Physics Forums (http://www.physicsforums.com/index.php)
-   -   Dirac delta function and Heaviside step function (http://www.physicsforums.com/showthread.php?t=234774)

 pedroobv May12-08 04:46 PM

[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 May12-08 05:04 PM

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 May12-08 05:16 PM

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

 All times are GMT -5. The time now is 12:53 PM.