- #1

CAF123

Gold Member

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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!