Integral of Shifted Dirac Delta Function by Continuous Function over Reals

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Theorem

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

Let $g$ be a continuous real function.


Then:

$\ds \int_{-\infty}^{+ \infty} \map {\delta} {x - s} \map g x \rd x = \map g s$


Proof

We have that:

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

where:

$\map {F_\epsilon} x = \begin{cases} 0 & : x < -\epsilon + s \\ \dfrac 1 {2 \epsilon} & : -\epsilon + s \le x \le \epsilon + s \\ 0 & : x > \epsilon + s \end{cases}$


We have that:

\(\ds \int_{-\infty}^{+ \infty} \map {F_\epsilon} x \map g x \rd x\) \(=\) \(\ds \int_{-\infty}^{-\epsilon + s} 0 \times \map g x \rd x + \int_{-\epsilon + s}^{\epsilon + s} \dfrac 1 {2 \epsilon} \map g x \rd x + \int_{\epsilon + s}^\infty 0 \times \map g x \rd x\) Definition of $F_\epsilon$
\(\ds \) \(=\) \(\ds \lim_{L \mathop \to -\infty} \int_L^{-\epsilon + s} 0 \times \map g x \rd x + \int_{-\epsilon + s}^{\epsilon + s} \dfrac 1 {2 \epsilon} \map g x \rd x + \lim_{L \mathop \to \infty} \int_{\epsilon + s}^L 0 \times \map g x \rd x\) Definition of Improper Integral
\(\ds \) \(=\) \(\ds \lim_{L \mathop \to -\infty} \int_L^{-\epsilon + s} 0 \rd x + \dfrac 1 {2 \epsilon} \int_{-\epsilon + s}^{\epsilon + s} \map g x \rd x + \lim_{L \mathop \to \infty} \int_{\epsilon + s}^L 0 \rd x\)
\(\ds \) \(=\) \(\ds \lim_{L \mathop \to -\infty} \paren {0 \times \paren {-\epsilon + s - L} } + \dfrac 1 {2 \epsilon} \int_{-\epsilon + s}^{\epsilon + s} \map g x \rd x + \lim_{L \mathop \to \infty} \paren {0 \times \paren {L - \epsilon - s} }\) Definite Integral of Constant
\(\ds \) \(=\) \(\ds \lim_{L \mathop \to -\infty} 0 + \dfrac 1 {2 \epsilon} \int_{-\epsilon + s}^{\epsilon + s} \map g x \rd x + \lim_{L \mathop \to \infty} 0\) simplification
\(\ds \) \(=\) \(\ds \dfrac 1 {2 \epsilon} \int_{-\epsilon + s}^{\epsilon + s} \map g x \rd x\)


From Darboux's Theorem:

$\ds m \paren {\epsilon + s - \paren {-\epsilon + s} } \le \int_{-\epsilon + s}^{\epsilon + s} \map g x \rd x \le M \paren {\epsilon + s - \paren {-\epsilon + s} }$

where:

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

on $\closedint {-\epsilon + s} {\epsilon + s}$.

Hence:

$\ds 2 m \epsilon \le \int_{-\epsilon + s}^{\epsilon + s} \map g x \rd x \le 2 M \epsilon$

and so dividing by $2 \epsilon$:

$\ds m \le \dfrac 1 {2 \epsilon} \int_{-\epsilon + s}^{\epsilon + s} \map g x \rd x \le M$


Then:

$\ds \lim_{\epsilon \mathop \to 0} M = m = \map g s$

and so by the Squeeze Theorem:

$\ds \lim_{\epsilon \mathop \to 0} \int_{-\infty}^{+ \infty} \map {F_\epsilon} x \map g x \rd x = \map g s$

But by definition of the Dirac delta function:

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


Hence the result.

$\blacksquare$


Sources

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