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

• MHB
Gold Member
MHB
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 $u=\dfrac{x}{\sqrt{2}}$

Gold Member
MHB
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*}.

Gold Member
MHB
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 $$\displaystyle I= \int_0^\infty e^{-\frac{x^2}{2}} dx$$.
Then $$\displaystyle I= \int_0^\infty e^{-\frac{y^2}{2}} dy$$ since that is just changing the name of the "dummy variable".

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

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 $$\displaystyle \theta$$ goes from 0 to $$\displaystyle \frac{\pi}{2}$$. $$\displaystyle x^2+ y^2= r^2$$ and $$\displaystyle dxdy= rdrd\theta$$.

The integral becomes $$\displaystyle \int_0^{2\pi}\int_0^\infty e^{-\frac{r^2}{2}}rdrd\theta$$
$$\displaystyle = 2\pi\int_0^\infty e^{-\frac{r^2}{2}}rdr$$

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