Integral to Infinity of Dirac Delta Function
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Theorem
Let $\map \delta x$ denote the Dirac delta function.
Then:
- $\ds \int_0^{+ \infty} \map \delta x \rd x = 1$
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:
| \(\ds \int_0^{+ \infty} \map {F_\epsilon} x \rd x\) | \(=\) | \(\ds \int_0^\epsilon \dfrac 1 \epsilon \rd x + \int_\epsilon^\infty 0 \rd x\) | Definition 1 of $\map {F_\epsilon} x$ | |||||||||||
| \(\ds \) | \(=\) | \(\ds \int_0^\epsilon \dfrac 1 \epsilon \rd x + \lim_{L \mathop \to \infty} \int_\epsilon^L 0 \rd x\) | Definition of Improper Integral | |||||||||||
| \(\ds \) | \(=\) | \(\ds \dfrac 1 \epsilon \paren {\epsilon - 0} + \lim_{L \mathop \to \infty} \paren {0 \times \paren {L - \epsilon} }\) | Definite Integral of Constant | |||||||||||
| \(\ds \) | \(=\) | \(\ds 1 + \lim_{L \mathop \to \infty} 0\) | simplification | |||||||||||
| \(\ds \) | \(=\) | \(\ds 1\) |
Then:
| \(\ds \int_0^{+ \infty} \map \delta x \rd x\) | \(=\) | \(\ds \lim_{\epsilon \mathop \to 0} \int_0^{+ \infty} \map {F_\epsilon} x \rd x\) | Definition 1 of Dirac Delta Function | |||||||||||
| \(\ds \) | \(=\) | \(\ds \lim_{\epsilon \mathop \to 0} 1\) | from above | |||||||||||
| \(\ds \) | \(=\) | \(\ds 1\) |
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: $\text {VIII}$. The Unit Impulse function or Dirac delta function: $1$.