Derivative of an integral containing a Dirac delta

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  • #1
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If I had a function g(x) defined by

[tex]g(x) = \int_{-\infty}^{\infty} f(x) \delta(x) dx[/tex]

where [tex]\delta(x)[/tex] is the dirac delta function, what would dg(x)/dx be? The fundamental theorem of calculus requires that [tex]f(x) \delta(x)[/tex] 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.
 

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  • #2
HallsofIvy
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[tex]g(x) = \int_{-\infty}^{\infty} f(x) \delta(x) dx[/tex]
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
[tex]g(x)= \int_{-\infty}^x f(t)\delta(t)dt[/tex]
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 [itex]\delta(x)[/itex] really is, that is differentiable at 0: the derivative of g(x) is [itex]\delta(x)f(x)[/itex].)
 
  • #3
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Hmm...

The Heaviside function,

[tex] H(x) = \int_{-\infty}^x \delta(t) \: dt [/tex]

shows what you mean, but how is it that many people define [tex] H(0) = 1/2 [/tex]? I think I'm missing the point...
 
  • #4
lurflurf
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roast said:
Hmm...

The Heaviside function,

[tex] H(x) = \int_{-\infty}^x \delta(t) \: dt [/tex]

shows what you mean, but how is it that many people define [tex] H(0) = 1/2 [/tex]? 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
 
  • #5
HallsofIvy
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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.
 

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