Integral of Shifted Dirac Delta Function by Continuous Function over Reals

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:

From Upper and Lower Bounds of Integral:


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

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.