Integral to Infinity of Dirac Delta Function by Continuous Function

Theorem
Let $\map \delta x$ denote the Dirac delta function.

Let $g$ be a continuous real function.

Then:


 * $\ds \int_0^{+ \infty} \map \delta x \map g x \rd x = \map g 0$

Proof
We have that:


 * $\map \delta x = \ds \lim_{\epsilon \mathop \to 0} \map {F_\epsilon} x$

where:


 * $\map {F_\epsilon} x = \begin{cases} 0 & : x < 0 \\ \dfrac 1 \epsilon & : 0 \le x \le \epsilon \\ 0 & : x > \epsilon \end{cases}$

We have that:

From Upper and Lower Bounds of Integral:


 * $\ds m \paren {\epsilon - 0} \le \int_0^\epsilon \map g x \rd x \le M \paren {\epsilon - 0}$

where:
 * $M$ is the maximum of $\map g x$
 * $m$ is the minimum of $\map g x$

on $\closedint 0 \epsilon$.

Hence:


 * $\ds m \epsilon \le \int_0^\epsilon \map g x \rd x \le M \epsilon$

and so dividing by $\epsilon$:


 * $\ds m \le \dfrac 1 \epsilon \int_0^\epsilon \map g x \rd x \le M$

Then:
 * $\ds \lim_{\epsilon \mathop \to 0} M = m = \map g 0$

and so by the Squeeze Theorem:


 * $\ds \lim_{\epsilon \mathop \to 0} \int_0^{+\infty} \map {F_\epsilon} x \map g x \rd x = \map g 0$

But by :


 * $\ds \lim_{\epsilon \mathop \to 0} \int_0^{+\infty} \map {F_\epsilon} x \map g x \rd x = \int_0^{+\infty} \map \delta x \map g x \rd x$

Hence the result.