Mellin Transform of Dirac Delta Function
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Theorem
Let $c \in \R_{>0}$ be a (strictly) positive real number.
Let $\map {\delta_c} t$ be the Dirac delta function.
Let $\MM$ be the Mellin transform.
Then:
- $\map {\MM \set {\map {\delta_c} t} } s = c^{s - 1}$
Proof
\(\ds \map {\MM \set {\map {\delta_c} t} } s\) | \(=\) | \(\ds \int_0^{\to +\infty} t^{s - 1} \map {\delta_c} t \rd t\) | Definition of Mellin Transform | |||||||||||
\(\ds \) | \(=\) | \(\ds \int_{c^-}^{c^+} t^{s - 1} \map {\delta_c} t \rd t\) | Definition of Dirac Delta Function: integrand is elsewhere zero | |||||||||||
\(\ds \) | \(=\) | \(\ds \int_{c^-}^{c^+} c^{s - 1} \map {\delta_c} t \rd t\) | $t$ is constant in interval $\closedint {c^-} {c^+}$ | |||||||||||
\(\ds \) | \(=\) | \(\ds c^{s - 1} \int_{c^-}^{c^+} \map {\delta_c} t \rd t\) | Primitive of Constant Multiple of Function | |||||||||||
\(\ds \) | \(=\) | \(\ds c^{s - 1}\) | Definition of Dirac Delta Function |
$\blacksquare$