Question

In the Table of Laplace Transforms (end of Chapter 8, page 469), verify$\mathit{\Gamma }$the function results for L5 and L6. Also show that$\mathit{L}\mathbf{(}\mathbf{1}\mathbf{/}\sqrt{\mathbf{t}}\mathbf{)}\mathbf{=}\sqrt{\mathbf{\pi }\mathbf{/}\mathbf{p}}$.

Short answer

The following are proved:

(a)The Gamma function$\Gamma$ results for $L5,L6$.

(b)$L(1/\sqrt{t})=\sqrt{\pi /p}$

Step-by-step solution

Given Information

The value of a common Gamma function of $\frac{1}{2}$.

$\Gamma \left(\frac{1}{2}\right)=\sqrt{\pi }$

Definition of a Gamma function

The Gamma Function is defined as
$\mathrm{\Gamma }\left(p\right)={\int }_0^{\infty }{x}^{p-1}{e}^{-x}dx,p>0$

Begin with the definition of the Gamma function.

Start with the definition of a Gamma Function.

$\mathbf{\Gamma }\mathbf{\left(}\mathit{p}\mathbf{\right)}\mathbf{=}{\mathbf{\int }}_{\mathbf{0}}^{\mathbf{\infty }}{\mathit{x}}^{\mathbf{p}\mathbf{-}\mathbf{1}}{\mathit{e}}^{\mathbf{-}\mathbf{x}}\mathit{d}\mathit{x}\mathbf{,}\mathbf{}\mathbf{}\mathbf{}\mathit{p}\mathbf{>}\mathbf{0}$

Make the following substitutions in this definition.

$\begin{array}{l}u=pt\\ t=\frac{u}{p}\\ dt=\frac{du}{p}\end{array}$

Simplify after substitutions.

$\begin{array}{c}\underset{0}{\overset{\infty }{\int }}{\left(\frac{u}{p}\right)}^k{e}^{-u}\frac{du}{p}=\frac{1}{p^{k+1}}\underset{0}{\overset{\infty }{\int }}{u}^k{e}^{-u}du\\ =\frac{1}{p^{k+1}}\Gamma (k+1)\end{array}$

Repeat the process for different substitutions

Make the following substitutions in the definition.

$\begin{array}{l}u=(a+p)t\\ t=\frac{u}{a+p}\\ dt=\frac{du}{(a+p)}\end{array}$

Simplify after making the substitutions.

$\begin{array}{c}\underset{0}{\overset{\infty }{\int }}{\left(\frac{u}{a+p}\right)}^k{e}^{-u}\frac{du}{a+p}=\frac{1}{{(a+p)}^{k+1}}\underset{0}{\overset{\infty }{\int }}{u}^k{e}^{-u}du\\ =\frac{\Gamma (k+1)}{{(a+p)}^{k+1}}\end{array}$

Repeat the process again for another substitution

Make the following substitutions in the definition.

$\begin{array}{l}u=pt\\ t=\frac{u}{p}\\ dt=\frac{du}{p}\end{array}$

Simplify after making the substitutions.

$\begin{array}{c}\underset{0}{\overset{\infty }{\int }}{\left(\frac{u}{p}\right)}^{-1/2}{e}^{-u}\frac{du}{p}=\underset{0}{\overset{\infty }{\int }}{u}^{-1/2}{e}^{-u}du\\ =\frac{\Gamma \left(\frac{1}{2}\right)}{p^{1/2}}\\ =\sqrt{\frac{\pi }{p}}\end{array}$

Hence the given statement is proved.