Integral of Dirac times Heaviside

[wil3]

I am trying to solve the integral

[itex]\int_{-\infty}^\infty H(x) \delta(x) dx[/itex]

Where H(x) is a unit step and d(x) is a standard Dirac delta. Mathematica chokes on this, but I'm pretty sure that the value is

[itex]\int_{-\infty}^\infty H(x) \delta(x) dx = \dfrac12 \left(H(0^+) + H(0^-) \right) = 1/2[/itex]

However, I am having trouble proving that my intuition is correct. Is this claim correct, and, if so, how can I show it?

Thank you in advance.

 

[pwsnafu]

The intuitive way would be...in order for ##\int_{-\infty}^\infty f(x) \delta(x) \, dx## to be well defined, ##f(x)## needs to have the domain to be all of the reals. That means ##H(0)## needs to be defined. A lot of books choose ##H(0) = 1/2## as a convention, but it is just a convention. You could theoretically choose any real number, and indeed the notation ##H_c (x)## is the Heaviside step function with ##H(0) = c##.

Anyway, ##\int_{-\infty}^\infty H(x) \delta(x) dx = H(0)##, so the question is what have you chosen ##H(0)## to be?

Disclaimer: it's worth pointing out that in the strict definition of Dirac delta, H(x) is not a valid test function. Mathematica "chokes" because that is the correct response. What you are asking for is an extension of Dirac to include discontinuous functions. There is no "correct" way to do this. You can use what you wrote as a new definition of Dirac, and then demonstrate that it is consistent with the standard definition. I personally wouldn't do it that way because the value is now dependent on the neighborhood of 0, rather than 0 itself.

 

[Office_Shredder]

One standard way that the Dirac delta function is extended is to define
[tex] \int_{-\infty}^{\infty} \delta(x) h(x) dx = \lim_{a\to 0} \frac{1}{\sqrt{2\pi} a} \int_{-\infty}^{\infty} e^{-x^2/(2a)} h(x) dx [/tex]
for h(x) any distribution. This satisfies the definition for [itex] \delta(x)[/itex] when h(x) is continuous and integrable (the Gaussian function defined has a peak at 0 that goes to infinity as a goes to zero, so this is not too hard to prove). and for h(x) the Heaviside function you will in fact get 1/2.

 

[wil3]

These were both great answers, I agree that using this trick is sketchy in a proof but it seems to work for my application. Thank you!

 

[davidmoore63@y]

Another method

Doesn't this integrate directly to 1/2 H^2(x) between -∞ and ∞, which equals 1/2?

This observation seems to have no dependence on how H(0) is defined, other than what is implied from the statement that d/dx of H(x) is δ(x).

 

[pwsnafu]

Doesn't this integrate directly to 1/2 H^2(x) between -∞ and ∞, which equals 1/2?

No, because Dirac delta is not a real function but a distribution. That integral that is written is not actually an integral. We just write the evaluation in that notation.

This observation seems to have no dependence on how H(0) is defined, other than what is implied from the statement that d/dx of H(x) is δ(x).

Dirac delta is the distributional derivative of Heaviside. Integration by substitution requires continuous differentiability, which Heaviside does not have.

 

[davidmoore63@y]

interesting thanks