Mellin Transform of Exponential

Theorem

Let $a$ be a complex constant and $e^t$ be the complex exponential.

Let $\MM$ be the Mellin transform.

Then:

$\map {\MM \set {e^{-a t} } } s = a^{-s} \, \map \Gamma s$

where $\map \Re a, \map \Re s > 0$

Proof

\(\ds \map {\MM \set {e^{-a t} } } s\)

\(=\)

\(\ds \int_0^{\to +\infty} t^{s - 1} e^{-a t} \rd t\)

Definition of Mellin Transform

\(\ds \)

\(=\)

\(\ds \int_0^{\to +\infty} \paren {\dfrac t a}^{s - 1} e^{-a \paren {\frac t a} } \frac {\d t} a\)

Integration by Substitution, $t \mapsto \dfrac t a$, $\d t \mapsto \dfrac {\d t} a$

\(\ds \)

\(=\)

\(\ds a^{-s} \int_0^{\to +\infty} t^{s - 1} e^{-t} \rd t\)

Primitive of Constant Multiple of Function

\(\ds \)

\(=\)

\(\ds a^{-s} \, \map \Gamma s\)

Definition of Gamma Function

$\blacksquare$

Also see

• Mellin Transform of Higher Order Exponential