Mellin Transform of Dirac Delta Function by Function

Theorem

Let $f: \R \to \R$ be a function.

Let $c \in \R_{>0}$ be a positive constant real number.

Let $\map {\delta_c} t$ be the Dirac delta function.

Let $\MM$ be the Mellin transform.

Then:

$\map {\MM \set {\map {\delta_c} t \map f t} } s = c^{s - 1} \map f c$

Proof

\(\ds \map {\MM \set {\map {\delta_c} t \map f t} } s\)

\(=\)

\(\ds \int_0^{\to +\infty} t^{s - 1} \map {\delta_c} t \map f t \rd t\)

Definition of Mellin Transform

\(\ds \)

\(=\)

\(\ds \int_{c^-}^{c^+} t^{s - 1} \map {\delta_c} t \map f t \rd t\)

Definition of Dirac Delta Function: integrand is elsewhere zero

\(\ds \)

\(=\)

\(\ds \int_{c^-}^{c^+} c^{s - 1} \map {\delta_c} t \map f c \rd t\)

$t$ is constant in interval $\closedint {c^-} {c^+}$

\(\ds \)

\(=\)

\(\ds c^{s - 1} \map f c \int_{c^-}^{c^+} \map {\delta_c} t \rd t\)

Primitive of Constant Multiple of Function

\(\ds \)

\(=\)

\(\ds c^{s - 1} \map f c\)

Definition of Dirac Delta Function

$\blacksquare$