I know that $\int_{-\epsilon}^\infty f(x)\delta(x)dx=f(0)$ but what about $\int_0^\infty f(x)\delta(x)dx$? I suppose we have to do this by definition since the lower limit is bang on $0$?
Dirac Delta function
9
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integration
distribution-theory
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1See http://math.stackexchange.com/questions/30146/delta-function-integrated-from-zero/30150#30150. – 2012-03-22
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0@joriki: Thanks! So it is just 1? – 2012-03-22
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2As Carl's answer and the (partly contradictory) answers to the question I linked to indicate, it's not well-defined. If you tell us the context in which you're trying to use this, more might be said about what the appropriate definition might be for that context. – 2012-03-22
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1Joriki's comment is right. My answer uses the measure-theoretic interpretation of $\delta$ as a point mass measure. But if you used the interpretation via test functions instead, you should get $f(0)/2$ instead of $f(0)$. – 2012-03-22
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0See [this answer](http://math.stackexchange.com/a/56684/6179). – 2012-03-24