Integral to Infinity of Dirac Delta Function by Continuous Function

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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:

\(\ds \int_0^{+ \infty} \map {F_\epsilon} x \map g x \rd x\) \(=\) \(\ds \int_0^\epsilon \dfrac 1 \epsilon \map g x \rd x + \int_\epsilon^\infty 0 \times \map g x \rd x\) Definition 1 of Dirac Delta Function
\(\ds \) \(=\) \(\ds \int_0^\epsilon \dfrac 1 \epsilon \map g x \rd x + \lim_{L \mathop \to \infty} \int_\epsilon^L 0 \times \map g x \rd x\) Definition of Improper Integral
\(\ds \) \(=\) \(\ds \dfrac 1 \epsilon \int_0^\epsilon \map g x \rd x + \lim_{L \mathop \to \infty} \int_\epsilon^L 0 \rd x\)
\(\ds \) \(=\) \(\ds \dfrac 1 \epsilon \int_0^\epsilon \map g x \rd x + \lim_{L \mathop \to \infty} \paren {0 \times \paren {L - \epsilon} }\) Definite Integral of Constant
\(\ds \) \(=\) \(\ds \dfrac 1 \epsilon \int_0^\epsilon \map g x \rd x + \lim_{L \mathop \to \infty} 0\) simplification
\(\ds \) \(=\) \(\ds \dfrac 1 \epsilon \int_0^\epsilon \map g x \rd x\)


From Darboux's Theorem:

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

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 Definition 1 of Dirac Delta Function:

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


Hence the result.

$\blacksquare$


Sources

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