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: {{begin-eqn}} {{eqn | l = u       | r = s t }} {{eqn | r = \paren {\sigma + i \omega} t }} {{eqn | ll= \leadsto       | l = s \rd t       | r = \rd u }} {{eqn | l = u       | r = \paren {\sigma + i \omega} \varepsilon }} {{eqn | ll= \leadsto       | l = t       | r = \varepsilon }} {{eqn | l = u       | r = \paren {\sigma + i \omega} L }} {{eqn | ll= \leadsto       | l = t       | r = L }} {{eqn | l = \int_\varepsilon^L t^q e^{-s t} \rd t       | r = \int_{\sigma \varepsilon \mathop + i \omega \varepsilon}^{\sigma L \mathop + i \omega L} \paren {\frac u s}^q e^{-u} \rd \frac u s }} {{eqn | r = \frac 1 {s^{q + 1} } \int_{\sigma \varepsilon \mathop + i \omega \varepsilon}^{\sigma L \mathop + i \omega L} u^q e^{-u} \rd u }} {{end-eqn}} 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: :600 px 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 $L \to +\infty$: {{begin-eqn}} {{eqn | l = \cmod {I_{C_2} }       | r = \cmod { \int_{L \omega}^{i \sigma \mathop + i L\omega} u^q e^{-u} \rd u} }} {{eqn | o = \le       | r = \max_{u \mathop \in C_2} \cmod {u^q e^{-u} } L       | c = Estimation Lemma }} {{eqn | r = \max_{u \mathop \in C_2} \cmod {u^q} \, \map \exp{-\map \Re u} L       | c = Modulus of Exponential is Exponential of Real Part }} {{eqn | o = \sim       | r = \cmod {L^q} e^{- L} L       | c = asymptotic equivalence for $\cmod u$ sufficiently large }} {{eqn | o = \le       | r = k e^{3^{-1} L} e^{-L}       | c = $L^q$ is of exponential order $3^{-1}$ for some constant $k > 0$, because $\map \Re q > -1$ }} {{eqn | r = \frac {k L} {\map \exp {\paren {1 - 3^{-1} } L} } }} {{eqn | o = \to       | r = 0       | rr= L \to +\infty       | c = Limit at Infinity of Polynomial over Complex Exponential }} {{end-eqn}} Now consider $I_{C_4}$ for $\varepsilon \to 0^+$: {{begin-eqn}} {{eqn | l = \cmod {I_{C_4} }       | r = \cmod {\int_{i \varepsilon}^{\varepsilon \sigma} u^q e^{-u} \rd u } }} {{eqn | o = \le       | r = \max_{u \mathop \in C_4} \cmod {u^q e^{-u} } \varepsilon \frac {\pi} 2       | c = Estimation Lemma }} {{eqn | r = \max_{u \mathop \in C_4} \cmod {u^q}  e^{-\map \Re u} \varepsilon \frac {\pi} 2       | c = Modulus of Exponential is Exponential of Real Part }} {{eqn | o = \sim       | r = \cmod {\varepsilon^{q + 1} } e^{-\varepsilon} \frac {\pi} 2       | c = asymptotic equivalence for $\cmod u$ sufficiently small }} {{eqn | r = \frac {\varepsilon^{\map \Re {q + 1} } } {e^\varepsilon} \frac {\pi} 2       | c = Modulus of Positive Real Number to Complex Power is Positive Real Number to Power of Real Part }} {{eqn | o = \to       | r = 0       | c = as $\varepsilon \to 0^+$, because $\map \Re q > -1$ }} {{end-eqn}} 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: {{begin-eqn}} {{eqn | l = \lim_{\varepsilon \mathop \to 0^+} \lim_{L \mathop \to +\infty} \int_{\varepsilon s}^{L s} u^q e^{-u} \rd u       | r = \lim_{\varepsilon \mathop \to 0^+} \lim_{L \mathop \to +\infty} \int_{\varepsilon \sigma}^{L \sigma} u^q e^{-u} \rd u }} {{eqn | r = \int_{\text{positive real axis} } u^s e^{-u} \rd u }} {{eqn | r = \int_0^{\to +\infty} t^{\paren {s + 1}-1} e^{-t} \rd t       | c = Complex Riemann Integral is Contour Integral }} {{end-eqn}} Thus: {{begin-eqn}} {{eqn | l = \laptrans {t^q}       | r = \lim_{\varepsilon \mathop \to 0^+} \lim_{L \mathop \to +\infty} \map I {\varepsilon, L} }} {{eqn | r = \frac 1 {s^{q + 1} } \int_0^{\to +\infty} t^{\paren {s + 1} - 1} e^{-t} \rd t  }} {{eqn | r = \frac {\map \Gamma {q + 1} } {s^{q + 1} } }} {{end-eqn}} and the Ansatz is proved correct. {{qed}} Category:Laplace Transforms kw5iaugt7lke91v25p8qt2707zvclww