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Dirac delta function and Heaviside step function

  • Thread starter pedroobv
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[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.
 

Answers and Replies

  • #2
Dick
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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.
 
  • #3
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Ok, thank you very much. That answers my question.
 

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