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Consider a smooth enough function $f(x)$, we have the following result. \begin{align} I(a, b) = \int_{a}^{b} f(x) \, \delta(x - x_{0}) \, dx = \begin{cases} f(x_{0}) \qquad & \mathrm{if} \; a < x_{0} < b \\ 0 \qquad & \mathrm{if} \; x_{0} < a \quad \mathrm{or} \quad x_{0} > b \end{cases} \end{align} with $a < b$.

Question: What happen to the integral if the singular point $x_{0}$ is right at the integration limit ? Say, $b = x_{0}$. I attempt to answer this question by shifting the upper limit by $\epsilon > 0$, but there is a problem as shown below.

Case 1: Shift the upper limit to a slightly larger value \begin{align} I_{+}(a, x_{0}) \; = \; \lim_{\epsilon \to 0} \int_{a}^{x_{0}+\epsilon} f(x) \, \delta(x-x_{0})\, dx \; = \;f(x_{0}) \end{align}

Case 2: Shift the upper limit to a slightly smaller value \begin{align} I_{-}(a, x_{0}) \; = \; \lim_{\epsilon \to 0} \int_{a}^{x_{0}-\epsilon} f(x) \, \delta(x-x_{0})\, dx \; = \; 0 \end{align}

The left-hand limit does not agree with the right-hand limit, $I_{+}(a, x_{0}) \neq I_{-}(a, x_{0})$. So is it true that integral like $I(a, x_{0})$ cannot be defined ?

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    $\begingroup$ Even worse, the expression $\int_a^b f(x) \delta(x-x_0)$ is not well defined in the first place. There is no function with such a property (i.e. which after integration with f returns $f$ at $x_0$), it's just a weird notation for the delta distribution $l(f) := f(x_0)$ (which is well defined on suitable function spaces). $\endgroup$ – Thomas Sep 12 '17 at 9:36
  • $\begingroup$ It can, but you decide what is the definition (as you decided to use the nonstandard $I_+$ and $I_-$). In fact, what you ask has been defined for finite order distributions for manifolds with boundary (even Stokes theorem has been reproved in this setting...). $\endgroup$ – John B Sep 12 '17 at 9:38
  • $\begingroup$ Your problem is not even the one in the title, but rather that you seem to be lacking any working definition of the Dirac delta distribution (which is not a function). So... please explain, what is $\delta$ to you? $\endgroup$ – Did Sep 12 '17 at 9:40
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You need to think about what your delta function is defined to be right from the start. You are right that the definition you are using does not naturally extend to an obvious answer, but its only the bad integral notation for the delta combined with ones intuition with manipulating integrals that causes the confusion!

Usually the formal mathematical way of defining the delta is as a distribution.

On a regular domain $\Omega \subset \mathbb{R}^n$ a distribution is a linear functional defined on $C^\infty_c(\Omega)$, where $C^\infty_c(\Omega)$ denotes the space of smooth functions whose support is a compact set. The Dirac delta about the point $x_0\in\Omega$ is then defined via $$ \delta_{x_0}\colon C^\infty_c(\Omega)\to\mathbb{R} $$ $$ \delta_{x_0}(f)=f(x_0) .$$ If this is the first time you are seeing this, note that this is precisely the same definition you were using. You just used the notation of the integral sign for some convenience, but the integrand was never actually an integrable function (it was never a function at all!) so you were kidding yourself that the operation being performed was a classical integral :p

Note that we only have here the delta defined on functions which vanish near the boundary $\partial \Omega$! (One can of course then extend the definition to spaces in which $C^\infty_c(\Omega)$ is appropriately dense, however the condition of vanishing at the boundary is crucial to the whole theory of distributions as it relies on defining duality by analogy with integration by parts). In the end, in your situation the answer depends whether the underlying space you are working over is in fact $[a,b]$ or if you are working on $\mathbb{R}$. If it is the former, then your answer won't be defined unless $f(a)=f(b)=0$. If it is the latter, then the answer is $f(x_0)$, straight from the definition!

I would guess the case is probably the latter, if you are dealing with functions on $\mathbb{R}$. To avoid confusion with the integral notation you should perhaps try and avoid this functional $I[a,b]$ that you are trying to define, and instead always define the dirac with 'limits' over the whole line, writing $$ \delta_{x_0}(f)= \int_{-\infty}^\infty f(x) \delta_{x_0}(x) dx = \int_{-\infty}^\infty f(x) \delta(x-x_0) dx = f(x_0) .$$ Then you will safely never make mistakes. But I hope this motivates you to read a bit about distributions anyway :)

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    $\begingroup$ Thanks! I finally got some time to understand your answer. I come from a physics background, and so never encounter such behaviour of delta function. I realized this "strange" integral when reading some stochastic dynamics papers, involving Gaussian white noise. $\endgroup$ – K_inverse Sep 19 '18 at 2:50

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