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?

  • $\begingroup$ So why everywhere we can find this version: en.wikipedia.org/wiki/Dirac_delta_function $\endgroup$ – Rafał Apriasz Dec 5 '15 at 10:17
  • 1
    $\begingroup$ 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. $\endgroup$ – Batominovski Dec 5 '15 at 10:24
  • $\begingroup$ use the function $f(x) = 1$ $\endgroup$ – user265328 Dec 5 '15 at 10:33
  • $\begingroup$ 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: $$ \int_{-\infty}^{0} \delta(x)dx + \int_{0}^{-\infty} \delta(-x)(-dx) = 2 \int_{-\infty}^{0} \delta(x)dx = 1 $$ $\endgroup$ – Rafał Apriasz Dec 5 '15 at 11:48
  • $\begingroup$ You could have a look at some representations of the delta function $\lim_{\epsilon\rightarrow0}\delta_{\epsilon}$, they are always even functions $\endgroup$ – tired Dec 5 '15 at 14:28

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