Laplace Transform of Complex Power

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:

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 $L \to +\infty$:

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

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:

Thus:

and the Ansatz is proved correct.