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I was calculating one physical problem and I stopped at one thing. By definition Dirac delta is given by that expression: $$ \int_{-\infty}^{\infty}\delta(x)dx = 1 $$ and additionaly there is one property which is correct (I sure that, because I've proven it 2 years ago but I can't find a book). I am talking about: $$ \int_{-\infty}^{0}\delta(x)dx = \int_{0}^{\infty}\delta(x)dx = \frac{1}{2} $$

I need to prove it again (yes I was looking for that in the Internet), I was trying use method by substitution like this: $$ \int_{-\infty}^{\infty}f(x)\delta(x-a)dx = f(a) $$ $$ x-a =t / dx = dt$$

But It doesn't work. Any sugestions?

Rafał Apriasz
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  • So why everywhere we can find this version: https://en.wikipedia.org/wiki/Dirac_delta_function – Rafał Apriasz Dec 05 '15 at 10:17
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    Sorry, misunderstood what you wanted to prove. Anyway, you can prove that the Dirac delta is an even distribution. Hence, its integral on the positive reals must be the same as the integral on the negative reals. – Batominovski Dec 05 '15 at 10:24
  • use the function $f(x) = 1$ –  Dec 05 '15 at 10:33
  • Is my thought is right? $$ \int_{-\infty}^{\infty} \delta(x)dx = \int_{-\infty}^{0} \delta(x)dx + \int_{0}^{\infty} \delta(x)dx = 1 $$ in the second integral let change $$ x \rightarrow -x$$ So we have: $$ \int_{-\infty}^{0} \delta(x)dx + \int_{0}^{-\infty} \delta(-x)(-dx) = 2 \int_{-\infty}^{0} \delta(x)dx = 1 $$ – Rafał Apriasz Dec 05 '15 at 11:48
  • You could have a look at some representations of the delta function $\lim_{\epsilon\rightarrow0}\delta_{\epsilon}$, they are always even functions – tired Dec 05 '15 at 14:28

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