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

karush

Well-known member
Jan 31, 2012
3,179
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
Jan 26, 2012
1,469
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 $\displaystyle \begin{align*} \int_0^{\infty}{ \mathrm{e}^{-\frac{x^2}{2}}\,\mathrm{d}x } = \sqrt{2}\cdot \frac{\sqrt{\pi}}{2} = \frac{\sqrt{2\,\pi}}{2}\end{align*}$.
 

karush

Well-known member
Jan 31, 2012
3,179
actually i didn't know that...
but sure helps in solving the problem
much Mahalo