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I know that: $$\int_{-\infty}^{+\infty} \mathrm{d}t \, f(t) \delta(t) = f(0)$$

However, I cannot figure out the result of the integral below:

$$\int_{0}^{+\infty} \mathrm{d}t \, f(t) \delta(t) = ?$$

Is it $$\int_{0}^{+\infty} \mathrm{d}t \, f(t) \delta(t) = \frac{f(0)}{2}?$$

Please provide a source for the answer too.

MOON
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  • It is not true, because: $\int_{0}^{\infty}f(t)\delta(t)\space\text{d}t=-f(0)\left(\theta(0)-1\right)$ – Jan Eerland Jun 01 '16 at 15:20
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    First you need a definition for the delta function. To begin with mathematically, it is a linear functional on a certain space of $C^\infty$ functions. Since the characteristic function of $[0,\infty)$ is not $C^\infty$, you will have to tell us what extension of the notion of the delta function you are talking about. In fact, you may need to specify whether your integral is over the set $[0,+\infty)$ or $(0,+\infty)$, or something else... – GEdgar Jun 01 '16 at 15:53
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    @JanEerland How can this comment be useful to the OP? (Not to mention its dubious mathematical accuracy.) – Did Jun 01 '16 at 16:15

2 Answers2

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The identity $$ \int_{-\infty}^{+\infty}dt\ f(t) \delta(t) = f(0)\tag{*} $$ is meaningless without context. Also this notation is a convenient abuse of notation, and not a standard (Riemann or Lebesgue) integral.

Let's say you are considering $\delta:\mathcal{S}(\mathbb{R})\to\mathbb{R}$ as a tempered distribution on the Schwartz class $\mathcal{S}(\mathbb{R})$. Then $(*)$ means nothing but the definition of $\delta$: $$ \delta(f)=f(0)\quad f\in\mathcal{S}(\mathbb{R}). $$ In this setting, $\int_{0}^{+\infty} \mathrm{d}t \, f(t) \delta(t) $ is not even a well-define notation.

Your question is a nice example demonstrating that it could be dangerous to think $\delta$ as a function of real variables.

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    That all said, if $\delta$ is regarded as the Dirac measure(which can be obtained from the distribution by the Riesz representation theorem), $\int_0^{+\infty}f(x),d\delta(x) = f(0)$. And it so happens that in certain research areas people actually have the tradition to write $\int \varphi(x) f(x),dx$ for positive measures $f$, whether or not absolutely continuous. – Ningxin Jun 01 '16 at 16:46
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    @Qiyu: in those fields, how does one write the integral over the set $(0,\infty)$? (If my measure is atomic, I'd be awfully precise about exactly what I am integrating over, and not use the ambiguous classical notation.) – Willie Wong Aug 01 '16 at 04:02
  • @WillieWong You use the standard notation $\int_{(0,\infty)} \phi(x)f(x),dx$. – Ningxin Aug 01 '16 at 08:12
  • I don't see how this answer is helpful to the OP. The integral in question may not satisfy mathematical rigour, however it can be repaired without difficulty so that it does become valid. The integration can be easily extended to negative values, by multiplying the integrand with the Heaviside function. The integral then satisfies (*) so it can be performed, yielding $f(0)/2$, as correctly assumed by the OP. – M. Wind Jun 15 '20 at 04:45
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See the question Delta function integrated from zero and Andrew's answer to the question.

According to both Bracewell and Andrew, your guess of $f(0)/2$ is right. See also my own question "Is distribution theory necessary for most users of the Dirac delta function?" and my answer to the question "Proof of Dirac Delta's sifting property".

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