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

[karush]

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]

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]

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

 

[HOI]

Here is the rather clever way that integral can be done.
Let [math]I= \int_0^\infty e^{-\frac{x^2}{2}} dx[/math].
Then [math]I= \int_0^\infty e^{-\frac{y^2}{2}} dy[/math] since that is just changing the name of the "dummy variable".

So [math]I^2= \left( \int_0^\infty e^{-\frac{x^2}{2}} dx\right)\left(\int_0^\infty e^{-\frac{y^2}{2}} dy\right)[/math]
[math]I^2= \int_0^\infty\int_0^\infty e^{-6\frac{x^2+ y^2}{2}} dxdy[/math].

To integrate that, change to polar coordinates! Since x goes from 0 to infinity and y goes from 0 to infinity, covering the first quadrant, in polar coordinates r goes from 0 to infinity and [math]\theta[/math] goes from 0 to [math]\frac{\pi}{2}[/math]. [math]x^2+ y^2= r^2[/math] and [math]dxdy= rdrd\theta[/math].

The integral becomes [math]\int_0^{2\pi}\int_0^\infty e^{-\frac{r^2}{2}}rdrd\theta[/math]
[math]= 2\pi\int_0^\infty e^{-\frac{r^2}{2}}rdr[/math]

Now let [math]u= \frac{r^2}{2}[/math] so that [math]du= r dr[/math] so the integral becomes
[math]2\pi\int_0^\infty e^{-u}du= 2\pi\left[-e^{-u}\right]_0^\infty= 2\pi\left[-(0- 1)\right]= 2\pi[/math].