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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 bound?

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    $\begingroup$ Is that a question? I'm not sure what you are asking. $\endgroup$ – User8128 May 6 '16 at 14:42
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    $\begingroup$ No. I still don't know what you're asking. What does it mean to integrate from $c$ to $d$ with $c < a < d$ but also say $x$ can't equal $a$? This doesn't seem meaningful to me. When we integrate with respect to $x$, the variable $x$ isn't something we can choose or put restrictions on (we can;t just say: $x \neq a$); instead, $x$ is some dummy variable that takes values in the range of integration. $\endgroup$ – User8128 May 6 '16 at 18:09
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    $\begingroup$ Often it's OK to treat $\delta(x)$ like a function in a hand-waving manner and not worry about distribution theory too much, but if you want to get into subtle points like $a$ being one of the integration limits, you should state more explicitly what sort of setting you're operating in, e.g., what function space are you considering? $\endgroup$ – joriki May 6 '16 at 20:41
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    $\begingroup$ @asaa: What does that mean? $\endgroup$ – joriki May 6 '16 at 21:57
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    $\begingroup$ No, it's not clear at all. Please elaborate. $\endgroup$ – joriki May 6 '16 at 22:22
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My guess is that if $\delta(t-a) = 0$ for any $t\neq a$, hence, if the integration domain does not include $a$ in the integration range, it is the same as multiplying $f(x)$ by zero, no matter how small you consider the $\epsilon$. Then the integral result is zero.

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    $\begingroup$ This is correct $\endgroup$ – Ross Millikan Apr 11 '18 at 13:56

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