Integral to Infinity of Shifted 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.
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:
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
- 1965: Murray R. Spiegel: Theory and Problems of Laplace Transforms ... (previous) ... (next): Chapter $1$: The Laplace Transform: Some Special Functions: $\text {VII}$. The Unit Step function: $3$.