# Evaluating and integral including unit step function

1. Feb 17, 2014

### user3

how can this integral be calculated:

∫[e^(−2mx) θ^2(x)+2θ(x)θ(−x)+e^(−2mx)θ^2(−x)]dx from -∞ to ∞

where θ(x) is the unit step function with its amplitude 0 everywhere before x=0 and θ(−x) is the unit step function with its amplitude 0 everywhere after x=0

In Introduction to Quantum Mechanics by Griffiths there is this problem: ψ(x)=e^(−m|x|) (there are other constants in the ψ that I think are not necessary to write). we need to find ∫ψ d^2ψ/dx^2 from -∞ to ∞

In the book, what the author does is first get

dψ/dx = −me^(−mx) if x >0 and me^(mx) if x <0 =−mθ(x)e^(−mx)+mθ(−x)e^(mx)

then he gets
d^2ψ/dx^2=−mδ(x)e^(−mx)+m^2 θ(x)e^(−mx) −mδ(−x)e^(mx)+m^2 θ(−x)emx=m∗(−2δ+me−m|x|)

where δ(x) is the dirac delta function (0 everywhere and infinite at x=0)

the integral we wanted is easy to get from there. But I wanted to try a new approach: get the second derivative in the piecewise form, then transform it to unit step functions and evaluate the integral.

I thought that if the author's method worked, then the other method must also work because it's logical. Is there something wrong with my reasoning ?

here's a cleaner statement of the question: http://math.stackexchange.com/quest...on-squared?noredirect=1#comment1427483_679623

2. Feb 17, 2014

### strangerep

It might seem logical, but I suspect that's because you don't understand distribution theory.

You've already been given a partial answer on stackexchange. The term $\theta(x)\theta(-x)$ complicates matters more than you might think -- because of the lurking discontinuity at 0.

Stick with the original method, where the derivative is implicitly understood as a weak derivative.