Question

Assuming that x is real, show the following relation between the error function and the Fresnel integrals.

$\mathit{e}\mathit{r}\mathit{f}\mathbf{(}\frac{\mathbf{1}\mathbf{-}\mathbf{i}}{\sqrt{\mathbf{2}}}\mathbf{.}\mathbf{x}\mathbf{)}\mathbf{=}\mathbf{(}\mathbf{1}\mathbf{-}\mathbf{i}\mathbf{)}\sqrt{\frac{\mathbf{2}}{\mathbf{\pi }}}\underset{\mathbf{0}}{\overset{\mathbf{x}}{\mathbf{\int }}}\mathbf{(}\mathbf{c}\mathbf{o}\mathbf{s}\mathbf{}{\mathbf{u}}^{\mathbf{2}}\mathbf{+}\mathbf{i}\mathbf{}\mathbf{s}\mathbf{i}\mathbf{n}\mathbf{}{\mathbf{u}}^{\mathbf{2}}\mathbf{)}\mathit{d}\mathit{u}$

Short answer

The relation between error function and Fresnel identity is established as $erf(\frac{1-\mathrm{i}}{\sqrt{2}}.\mathrm{x})=(1-\mathrm{i})\sqrt{\frac{2}{\mathrm{\pi }}}\underset{0}{\overset{\mathrm{x}}{\int }}(\cos {\mathrm{u}}^2+\mathrm{i}\sin {\mathrm{u}}^2)du$.

Step-by-step solution

Given information

It is given that x is real in error function.

Formula of the error function

The formula for the error function is given as$erf\left(\mathrm{x}\right)=\frac{2}{\sqrt{\mathrm{\pi }}}\underset{0}{\overset{\mathrm{x}}{\int }}{e}^{-{\mathrm{t}}^2}dt$

Establish the relation

Use the formula for the error function and substitute x by$\frac{1-i}{\sqrt{2}}$we get role="math" localid="1664350246689" $\frac{2}{\sqrt{\mathrm{\pi }}}\underset{0}{\overset{\frac{1-i}{\sqrt{2}}.x}{\int }}{e}^{-t^2}dt$.

Then put $t=\frac{1-i}{\sqrt{2}}u$.

Which implies, $dt=\frac{1-i}{\sqrt{2}}du$.

Integral becomes as $\frac{2}{\sqrt{\mathrm{\pi }}}\underset{0}{\overset{x}{\int }}{e}^{-\left(\frac{{\left(1-i\right)}^2}{2}.u^2\right)}\frac{1-i}{\sqrt{2}}du$.

$\begin{array}{rcl}{\left(1-i\right)}^2& =& 1-2i-1\\ {\left(1-i\right)}^2& =& -2i\\ -\left(\frac{{\left(1-i\right)}^2}{2}\right)& =& -\frac{-2i}{2}=i\end{array}$

Now the integral becomes as role="math" localid="1664350398123" width="148" height="67">1-i2π∫0xe-iu2du.

Use Euler identity

The integral after using Euler identity is mentioned below.

$\left(1-i\right)\frac{2}{\sqrt{\mathrm{\pi }}}\underset{0}{\overset{x}{\int }}{e}^{-i{u}^2}du=\left(1-i\right)\frac{2}{\sqrt{\mathrm{\pi }}}\underset{0}{\overset{x}{\int }}\left(\cos u^2+i\sin u^2\right)du$

Hence proved.