Question

Express the complementary error function erfc(x)as an incomplete$\mathit{\Gamma }\mathbf{\text{'}}$function (see Problem 2) and use your result in Problem 2to obtain (again) the asymptotic expansion oferfc(x) as in (10.4) .

Short answer

The several terms of asymptotic series is mentioned below.$erfc\left(x\right)=\frac{e^{-x^2}}{x\sqrt{\mathrm{\pi }}}\left(1-\frac{1}{2x^2}+\frac{3}{8x^4}-\frac{3.5}{16x^6}+...\right)$.

Step-by-step solution

Given Information

The complementary error function is erfc(x).

Definition of Gamma function.

Gamma function behaves similarly to a factorial for natural numbers (a discrete set), its extension to positive real numbers (a continuous set) allows it to be used to model situation involving continuous change.

Gamma function$\mathit{\Gamma }\mathbf{\text{'}}\mathbf{\left(}\mathit{p}\mathbf{\right)}\mathbf{=}\underset{\mathbf{0}}{\overset{\mathbf{\infty }}{\mathbf{\int }}}{\mathit{x}}^{\mathbf{p}\mathbf{-}\mathbf{1}}{\mathit{e}}^{\mathbf{-}\mathbf{x}}\mathit{d}\mathit{x}$.

Find the terms of asymptotic series.

The incomplete gamma function is $\underset{0}{\overset{\infty }{\int }}{t}^{p-1}{e}^{-t}dt=\Gamma \text{'}\left(p,x\right)$.

Substitute the values given below in the function.

$\begin{array}{rcl}t& =& u^2\\ dt& =& 2udu\end{array}$

The function becomes as follows.

$\begin{array}{rcl}\Gamma \text{'}(p.x)& =& \underset{\sqrt{x}}{\overset{\infty }{\int }}{u}^{2p-2}{e}^{-u^2}du\\ & =& 2\underset{\sqrt{x}}{\overset{\infty }{\int }}{u}^{2p-1}{e}^{-u^2}du\end{array}$

For $\begin{array}{rcl}p& =& \frac{1}{2}\end{array}$, The gamma function becomes as follows.

$\begin{array}{rcl}\Gamma \text{'}\left(\frac{1}{2},x\right)& =& 2\underset{\sqrt{x}}{\overset{\infty }{\int }}{u}^0{e}^{-u^2}du\\ & =& 2\underset{\sqrt{x}}{\overset{\infty }{\int }}{e}^{-u^2}du\end{array}$

Express the gamma function as complementary error function.

$\begin{array}{rcl}\frac{1}{\sqrt{\mathrm{\pi }}}\Gamma \text{'}\left(\frac{1}{2},x^2\right)& =& \frac{2}{\sqrt{\mathrm{\pi }}}\underset{\sqrt{x}}{\overset{\infty }{\int }}{e}^{-u^2}du\\ & =& erfc\left(x\right)\end{array}$

Formula states the equation mentioned below.

$\begin{array}{rcl}\Gamma \text{'}(p.x)& =& x^{p-1}{e}^{-x}\left(1+\frac{\left(p-1\right)}{x}+\frac{\left(p-1\right)\left(p-2\right)}{x^2}+\frac{\left(p-1\right)\left(p-2\right)\left(p-3\right)}{x^3}+...\right)\end{array}$

Substitute $\begin{array}{rcl}p& =& \frac{1}{2}\end{array}$in the equation mentioned above.

$\begin{array}{rcl}\Gamma \text{'}\left(\frac{1}{2},x^2\right)& =& \frac{e^{-x}}{x}\left(1-\frac{1}{2x^2}+\frac{3}{8x^3}+...\right)\end{array}$

Substitute the values mentioned above in the equation mentioned below.

$\begin{array}{rcl}erfc\left(x\right)& =& \frac{1}{\sqrt{\mathrm{\pi }}}\Gamma \text{'}\left(\frac{1}{2},x^2\right)\\ & =& \frac{e^{-x}}{x\sqrt{\mathrm{\pi }}}\left(1-\frac{1}{2x^2}+\frac{3}{8x^3}+...\right)\end{array}$

Hence, the several terms of asymptotic series is mentioned below.

$\begin{array}{rcl}erfc\left(x\right)& =& \frac{e^{-x}}{x\sqrt{\mathrm{\pi }}}\left(1-\frac{1}{2x^2}+\frac{3}{8x^3}-\frac{3.5}{16x^6}...\right)\end{array}$.