Integral to Infinity of Shifted Dirac Delta Function by Continuous Function

From ProofWiki
Jump to navigation Jump to search

Theorem

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

Let $g$ be a continuous real function.

Let $a \in \R_{\ge 0}$ be a positive real number.


Then:

$\displaystyle \int_0^{+ \infty} \map \delta {x - a} \, \map g x \rd x = \map g a$


Proof

We have that:

$\map \delta {x - a} = \displaystyle \lim_{\epsilon \mathop \to 0} \map {F_\epsilon} x$

where:

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


We have that:

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


From Upper and Lower Bounds of Integral:

$\displaystyle m \paren {\paren {a + \epsilon} - a} \le \int_a^{a + \epsilon} \map g x \rd x \le M \paren {\paren {a + \epsilon} - 0}$

where:

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

on $\closedint a {a + \epsilon}$.

Hence:

$\displaystyle m \epsilon \le \int_a^{a + \epsilon} \map g x \rd x \le M \epsilon$

and so dividing by $\epsilon$:

$\displaystyle m \le \dfrac 1 \epsilon \int_a^{a + \epsilon} \map g x \rd x \le M$


Then:

$\displaystyle \lim_{\epsilon \mathop \to 0} M = m = \map g a$

and so by the Squeeze Theorem:

$\displaystyle \lim_{\epsilon \mathop \to 0} \int_0^{+ \infty} \map {F_\epsilon} x \, \map g x \rd x = \map g a$

But by definition of the Dirac delta function:

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


Hence the result.

$\blacksquare$

Sources

Retrieved from "https://proofwiki.org/w/index.php?title=Integral_to_Infinity_of_Shifted_Dirac_Delta_Function_by_Continuous_Function&oldid=405938"