Laplace Transform of Complex Power

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Theorem

Let $q$ be a constant complex number with $\map \Re q > -1$

Let $t^q: \R_{>0} \to \C$ be a branch of the complex power multifunction chosen such that $f$ is continuous on the half-plane $\map \Re s > 0$.


Then $f$ has a Laplace transform given by:

$\laptrans {t^q} = \dfrac {\map \Gamma {q + 1} } {s^{q + 1} }$

where $\Gamma$ denotes the gamma function.


Proof

By definition of Laplace transform for a function not continuous at zero

$\displaystyle \laptrans {t^q} = \lim_{\varepsilon \mathop \to 0^+} \lim_{L \mathop \to +\infty} \map I {\varepsilon, L}$

where:

$\displaystyle \map I {\varepsilon, L} = \int_\varepsilon^L t^q e^{-s t} \rd t$


Let $n \in \Z_{>0}$.

From Laplace Transform of Positive Integer Power:

$\laptrans {t^n} = \dfrac {n!} { s^{n + 1} }$

From Gamma Function Extends Factorial, a reasonable Ansatz is:

$\laptrans {t^q} = \dfrac {\map \Gamma {q + 1} } { s^{q + 1} }$

From Poles of Gamma Function, $\map \Gamma {-1}$ is undefined.

This suggests that having $q > -1$ would be a good idea for $q$ wholly real.

Reasons for insisting $\map \Re q > -1$ for complex $q$ will become apparent during the course of the proof.


With the aim of expressing $\map I {\varepsilon, L}$ in a form similar to the integral defining $\Gamma$, we use Complex Riemann Integral is Contour Integral to express $\map I {\varepsilon, L}$ as a contour integral.

Write $s = \sigma + i \omega$ for $\sigma, \omega \in \mathbb R$ and $\sigma > 0$.

From Complex Power by Complex Exponential is Analytic, the integrand for $\map I {\varepsilon, L}$ is analytic.

Thus the conditions for Contour Integration by Substitution are satisfied for $\map \Re s > 0$.

Substitute:

\(\displaystyle u\) \(=\) \(\displaystyle s t\)
\(\displaystyle \) \(=\) \(\displaystyle \paren {\sigma + i \omega} t\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle s \rd t\) \(=\) \(\displaystyle \rd u\)
\(\displaystyle u\) \(=\) \(\displaystyle \paren {\sigma + i \omega} \varepsilon\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle t\) \(=\) \(\displaystyle \varepsilon\)
\(\displaystyle u\) \(=\) \(\displaystyle \paren {\sigma + i \omega} L\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle t\) \(=\) \(\displaystyle L\)
\(\displaystyle \int_\varepsilon^L t^q e^{-s t} \rd t\) \(=\) \(\displaystyle \int_{\sigma \varepsilon \mathop + i \omega \varepsilon}^{\sigma L \mathop + i \omega L} \paren {\frac u s}^q e^{-u} \rd \frac u s\)
\(\displaystyle \) \(=\) \(\displaystyle \frac 1 {s^{q + 1} } \int_{\sigma \varepsilon \mathop + i \omega \varepsilon}^{\sigma L \mathop + i \omega L} u^q e^{-u} \rd u\)


Denote:

$I_C := \displaystyle \int_C u^q e^{-u} \rd u$

Consider the contour:

$C = C_1 + C_2 - C_3 - C_4$

where:

$C_1$ is the line segment connecting $\sigma \varepsilon$ to $L \sigma$
$C_2$ is the line segment connecting $L\sigma$ to $L\sigma + i L \omega$
$C_3$ is the line segment connecting $i \varepsilon$ to $L\sigma + i L \omega$
$C_4$ is the circular arc connecting $i \varepsilon$ to $\sigma \varepsilon$ whose center is at the origin.


Note that $C_1$ and $C_2$ are positively oriented while $C_3$ and $C_4$ are negatively oriented.

The contours are illustrated in the following graph:

LaplaceTransformGamma.png


By the Cauchy-Goursat Theorem:

$I_C = I_{C_1} + I_{C_2} - I_{C_3} - I_{C_4} = 0$

Consider $I_{C_2}$ as $L \to +\infty$:

\(\displaystyle \cmod {I_{C_2} }\) \(=\) \(\displaystyle \cmod { \int_{L \omega}^{i \sigma \mathop + i L\omega} u^q e^{-u} \rd u}\)
\(\displaystyle \) \(\le\) \(\displaystyle \max_{u \mathop \in C_2} \cmod {u^q e^{-u} } L\) Estimation Lemma
\(\displaystyle \) \(=\) \(\displaystyle \max_{u \mathop \in C_2} \cmod {u^q} \, \map \exp{-\map \Re u} L\) Modulus of Exponential is Exponential of Real Part
\(\displaystyle \) \(\sim\) \(\displaystyle \cmod {L^q} e^{- L} L\) asymptotic equivalence for $\cmod u$ sufficiently large
\(\displaystyle \) \(\le\) \(\displaystyle k e^{3^{-1} L} e^{-L}\) $L^q$ is of exponential order $3^{-1}$ for some constant $k > 0$, because $\map \Re q > -1$
\(\displaystyle \) \(=\) \(\displaystyle \frac {k L} {\map \exp {\paren {1 - 3^{-1} } L} }\)
\(\displaystyle \) \(\to\) \(\displaystyle 0\) \(\displaystyle L \to +\infty\) Limit at Infinity of Polynomial over Complex Exponential


Now consider $I_{C_4}$ for $\varepsilon \to 0^+$:

\(\displaystyle \cmod {I_{C_4} }\) \(=\) \(\displaystyle \cmod {\int_{i \varepsilon}^{\varepsilon \sigma} u^q e^{-u} \rd u }\)
\(\displaystyle \) \(\le\) \(\displaystyle \max_{u \mathop \in C_4} \cmod {u^q e^{-u} } \varepsilon \frac {\pi} 2\) Estimation Lemma
\(\displaystyle \) \(=\) \(\displaystyle \max_{u \mathop \in C_4} \cmod {u^q} e^{-\map \Re u} \varepsilon \frac {\pi} 2\) Modulus of Exponential is Exponential of Real Part
\(\displaystyle \) \(\sim\) \(\displaystyle \cmod {\varepsilon^{q + 1} } e^{-\varepsilon} \frac {\pi} 2\) asymptotic equivalence for $\cmod u$ sufficiently small
\(\displaystyle \) \(=\) \(\displaystyle \frac {\varepsilon^{\map \Re {q + 1} } } {e^\varepsilon} \frac {\pi} 2\) Modulus of Positive Real Number to Complex Power is Positive Real Number to Power of Real Part
\(\displaystyle \) \(\to\) \(\displaystyle 0\) as $\varepsilon \to 0^+$, because $\map \Re q > -1$

Thus $I_{C_4} \to 0$ as $\varepsilon \to 0^+$ and $I_{C_2} \to 0$ as $L \to +\infty$.


Thus, taking limits on $I_C = 0$:

$I_C = I_{C_1} - I_{C_3} = 0$


That is, $I_{C_1} = I_{C_3}$ in the limit:

\(\displaystyle \lim_{\varepsilon \mathop \to 0^+} \lim_{L \mathop \to +\infty} \int_{\varepsilon s}^{L s} u^q e^{-u} \rd u\) \(=\) \(\displaystyle \lim_{\varepsilon \mathop \to 0^+} \lim_{L \mathop \to +\infty} \int_{\varepsilon \sigma}^{L \sigma} u^q e^{-u} \rd u\)
\(\displaystyle \) \(=\) \(\displaystyle \int_{\text{positive real axis} } u^s e^{-u} \rd u\)
\(\displaystyle \) \(=\) \(\displaystyle \int_0^{\to +\infty} t^{\paren {s + 1}-1} e^{-t} \rd t\) Complex Riemann Integral is Contour Integral

Thus:

\(\displaystyle \laptrans {t^q}\) \(=\) \(\displaystyle \lim_{\varepsilon \mathop \to 0^+} \lim_{L \mathop \to +\infty} \map I {\varepsilon, L}\)
\(\displaystyle \) \(=\) \(\displaystyle \frac 1 {s^{q + 1} } \int_0^{\to +\infty} t^{\paren {s + 1} - 1} e^{-t} \rd t\)
\(\displaystyle \) \(=\) \(\displaystyle \frac {\map \Gamma {q + 1} } {s^{q + 1} }\)

and the Ansatz is proved correct.

$\blacksquare$