Laplace Transform of Complex Power

Theorem

Let $q$ be a constant complex number with $\operatorname{Re} q > -1$

Let:

$t^q: \left({0\,.\,.\,\to}\right) \to \C$

be a branch of the complex power multifunction chosen such that $f$ is continuous on the half-plane $\operatorname{Re} s > 0$.

Then $f$ has a Laplace transform given by:

$\mathcal L\left\{{t^q}\right\}(s) = \dfrac{\Gamma\left({q+1}\right)}{s^{q+1}}$

where $\Gamma$ denotes the gamma function.

Proof

$\displaystyle \mathcal L\left\{{t^q}\right\} = \lim_{\varepsilon \mathop \to 0^+}\lim_{L \mathop \to +\infty}I\left({\varepsilon,L}\right)$

where:

$\displaystyle I\left({\varepsilon,L}\right) = \int_{\varepsilon}^{L} t^q e^{-s t} \rd t$.

Recall the case where $q$ is a positive integer:

$\displaystyle \mathcal L \left\{ {t^n} \right\} = \frac {n!} { s^{n+1} }$

Because the gamma function extends the factorial, a reasonable Ansatz is:

$\displaystyle \mathcal L \left\{ {t^q} \right\} = \frac {\Gamma\left({q+1}\right)} { s^{q+1} }$

From the poles of the gamma function, $\Gamma\left({-1}\right)$ is undefined.

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

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

To the end of expressing $\displaystyle I \left({\varepsilon,L}\right)$ to a form similar to the integral defining $\Gamma$, we view $I\left({\varepsilon,L}\right)$ as a contour integral.

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

By the lemma, the integrand for $I\left({\varepsilon,L}\right)$ is analytic.

Thus the conditions for integration by substitution are satisfied for $\operatorname{Re}s > 0$.

Substitute:

 $\displaystyle u$ $=$ $\displaystyle st$ $\quad$ $\quad$ $\displaystyle$ $=$ $\displaystyle \left({\sigma + i \omega}\right)t$ $\quad$ $\quad$ $\displaystyle \implies \ \$ $\displaystyle s \rd t$ $=$ $\displaystyle \rd u$ $\quad$ $\quad$ $\displaystyle u$ $=$ $\displaystyle \left({\sigma + i \omega}\right)\varepsilon$ $\quad$ $\quad$ $\displaystyle \implies \ \$ $\displaystyle t$ $=$ $\displaystyle \varepsilon$ $\quad$ $\quad$ $\displaystyle u$ $=$ $\displaystyle \left({\sigma + i \omega}\right)L$ $\quad$ $\quad$ $\displaystyle \implies \ \$ $\displaystyle t$ $=$ $\displaystyle L$ $\quad$ $\quad$ $\displaystyle \int_{\varepsilon}^{L} t^q e^{-s t} \rd t$ $=$ $\displaystyle \int_{\sigma \varepsilon \mathop + i \omega \varepsilon}^{\sigma L \mathop + i \omega L} \left({\frac u s}\right)^q e^{-u} \rd \frac{u}{s}$ $\quad$ $\quad$ $\displaystyle$ $=$ $\displaystyle \frac{1}{ s^{q+1} }\int_{\sigma \varepsilon \mathop + i \omega \varepsilon}^{\sigma L \mathop + i \omega L} u^q e^{-u} \rd u$ $\quad$ $\quad$

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 a line segment connecting $\sigma \varepsilon$ to $L\sigma$
$C_2$ is a line segment connecting $L\sigma$ to $L\sigma + i L \omega$
$C_3$ is a line segment connecting $i \varepsilon$ to $L\sigma + i L \omega$
$C_4$ is a circular arc connecting $i\varepsilon$ to $\sigma \varepsilon$

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:

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 $L \to +\infty$:

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

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

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

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:

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

Thus:

 $\displaystyle \mathcal L \{t^q\}$ $=$ $\displaystyle \lim_{\varepsilon \mathop \to 0^+} \lim_{L \mathop \to +\infty} I \left({\varepsilon, L}\right)$ $\quad$ $\quad$ $\displaystyle$ $=$ $\displaystyle \frac 1 { s^{q+1} } \int_0^{\to +\infty} t^{\left({s+1}\right)-1} e^{-t} \rd t$ $\quad$ $\quad$ $\displaystyle$ $=$ $\displaystyle \frac {\Gamma\left({q+1}\right)} {s^{q+1} }$ $\quad$ $\quad$

... and the Ansatz is proved correct.

$\blacksquare$