Main Page

From ProofWiki
Jump to navigation Jump to search

Welcome to $\mathsf{Pr} \infty \mathsf{fWiki}$


$\mathsf{Pr} \infty \mathsf{fWiki}$ is an online compendium of mathematical proofs! Our goal is the collection, collaboration and classification of mathematical proofs. If you are interested in helping create an online resource for math proofs feel free to register for an account. Thanks and enjoy!

If you have any questions, comments, or suggestions please post on the discussion page, or contact one of the administrators. Also, feel free to take a look at the frequently asked questions because you may not be the first with your idea.

To see what's currently happening in the community, visit the community portal.

24,222 Proofs 18,937 Definitions Help

Featured Proof

Laplace Transform of Complex Power


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

Let $t^q$ be the the principal branch of the $q$-th complex power whose domain contains the half-plane $\map \Re s > 0$.

Then $t^q$ has a Laplace transform given by:

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

where $\Gamma$ denotes the gamma function.


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

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


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

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 $\map \Re s > 0$.


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


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

Consider the contour:

$C = C_1 + C_2 - C_3 - C_4$


$C_1$ 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:


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 $L \to +\infty$:

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

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

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

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


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

and the Ansatz is proved correct.