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I've come across the equation $$\int_0^1 dx \frac{dA(x)}{dx} + B = C = \text{finite}$$ in my readings on a certain topic in physics and, in both articles i have read, the following step is taken $$\int_0^1 dx \left( \frac{dA(x)}{dx} + (B-C)\delta(1-x) \right) = \text{finite}$$

For the purposes of my question, I don't think the explicit form of the functions $A,B,C$ are of concern. The reason $\delta(1-x)$ is chosen is associated with the underlying physics which I also think is not important. It seems to me that the replacement $$1 = \int_0^1 \delta(1-x) dx$$ has been made to go from line $1$ to line $2$ in the above. $\mathbf{My\,questions\, are}$:

$1)$ Why is this replacement even correct? The zero of the delta function occurs at the edge of the interval so I don't see why the integral should evaluate to one.

$2)$ I put this integral into wolfram alpha. It returns $\int_0^1 dx \delta(1-x) = \theta(0)$, where $\theta(0)$ is the step function evaluated at zero. Upon subsequently putting $\theta(0)$ into wolfram alpha it returns $1$. So a) how is $\theta(0)$ obtained analytically and b) why is $\theta(0) = 1$?

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    Evaluation of the delta function at the endpoints of an interval is not well-defined by the mathematical framework of distribution theory. You do indeed essentially get $\theta(0)$ where $\theta$ is the unit step function, since $\theta'=\delta$ in the sense of distribution theory, but the value of the unit step function at zero does not really exist except by convention. – 2017-02-12
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    Commonly in physics you consider symmetric "nascent delta functions"; under this assumption the $\theta(0)$ that is recovered is $1/2$. But $\theta(0)$ can be anything strictly between $0$ and $1$ by choosing the right type of nascent delta functions (also known as an approximate identity). – 2017-02-12
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    @Ian Thanks! I get the result $1-\theta(0)$ with that replacement of $\theta' = \delta$ but would you know why wolfram alpha would give just $\theta(0)$ and why, when I subsequently put $\theta(0)$ into wolfram alpha, it returns $1$? Has it just made a particular choice of nascent delta functions that you mentioned in the background or something? – 2017-02-12
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    It is using the right-continuous version of the step function. At any rate the quantity you are seeking is not well-defined. – 2017-02-12
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    I see - so not well defined in that the value of $\theta(0)$ (with which the integral in question essentially evaluates to) does not take a value universally agreed upon? Instead, it is such that it may take on values from $[0,1]$ and the value chosen is the one appropriate at hand which in this case was one. Is that the right way to think about it? – 2017-02-12
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    Basically yes, though the fact that WA told you that $\theta(0)=1$ should not be regarded as evidence that your integral should be regarded as $1$. You really need more context: essentially, in what limit was your delta function realized? If there was no limit and you are just trying to formulate it in terms of distribution theory alone, then your question is meaningless. – 2017-02-13
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    Thanks! Here is the step shown explicitly with the physics context included (the only relevant equations are the third and fourth ones shown here https://www.ippp.dur.ac.uk/~krauss/Lectures/QuarksLeptons/QCD/DGLAP_2.html) He defines what is called the plus distribution in terms of a limit after these equations but I think that comes on top and not associated with the input of the delta. – 2017-02-13

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