Dirac delta function and Heaviside step function

  1. [SOLVED] Dirac delta function and Heaviside step function

    In Levine's Quantum Chemistry textbook the Heaviside step function is defined as:

    [tex]H(x-a)=1,x>a[/tex]
    [tex]H(x-a)=0,x<a[/tex]
    [tex]H(x-a)=\frac{1}{2},x=a[/tex]​

    Dirac delta function is:

    [tex]\delta (x-a)=dH(x-a) / dx[/tex]​

    Now, the integral:

    [tex]\int ^{\infty}_{-\infty}f(x)\delta (x-a)dx[/tex]​

    Is evaluated using integration by parts considering

    [tex]u=f(x), du=f'(x)[/tex]
    [tex]dv=\delta (x-a)dx, v=H(x-a)[/tex]​

    We have then:
    [tex]\int ^{\infty}_{-\infty}f(x)\delta (x-a)dx=f(x)H(x-a)|^{\infty}_{-\infty}-\int ^{\infty}_{-\infty}H(x-a)f'(x)dx[/tex]

    [tex]\int ^{\infty}_{-\infty}f(x)\delta (x-a)dx=f(\infty)-\int ^{\infty}_{-\infty}H(x-a)f'(x)dx[/tex]​

    Since [tex]H(x-a)[/tex] vanishes for [tex]x<a[/tex], the integral becomes:

    [tex]\int ^{\infty}_{-\infty}f(x)\delta (x-a)dx=f(\infty)-\int ^{\infty}_{a}H(x-a)f'(x)dx=f(\infty)-\int ^{\infty}_{a}f'(x)dx[/tex]​

    This is the point where my question arrives. [tex]H(x-a)[/tex] is considered to have a value of unity for all the integral and that's why it is pulled out of the integral as a constant, however the lower bound of the integral is [tex]a[/tex] and in this point [tex]H(x-a)=1/2[/tex]. Could you please tell me if the following explanation is correct?

    I think that because in all the integral, except in [tex]a[/tex], [tex]H(x-a)=1[/tex] and since the upper bound is infinity the value of the integral at the point [tex]a[/tex] can be ignored.

    If I'm wrong, any suggestion for correcting my explanation will be appreciated.
     
  2. jcsd
  3. Dick

    Dick 25,887
    Science Advisor
    Homework Helper

    You can always ignore the value of a function at a single point when you are integrating. If you remember how Riemann sums work, the contribution of the single point can be put into a rectangle of height 1/2 and width 0. So it makes no contribution.
     
  4. Ok, thank you very much. That answers my question.
     
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