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A problem occurs when I was solving an exersice of perturbative kind. The delta function has the fundamental property that

\begin{align} \int_{-\infty}^{\infty}f(x)\delta(x-a)dx=f(a) \end{align}

and, in fact, \begin{align} \int_{a-\epsilon}^{a+\epsilon}f(x)\delta(x-a)dx=f(a) \end{align} How change these formula if $a$ is not in the domain of integration? \begin{align} \int_{-\infty}^{a-\epsilon}f(x)\delta(x-a)dx + \int_{a+\epsilon}^{\infty}f(x)\delta(x-a)dx\end{align}

Also if $a$ is one of upper or lower bounds?

aaa
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    @asaa I've clarified the title to better reflect your comment, but you should edit your question to make it clear that that is what you're asking. (Comments on this site are temporary and only meant to help improve the question itself.) – E.P. May 06 '16 at 15:52
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    @asaa Don't thank me - do the work you're asked to do. Match the work you're expecting others to do for you with work of your own, and the clarity in the answers you're hoping for with clarity in your own question. The score of this question quite accurately reflects its current usefulness to anyone but yourself. – E.P. May 06 '16 at 16:03
  • @EmilioPisanty huh,just after editing title it starts to get lower vot. – aaa May 06 '16 at 16:11
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    @asaa No. It gets lower votes because it's a bad question, incomplete and badly phrased. Fix it and the score may improve. – E.P. May 06 '16 at 16:21

1 Answers1

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If you have an integral of the form $$ \int_\alpha^\beta f(x)\delta(x-a)\mathrm dx $$ with a finite (and fixed!, i.e. it does not depend on $a$) domain of integration, then $$ \int_\alpha^\beta f(x)\delta(x-a)\mathrm dx=\begin{cases} f(a)& \text{if }\alpha<a<\beta \\ 0& \text{otherwise}. \end{cases} $$

This is because $\delta(x-a)$ is zero for all $x=a$, so you're integrating something that is identically zero (and without any funky spikes), and that gives zero.

More rigorously, you can rephrase any integral with finite integration domain $(\alpha,\beta)$ in terms of the characteristic function $\chi_{(\alpha,\beta)}$ of the interval,

$$\chi_{(\alpha,\beta)}(x)=\begin{cases} 1& \text{if }\alpha<x<\beta, \\ 0& \text{otherwise}, \end{cases}$$

so that $$ \int_\alpha^\beta F(x)\mathrm dx=\int_{-\infty}^\infty\chi_{(\alpha,\beta)}(x)F(X)\:\mathrm dx $$ for any function $F$, and therefore \begin{align} \int_\alpha^\beta f(x)\delta(x-a)\mathrm dx & = \int_{-\infty}^\infty\chi_{(\alpha,\beta)}(x) f(x)\delta(x-a)\:\mathrm dx \\ & = \chi_{(\alpha,\beta)}(a) f(a) \\ & =\begin{cases} f(a)& \text{if }\alpha<a<\beta, \\ 0& \text{otherwise}. \end{cases} \end{align}

E.P.
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  • Could you please say the reason too! – aaa May 06 '16 at 15:52
  • It's by definition. – B. Pasternak May 06 '16 at 16:00
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    You can imagine the $\delta (x-a)$ function (a mathematician would cringe at this) as a function that is peaked at $x=a$ and with zero spread. So when you look at $f(x) \delta (x-a)$ it samples the region around $a$ and is zero elsewhere. – jim May 06 '16 at 16:06
  • @anon0909 Why not? It's simply the function (well it's not a function at all, it's a measure) which picks out that one value..where and why is any intuition needed here? I agree that intuition behind ideas is nice, but this is really straightforward. – B. Pasternak May 06 '16 at 16:12
  • What happens if $a=\alpha$ or $\beta$. In your answer you say that in this case the integrand is zero. Is this correct? I have seen other definitions as well. – Prahar May 06 '16 at 17:05
  • @prahar could you please explain more that definition? Question also! – aaa May 06 '16 at 17:43
  • In particular, we have $\int_0^\infty f(x) \delta(x) = \frac{1}{2}\lim\limits_{\epsilon\to 0^+} f(\epsilon)$. This is the answer that one will gets when a function is integrated against the Dirac Delta Distribution and then take the limit to retrieve the answer. – Prahar May 06 '16 at 17:54
  • @Prahar Delta functions at the interval boundary are not generally well-defined. Usually one takes the symmetric version (which evaluates as you indicate) but his needs to be indicated (example; eq. 5). In general, if you write $\delta(x)=\lim_{n\to\infty}\delta_n(x)$, the symmetric $\delta$ you describe correspond to all the $\delta_n$ being even, but you could also take asymmetric $\delta_n$s if you wanted. – E.P. May 06 '16 at 19:43
  • @asaa For further information on the definition of the delta function, see your favourite textbook. – E.P. May 06 '16 at 19:44