91Ó°ÊÓ

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

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

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

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

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

In the last integral make substitution and find the value.

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

Now, the integral becomes as given below.

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

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

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

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

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

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

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

It is the required result.