}}\Psi(x,0)dx

The Attempt at a Solution

I found that A=\sqrt(\frac{2a^{3}}{\pi}).
To find the momentum wave function, I haev
\Phi(p,0)=\frac{A}{\sqrt(2\pi\hbar)}\int_{-inf}^{inf} \frac{e^{-\frac{ipx}{\hbar}}}{x^{2}+a^{2}}dx
but frankly I have never seen an integral of this kind and have no clue on how to proceed. I've looked at my old calculus book, some tables of integrals and wolframalpha but they're yielding nothing. Am I missing something?

 

[Amok]

Take a look at a Fourier transform table, because that's what your integral is.

 

[fzero]

Expand 1/(x^2+a^2) in partial fractions using the imaginary roots. In each of the 2 terms you get, make a linear change of variables and then integrate directly.

 

[dextercioby]

Typically residues theorem application. This one is a case in which studying complex analysis before quantum mechanics really pays off. Or, as others suggested, you look it up in a table of Fourier transformations, eg on Wikipedia.

 

[arenaninja]

I searched some Fourier transform tables on the web (even saw the one in wkipedia) but frankly I couldn't discern the one that applied to this case (in fact I thought most of them do not look very similar).

fzero, thanks for the suggestion. I think I see where I could go with this. So the setup is:
\frac{A}{x+ia}+\frac{B}{x-ia}=1
from which I get
x(A+B)=0 and this implies A=-B
and
A(-ia)+(-A)ia=1
so I worked out that
A=\frac{-1}{2ia}
A weird number for sure (and complex to boot), but I suppose it's be expected in this situation.
So now I have
A \int_{-inf}^{inf} \frac{e^{-ipx/h}}{x+ia}dx - A \int_{-inf}^{inf} \frac{e^{-ipx/h}}{x-ia}dx
but I don't see how a change of variables will convert these two to functions that I can integrate.

Anything else I could add to my arsenal?

 

[fzero]

Try x = y \pm ia. The idea is to get an integral of the form

\int_{-\infty}^\infty \frac{e^{-i\alpha y}}{y} dy.

This is an integral that you can look up in a table of Fourier transforms. For instance, it's 309 (up to conventions) on http://en.wikipedia.org/wiki/Fourier_transform#Distributions

 

[Amok]

I searched some Fourier transform tables on the web (even saw the one in wkipedia) but frankly I couldn't discern the one that applied to this case (in fact I thought most of them do not look very similar).

fzero, thanks for the suggestion. I think I see where I could go with this. So the setup is:
\frac{A}{x+ia}+\frac{B}{x-ia}=1
from which I get
x(A+B)=0 and this implies A=-B
and
A(-ia)+(-A)ia=1
so I worked out that
A=\frac{-1}{2ia}
A weird number for sure (and complex to boot), but I suppose it's be expected in this situation.
So now I have
A \int_{-inf}^{inf} \frac{e^{-ipx/h}}{x+ia}dx - A \int_{-inf}^{inf} \frac{e^{-ipx/h}}{x-ia}dx
but I don't see how a change of variables will convert these two to functions that I can integrate.

Anything else I could add to my arsenal?

A while after answering your question I looked up this integral in a Fourier Transform table and I suspect this might involve having to use distributions, or am I wrong (anybody)?

Typically residues theorem application. This one is a case in which studying complex analysis before quantum mechanics really pays off. Or, as others suggested, you look it up in a table of Fourier transformations, eg on Wikipedia.

How would you compute such an integral using the residue theorem?

 

[vela]

How would you compute such an integral using the residue theorem?

Replace x by z and close the contour in the upper half-plane or lower half-plane depending on the sign of k. Show that the only part of the contour which contributes to the integral is the portion along the real axis. The integrand has two simple poles, so it's a straightforward calculation to find the residue at each pole.