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As far as I can tell the Dirac delta function in an integral picks the value of the multiplying function at the peak provided the peak is within the boundary, i.e. $$\int^{a+e}_{a-e} \delta (x-a) f(x)\ \mathrm{d}x = f(a).$$ But what happens if a peak is a boundary? $$\int^{a}_{a-e} \delta (x-a) f(x)\ \mathrm{d}x$$ My attempt is $$\lim_{u\to 0}\int^{a+u}_{a-e} \delta (x-a) f(x)\ \mathrm{d}x = \lim_{u\to 0}f(a)$$ but while this process is reasonably rigorous for a function with a primitive, it's not necessarily true for this.

ForgotALot
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AAM
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    The standard procedure is to take $\int_a^b \ \delta(x-a)f(x)=\color{red}{\frac{1}{2}} f(a)$, but there is no reason to expect that this will always lead to the "correct result". – AccidentalFourierTransform Apr 13 '16 at 07:05
  • @AccidentalFourierTransform indeed, I routinely work with situations where the correct result is $\int_a^b \delta(x-a)f(x)dx = f(a)$. I suppose this means an ideal answer would address the different possible conventions and any factors that might affect which one a person should choose. – David Z Apr 13 '16 at 07:34
  • Thanks, but how would I then know the correct result to use? – AAM Apr 13 '16 at 07:46
  • @AccidentalFouriertransform I would also stick at definition/property $\int_a^b \delta(x-a)f(x)dx = f(a)$ not $(1/2)f(a)$. It is the convention taken when one works with $\delta$ distribution with Laplace transform. – Jean Marie Apr 13 '16 at 09:32
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    @jeanmarie it really depends on context. In physics we usually think of $\delta$ as the "limit" of nascent deltas, which are usually even functions. In these cases, the factor 1/2 makes sense (take for example $f=1$, in which case the integral is just half the area of $\delta$, that is, 1/2). But its true that in some other contexts the factor 1/2 doesnt make much sense. – AccidentalFourierTransform Apr 13 '16 at 09:46
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    @AccidentalFourierTransform I agree – Jean Marie Apr 13 '16 at 09:49
  • it should never happen that you have $\int_0^1 \delta(x) dx$, what you wrote meant doesn't mean anything with the usual definition of distributions. – reuns Apr 13 '16 at 10:31
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    $\int_0^1 f(x) \delta(x) dx$, strictly speaking, should be understood as shorthand for $\int_{-\infty}^\infty (f(x) 1_{[0,1]}(x)) \delta(x) dx$. But this makes no sense unless $f(0)=0$ because otherwise $f 1_{[0,1]}$ is not continuous at zero, hence it is not in the domain of $\delta$. Any interpretation you come up with will not have real distribution-theoretic meaning because it will depend on how you mollify the delta function. – Ian Apr 15 '16 at 04:12
  • I see, thanks. For my case I did end up using just $f(a)$, it is in physics context, so I am guessing its not being used properly in terms of a distribution theory context, @David Z , what kind of situations would it apply? It doesn't seem quite obvious in my case as the actual calculation is a bit more complicated, I will look into it. – AAM Apr 22 '16 at 10:17
  • @AAM the example I'm familiar with is Feynman diagrams involving a gluon emitted with a fraction $1-\xi$ of the forward momentum of the parent particle. We integrate over $\xi$ to take into account all possible gluon momenta, and $\delta(1-\xi)$ represents the contribution of a diagram with no emitted gluon. In that case $\xi > 1$ is physically meaningless; it wouldn't make sense to consider our deltas as the limit of symmetric peaked functions, but they could be the limit of asymmetric functions (defined only on $\xi < 1$), for which $\int^1 f(\xi)d\xi = 1$. – David Z Apr 22 '16 at 10:35
  • In retrospect, it's possible this could have been written in a way that would make it a good question for [physics.SE]. Such a question would ask about the physical reasoning for using a particular convention for the integral of a delta function, in some specific situation. – David Z Apr 22 '16 at 10:37

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