Derivative of an integral containing a Dirac delta

[roast]

If I had a function g(x) defined by

$$g(x) = \int_{-\infty}^{\infty} f(x) \delta(x) dx$$

where $$\delta(x)$$ is the dirac delta function, what would dg(x)/dx be? The fundamental theorem of calculus requires that $$f(x) \delta(x)$$ needs to be a continuous and differentiable function before I can immediately say that dg(x)/dx = f(x) \delta(x), which is clearly not the case.

 

[HallsofIvy]

$$g(x) = \int_{-\infty}^{\infty} f(x) \delta(x) dx$$
is not a function of x! It's derivative is 0. In fact that's true of any definite integral of any integrable function! Perhaps a little more interesting would be:
What is the derivative of
$$g(x)= \int_{-\infty}^x f(t)\delta(t)dt$$
but not a whole lot more! If x< 0, g(x)= 0 and the derivative is 0. If x> 0, g(x)= f(0) and the derivative is 0. However, g(x) is not differentiable at 0.

(In terms of 'distributions' or 'generalized functions', which is what $\delta(x)$ really is, that is differentiable at 0: the derivative of g(x) is $\delta(x)f(x)$.)

 

[roast]

Hmm...

The Heaviside function,

$$H(x) = \int_{-\infty}^x \delta(t) \: dt$$

shows what you mean, but how is it that many people define $$H(0) = 1/2$$? I think I'm missing the point...

 

[lurflurf]

Hmm...

The Heaviside function,

$$H(x) = \int_{-\infty}^x \delta(t) \: dt$$

shows what you mean, but how is it that many people define $$H(0) = 1/2$$? I think I'm missing the point...

That is because it is nice if
f(x-)=f(x)=f(x+)
but if
f(x+)!=f(x-)
the next best thing is if
f(x)=(f(x-)+f(x+))/2

 

[HallsofIvy]

The value H(0)= 1/2 is convenient but really irrelevant. H(0) is still distcontinuous at 0. Since the crucial point with distributions is their integral properties values at individual values of x are not important.