91Ó°ÊÓ

The given statement to prove is $\underset{-∞}{\overset{∞}{∫}}{e}^{\frac{-{\mathrm{y}}^2}{2}}dt=\sqrt{2\mathrm{π}}$.

The error of the function is defined as $\mathit{e}\mathit{r}\mathit{f}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{=}\frac{\mathbf{2}}{\sqrt{\mathbf{π}}}\underset{\mathbf{0}}{\overset{\mathbf{x}}{\mathbf{∫}}}{\mathit{e}}^{\mathbf{-}{\mathbf{t}}^{\mathbf{2}}}\mathit{d}\mathit{t}$.

(a)

The function $\sqrt{2\mathrm{π}}\mathrm{ϕ}(∞)=\underset{-∞}{\overset{∞}{∫}}{e}^{\frac{-u^2}{2}}du$.

Solve further to get the required answer.

$\begin{array}{rcl}\sqrt{2\mathrm{π}}\mathrm{ϕ}(∞)& =& \underset{-∞}{\overset{∞}{∫}}{e}^{\frac{-u^2}{2}}du\\ \sqrt{2\mathrm{π}}\left[\frac{1}{2}+\frac{1}{2}erf\left(∞\right)\right]& =& \underset{-∞}{\overset{∞}{∫}}{e}^{\frac{-u^2}{2}}du\\ \sqrt{2\mathrm{π}}& =& \underset{-∞}{\overset{∞}{∫}}{e}^{\frac{-u^2}{2}}du\end{array}$

Or it can be said that role="math" localid="1664345288861" width="135" height="65">∫-∞∞e-y22dy=2π, hence proved.

Part (b)

Convert $\begin{array}{rcl}& & \underset{-∞}{\overset{∞}{∫}}{e}^{\frac{-u^2}{2}}du\end{array}$into gamma function to get the required results.

$\begin{array}{rcl}\underset{-∞}{\overset{∞}{∫}}{e}^{\frac{-u^2}{2}}du& =& \underset{0}{\overset{∞}{2∫}}{e}^{\frac{-u^2}{2}}du\end{array}$

Consider the following:

$$\begin{gathered} s=\frac{u^2}{2} \\ u=\sqrt{2s} \\ du=\frac{1}{\sqrt{2s}}ds \end{gathered}$$

Substitute the above value in $\begin{array}{rcl}& & \underset{0}{\overset{∞}{∫}}{e}^{\frac{-u^2}{2}}du\end{array}$.

$\frac{2}{\sqrt{2}}\begin{array}{rcl}& & \underset{0}{\overset{∞}{∫}}\end{array}\begin{array}{rcl}& & e\end{array}\begin{array}{rcl}& & -s\end{array}\begin{array}{rcl}& & s\end{array}\begin{array}{rcl}& & \frac{-1}{2}\end{array}\begin{array}{rcl}& & d\end{array}\begin{array}{rcl}& & s\end{array}\begin{array}{rcl}& =& \end{array}\begin{array}{rcl}& & \sqrt{2\mathrm{Γ}}\end{array}\begin{array}{rcl}& & \left(\frac{1}{2}\right)\end{array}\begin{array}{rcl}& & \end{array}\begin{array}{rcl}& & \frac{2}{\sqrt{2}}\end{array}\begin{array}{rcl}& & \underset{0}{\overset{∞}{∫}}\end{array}\begin{array}{rcl}& & e\end{array}\begin{array}{rcl}& & {}^{-s}\end{array}\begin{array}{rcl}& & s\end{array}\begin{array}{rcl}& & {}^{\frac{-1}{2}}\end{array}\begin{array}{rcl}& & d\end{array}\begin{array}{rcl}& & s\end{array}\begin{array}{rcl}& =& \end{array}\begin{array}{rcl}& & \sqrt{2\mathrm{Γ}}\end{array}$

Hence, $\sqrt{2\mathrm{π}}=\begin{array}{rcl}& & \underset{-∞}{\overset{∞}{∫}}\end{array}\begin{array}{rcl}& & e\end{array}\begin{array}{rcl}& & \frac{-u^2}{2}\end{array}\begin{array}{rcl}& & d\end{array}\begin{array}{rcl}& & u\end{array}\begin{array}{rcl}& & \end{array}\begin{array}{rcl}& & o\end{array}\begin{array}{rcl}& & r\end{array}\begin{array}{rcl}& & \end{array}\begin{array}{rcl}& & \underset{-∞}{\overset{∞}{∫}}\end{array}\begin{array}{rcl}& & e\end{array}\begin{array}{rcl}& & \frac{-y^2}{2}\end{array}\begin{array}{rcl}& & d\end{array}\begin{array}{rcl}& & y\end{array}\begin{array}{rcl}& =& \end{array}\begin{array}{rcl}& & \sqrt{2\mathrm{π}}\end{array}$equation has been proved.