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Evaluating and integral including unit step function

  1. Feb 17, 2014 #1
    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. jcsd
  3. Feb 17, 2014 #2


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    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.
    See also distribution (the section on differentiation).

    If you stick to such distributional methods, at least you don't miss anything, even though it might mean you must deal with Dirac deltas.
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