7.1.17 int e^{-\dfrac{x^2}{2}} dx from 0 to infty

karush

Well-known member
Evaluate $\displaystyle\int_0^\infty e^{-\dfrac{x^2}{2}} dx$

ok first reponse is use IBP but can we use $e^u$ where $u=-\dfrac{x^2}{2}$ ot $u=\dfrac{x}{\sqrt{2}}$

Prove It

Well-known member
MHB Math Helper
It's well known that \displaystyle \begin{align*} \int_{-\infty}^{\infty}{\mathrm{e}^{-x^2}\,\mathrm{d}x} = \sqrt{\pi} \end{align*}, and due to the evenness of this function, that means \displaystyle \begin{align*} \int_0^{\infty}{\mathrm{e}^{-x^2}\,\mathrm{d}x} = \frac{\sqrt{\pi}}{2} \end{align*}.

If we compare the functions \displaystyle \begin{align*} f\left( x \right) = \mathrm{e}^{-x^2} \end{align*} and \displaystyle \begin{align*} g\left( x \right) = \mathrm{e}^{-\left( \frac{x}{\sqrt{2}} \right) ^2 } \end{align*}, we can see that \displaystyle \begin{align*} g\left( x \right) \end{align*} is the image of \displaystyle \begin{align*} f\left( x \right) \end{align*} after a dilation by factor \displaystyle \begin{align*} \sqrt{2} \end{align*} from the \displaystyle \begin{align*} y \end{align*} axis. Therefore, their integrals are also dilated by factor \displaystyle \begin{align*} \sqrt{2} \end{align*}.

Therefore \displaystyle \begin{align*} \int_0^{\infty}{ \mathrm{e}^{-\frac{x^2}{2}}\,\mathrm{d}x } = \sqrt{2}\cdot \frac{\sqrt{\pi}}{2} = \frac{\sqrt{2\,\pi}}{2}\end{align*}.

• topsquark and karush

karush

Well-known member
actually i didn't know that...
but sure helps in solving the problem
much Mahalo