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CAF123
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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. 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##?
Thanks!
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. 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##?
Thanks!