Dirac delta function and Heaviside step function

[pedroobv]

[SOLVED] Dirac delta function and Heaviside step function

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

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

Dirac delta function is:

\delta (x-a)=dH(x-a) / dx​

Now, the integral:

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

Is evaluated using integration by parts considering

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

We have then:

\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

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

Since H(x-a) vanishes for x<a, the integral becomes:

\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​

This is the point where my question arrives. H(x-a) 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 a and in this point H(x-a)=1/2. Could you please tell me if the following explanation is correct?

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

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

 

[Dick]

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.

 

[pedroobv]

Ok, thank you very much. That answers my question.