Integral of Dirac Delta Function over Reals

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Theorem

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

Then:

$\ds \int_{-\infty}^{+\infty} \map \delta x \rd x = 1$


Proof

We have that:

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

where:

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


We have that:

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


Then:

\(\ds \int_{-\infty}^{+\infty} \map \delta x \rd x\) \(=\) \(\ds \lim_{\epsilon \mathop \to 0} \int_{-\infty}^{+\infty} \map {F_\epsilon} x \rd x\) Definition 2 of Dirac Delta Function
\(\ds \) \(=\) \(\ds \lim_{\epsilon \mathop \to 0} 1\) from above
\(\ds \) \(=\) \(\ds 1\)

Hence the result.

$\blacksquare$


Sources

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