Question

Using (5.3) with (3.4) and (4.1), find $\mathit{\Gamma }\mathbf{(}\mathbf{3}\mathbf{/}\mathbf{2}\mathbf{)}$,$\mathit{\Gamma }\mathbf{(}\mathbf{-}\mathbf{1}\mathbf{/}\mathbf{2}\mathbf{)}$, and$\mathit{\Gamma }\mathbf{(}\mathbf{-}\mathbf{3}\mathbf{/}\mathbf{2}\mathbf{)}$in terms of$\sqrt{\mathbf{\pi }}$.

Short answer

The solution of the given gamma functions are as follows:

$\begin{array}{l}\Gamma \left(\frac{3}{2}\right)=\frac{\sqrt{\pi }}{3}\\ \Gamma \left(\frac{-1}{2}\right)=-2\sqrt{\pi }\\ \Gamma \left(\frac{-3}{2}\right)=\frac{4}{3}\sqrt{\pi }\end{array}$

Step-by-step solution

Given information

The given Gamma functions are $\Gamma \left(\frac{3}{2}\right),\Gamma \left(-\frac{1}{2}\right)$ and $\Gamma \left(-\frac{3}{2}\right)$. Answers need to be expressed in terms of $\sqrt{\pi }$.

Definition of a Gamma function

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

Write the relevant relations for Gamma functions

Write the relations of a Gamma function.

$\begin{array}{c}\Gamma \left(\frac{1}{2}\right)=\sqrt{\pi }\\ \Gamma \left(x\right)=\frac{1}{x}\Gamma (x+1)\\ \Gamma (x+1)=p\Gamma \left(x\right)\end{array}$

Substitute appropriate values and find the value of Gamma functions

Substitute$\frac{3}{2}$ with into$\Gamma \left(x\right)=\frac{1}{x}\Gamma (x+1)$ and use the relations of gamma functions.

$\begin{array}{c}\Gamma \left(\frac{3}{2}\right)=\frac{1}{2}\Gamma \left(\frac{1}{2}\right)\\ =\frac{\sqrt{\pi }}{2}\end{array}$

Repeat the process by making different substitutions.

Substitute$x$ with$\frac{-1}{2}$ in the equation.

$\begin{array}{c}\Gamma \left(x\right)=\frac{1}{x}\Gamma (x+1)\\ \Gamma \left(\frac{-1}{2}\right)=\frac{1}{\frac{-1}{2}}\Gamma \left(\frac{1}{2}\right)\\ =-2\sqrt{\pi }\end{array}$

Similarly,Substitute x with$\frac{-3}{2}$ in the equation.

$\begin{array}{c}\Gamma \left(x\right)=\frac{1}{x}\Gamma (x+1)\\ \Gamma \left(\frac{-3}{2}\right)=\frac{1}{\frac{-3}{2}}\Gamma \left(\frac{-1}{2}\right)\\ =\frac{4}{3}\sqrt{\pi }\end{array}$

Thus, the required gamma functions are ,$\Gamma \left(\frac{1}{2}\right)=\frac{\sqrt{\pi }}{3}$,$\Gamma \left(\frac{-1}{2}\right)=-2\sqrt{\pi }$ and $\Gamma \left(\frac{-3}{2}\right)=\frac{4}{3}\sqrt{\pi }$.