Values of Dirac Delta Function over Reals

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Theorem

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

Then:

$\map \delta x := \begin {cases} \infty & : x = 0 \\ 0 & : x \ne 0 \end {cases}$


Proof

We have that:

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

where:

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


Therefore:

\(\ds \map \delta 0\) \(=\) \(\ds \ds \lim_{\epsilon \mathop \to 0} \map {F_\epsilon} 0\)
\(\ds \) \(=\) \(\ds \dfrac 1 {2 \times 0 }\)
\(\ds \) \(=\) \(\ds \infty\)

and

\(\ds \map \delta {x \ne 0}\) \(=\) \(\ds \ds \lim_{\epsilon \mathop \to 0} \map {F_\epsilon} {x \ne 0}\)
\(\ds \) \(=\) \(\ds 0\)

Therefore:

$\map \delta x := \begin {cases} \infty & : x = 0 \\ 0 & : x \ne 0 \end {cases}$


Warning

Note that while the Dirac delta function $\map \delta x$ is usually so referred to as a function and treated as a function, it is generally considered not actually to be a function at all.