How to prove $\int_{-\infty}^\infty \delta(x)dx=1?$ Dirac delta function is defined as-
$\delta(x)=\infty$ when $x=0,$ and
$\delta(x)=0$ when $x\neq0$.
How to prove $\int_{-\infty}^\infty \delta(x)dx=1?$
(I get $\int_{-\infty}^\infty \delta(x)dx=0$, because $0.\infty=0$ )
 A: Your definition does not work.


*

*the "equation" $\delta(x)=0$ has no meaning in the reals, and this does not define a function.

*the normalization condition which you want to prove belongs to the definition of the usual Dirac distribution, and being a convention, it cannot be proven (and certainly not with the given).
A: First off, $0.\infty$ is not $0$. The symbol $\infty$ does not represent a number, so you cannot do arithmetic with it. Some times we use $0.\infty$ as a classification of a limit, and in those cases it could result in anything (it's a so-called "indeterminate form"), but by itself, without the context of limits, $0.\infty$ is meaningless.
However, the bigger problem here is that you've got things the wrong way around. It is not that $\delta(0) = \infty$ and $\delta(x) = 0$ for all other numbers, and therefore $\int_{-\infty}^\infty \delta(x)\,dx = 1$. It's the other way around: We define the "function" (or rather "density" or "distribution") $\delta(x)$ so that if $a<0<b$ (including $\pm\infty$) then $\int_a^b\delta(x)\cdot f(x)\,dx = f(0)$ for all functions $f:\Bbb R\to \Bbb R$, and if $0<a<b$ or $a<b<0$, then the integral evaluates to $0$.
The actual value of $\delta$ at specific points isn't terribly important because it is almost exclusively used inside an integral when evaluated.
A: $\int_{-\infty}^\infty \delta(x)dx$ is notation for $\delta*1$ which is a distribution convolved with a $C^\infty$ function. Note for any $\phi \in C_c^\infty$, $\langle \delta*1 , \phi \rangle = \langle \delta, \phi*\tilde{1}\rangle = (\phi*\tilde{1})(0) = \int_{-\infty}^\infty \phi(x)dx$, so $\delta*1$ is the same, in the sense of distributions, as the locally integrable function that is identically $1$. Hence the equality $\int_{-\infty}^\infty \delta(x)dx = 1$.
