We know integral of any function over a null set is zero. But for Dirac delta function ($\delta=+\infty$ iff $x=0$ otherwise $\delta=0$) $$ \int_{-\infty}^{+\infty}\delta =\int_0^0\delta =1. $$ Is it a contradiction?
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The Dirac delta is not actually a function, so there is no contradiction. – Nate Eldredge May 19 '13 at 22:39
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No, the expression you wrote is not a Lebesgue or Riemann integral, and $\delta$ is not a function in the conventional sense. Look for distributions (generalized functions): http://en.wikipedia.org/wiki/Distribution_%28mathematics%29 – Andrés E. Caicedo May 19 '13 at 22:39
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See this question http://math.stackexchange.com/questions/395850/dirac-delta-or-dirac-delta-function – Baby Dragon May 19 '13 at 22:40
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Thank you, Nate, Andres and Baby Dragon. – Falang May 19 '13 at 22:50
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You can regard it as a measure, the Dirac measure, which gives mass $1$ to $\{0\}$ , and zero mass to other subsets of $\mathbb R$ which do not contain zero(let's say Borel sets). For a Borel set $A\subset \mathbb R$ define:
$$\mu (A)=\begin{cases}1,&0\in A\\0,&0\notin A\end{cases}$$ Therefore: $$\int _{-\infty }^\infty d\mu=\int_{\{0\}}1\cdot d\mu+\int_{\mathbb R\setminus\{0\}}d\mu=\mu(\{0\})+\mu(\mathbb R\setminus\{0\})=1+0=1$$
Dimitris
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