How Do You Calculate First Order Correction in a Perturbed Infinite Square Well?

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Homework Statement


I have the particle in the infinite square well and need to calculate the first order correction energy and the wave function. L is the width and the potential is:
1/2 mw2x2 in the -L/2 < x < L/2
and infinity in x <= -L/2 and x>=L/2

Homework Equations


H'=H-H0[/B]

The Attempt at a Solution


I have stated that the perturbed Hamiltonian is equal to 1/2 mw2x2.
I am confused by the integral itself; I am not sure which boundaries to use. I am assuming that since the potential is infinite in L/2 and -L/2 I am not supposed to use those boundary conditions?
 
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if the potential is infinite, then the wave function is zero
 
Yes, but which boundaries to use then when integrating? Is the perturbed Hamiltonian correct?
 
only integrate where the wave function is non-zero
 
That is what I'm saying. I don't know how to put boundaries since I can't use -L/2 and L/2. In all the other examples I've had, the <= and >= signs were where the potential isn't infinite, as it is in the general infinite square well problem.
 
the original wave function is valid between \pm L/2 so those are the limits of integration.
 
So, I approach this problem the same, regardless this < sign? It confuses me because it says than when x is + or -L/2, the potential is infinite, hence wave function is zero. How is it right to put those boundaries once the integral is solved?
 
it is a perturbation, it is supposed to be small, if the width of the well is not too large the perturbation at the bottom will only slightly affect the energies and wave functions.
 
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