Question

Carry through the algebra to get the equation

$\mathit{e}\mathit{r}\mathit{f}\mathit{c}\left(x\right)\mathbf{=}\mathbf{1}\mathbf{-}\mathit{e}\mathit{r}\mathit{f}\left(x\right)\mathbf{\square }\frac{{\mathbf{e}}^{\mathbf{-}{\mathbf{x}}^{\mathbf{2}}}}{\mathbf{x}\sqrt{\mathbf{\pi }}}\left(1-\frac{1}{2x^2}+\frac{1.3}{{\left(2x^2\right)}^2}-\frac{1.3.5}{{\left(2x^2\right)}^3}+.....\right)$

Short answer

The required result $erfc\left(\mathrm{x}\right)=1-erf\left(\mathrm{x}\right)\square \frac{{\mathrm{e}}^{-{\mathrm{x}}^2}}{\mathrm{x}\sqrt{\mathrm{\pi }}}(1-\frac{1}{2{\mathrm{x}}^2}+\frac{1.3}{{\left(2{\mathrm{x}}^2\right)}^2}-\frac{1.3.5}{{\left(2{\mathrm{x}}^2\right)}^3}+.....)$ is derived through algebra.

Step-by-step solution

Given Information

It is given to use the algebra to derive the required expression.

Concept and Formula

The formula of error function to be used is given belowresult.

$\mathit{e}\mathit{r}\mathit{f}\mathit{c}\left(x\right)\mathbf{=}\sqrt{\frac{\mathbf{4}}{\mathbf{\pi }}}\left(\frac{e^{-x^2}}{2x}-\frac{e^{-x^2}}{4x^3}+\frac{3}{4}\underset{x}{\overset{\infty }{\int }}{t}^{-4}{e}^{-t^2}dt\right)$

By the method of substitution find the values of integrals and then use above formula.

Make necessary substitution

Use the result $erfc\left(\mathrm{x}\right)=\sqrt{\frac{4}{\mathrm{\pi }}}(\frac{{\mathrm{e}}^{-{\mathrm{x}}^2}}{2\mathrm{x}}-\frac{{\mathrm{e}}^{-{\mathrm{x}}^2}}{4{\mathrm{x}}^3}+\frac{3}{4}\underset{\mathrm{x}}{\overset{\infty }{\int }}{\mathrm{t}}^{-4}{\mathrm{e}}^{-{\mathrm{t}}^2}\mathrm{dt})..........\left(1\right)$.

In the last integral make substitution and find the value.

$$\begin{gathered} \left(\underset{\mathrm{x}}{\overset{\infty }{\int }}{\mathrm{t}}^{-4}{\mathrm{e}}^{-{\mathrm{t}}^2}\mathrm{dt}\right)=-\frac{1}{2}\underset{x}{\overset{\infty }{\int }}{t}^{-5}\frac{d}{dt}{e}^{-t^2}dt \\ u=t^{-5} \\ du=-5t^{-6}dt \\ v=e^{-t^2} \\ v\text{'}=\frac{d}{dt}{e}^{-t^2} \end{gathered}$$

Now, the integral becomes as given below.

role="math" localid="1664358381158" $\begin{array}{rcl}\left(\underset{\mathrm{x}}{\overset{\infty }{\int }}{\mathrm{t}}^{-4}{\mathrm{e}}^{-{\mathrm{t}}^2}\mathrm{dt}\right)& =& -\frac{1}{2}\left[{\mathrm{t}}^{-4}{\mathrm{e}}^{-{\mathrm{t}}^2}\overline{){}_x^{\infty }}\underset{x}{\overset{\infty }{\int }}\frac{-5}{t^6}.e^{-t^2}dt\right]\\ & =& -\frac{1}{2}\left[-x^{-4}{\mathrm{e}}^{-x^2}+5\underset{x}{\overset{\infty }{\int }}\frac{1}{t^6}.e^{-t^2}dt\right]\end{array}$

Substitute the value of integral

Use equation 1 and substitute the value of integral $\begin{array}{rcl}& & \left(\underset{\mathrm{x}}{\overset{\infty }{\int }}{\mathrm{t}}^{-4}{\mathrm{e}}^{-{\mathrm{t}}^2}\mathrm{dt}\right)\end{array}$.

$erfc\left(x\right)=\sqrt{\frac{4}{\mathrm{\pi }}}\left(\frac{e^{-x^2}}{2x}-\frac{e^{-x^2}}{4x^3}+\frac{3e^{-x^2}}{8x^5}-\frac{3.5}{8}\underset{z}{\overset{\infty }{\int }}\frac{1}{t^6}{e}^{-t^2}dt\right)..........\left(2\right)$

Follow the same procedure to find $\begin{array}{rcl}& & \left(\underset{\mathrm{x}}{\overset{\infty }{\int }}\frac{1}{t^6}.{\mathrm{e}}^{-{\mathrm{t}}^2}\mathrm{dt}\right)\end{array}$.

$\begin{array}{rcl}\left(\underset{x}{\overset{\infty }{\int }}{t}^{-6^2}{e}^{-t^2}dt\right)& =& -\frac{1}{2}\left[t^{-7}{e}^{-t^2}\overline{){}_x^{\infty }}-\underset{x}{\overset{\infty }{\int }}\frac{-7}{t^8}.e^{-t^2}dt\right]\\ & =& -\frac{1}{2}\left[-x^{-7}{e}^{-x^2}+7\underset{x}{\overset{\infty }{\int }}\frac{-7}{t^8}.e^{-t^2}dt\right]\end{array}$

Use equation and substitute the value of integral $\begin{array}{rcl}& & \left(\underset{x}{\overset{\infty }{\int }}\frac{1}{t^6}.e^{-t^2}dt\right)\end{array}$.

$\begin{array}{rcl}erfc\left(x\right)& =& -1erf\left(x\right)\square \frac{e^{-x^2}}{\sqrt{\mathrm{\pi }}x}\left[1-\frac{1}{2x^2}+\frac{3}{4x^4}-\frac{3.5}{8x^6}+.....\right]\end{array}$

It is the required result.