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A Dirac
symbol can be interpreted as a linear functional, i.e. a linear mapping from a function space, consisting e.g. of certain real functions, to
(or
), having the property
One may think this as the inner product
of a function
and another "function"
, when the well-known formula
is true.\, Applying this to\,
,\, one gets
i.e. the Laplace transform
![{\displaystyle {\begin{matrix}{\mathcal {L}}\{\delta (t)\}\;=\;1.\end{matrix}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/116c90c50874961de0f74bd302df86d818cc9a5f)
By the delay theorem, this result may be generalised to
\\
When introducing a so-called "Dirac delta function", for example
![{\displaystyle {\begin{matrix}\eta _{\varepsilon }(t)\;:=\;{\begin{cases}{\frac {1}{\varepsilon }}\quad {\mbox{for}}\;\;0\leq t\leq \varepsilon ,\\0\quad {\mbox{for}}\qquad t>\varepsilon ,\end{cases}}\end{matrix}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7b657874e1554d101d022c89888cf450fb3f35bc)
as an "approximation" of Dirac delta, we obtain the Laplace transform
As the Taylor expansion shows, we then have
according to ref.(2).
The Dirac delta ,
, can be correctly defined as a linear functional, i.e. a linear mapping from a function space, consisting e.g. of certain real functions, to
(or
), having the property
One may think of this as an inner product
of a function
and another "function"
, when the well-known formula
holds.\, By applying this to \,
,\, one gets
i.e. the Laplace transform
![{\displaystyle {\begin{matrix}{\mathcal {L}}\{\delta (t)\}\;=\;1.\end{matrix}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/116c90c50874961de0f74bd302df86d818cc9a5f)
By the delay theorem, this result may be generalised to: