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 January 4th, 2017, 08:49 AM #1 Newbie   Joined: Mar 2016 From: India Posts: 14 Thanks: 0 Integration Can you please help me out for this integration ∫e^((〖-z〗^2)/2.)dz (this is integration of e power (-(square of Z) divided by 2))
 January 4th, 2017, 09:36 AM #2 Senior Member     Joined: Sep 2015 From: CA Posts: 594 Thanks: 318 https://en.wikipedia.org/wiki/Normal_distribution your integral is just this with $\mu=0,~\sigma=1$ without the normalizing factor. The integral of the normal distribution with the normalizing factor is 1. So what must your integral be?
 January 5th, 2017, 06:00 AM #3 Math Team   Joined: Jan 2015 From: Alabama Posts: 2,210 Thanks: 555 If you are trying to do this in a "closed form" way, you should realize that almost all elementary functions have integrals that cannot be written in terms of elementary functions! In particular, the integral of $e^{x^2/2}$ cannot be written in terms of elementary functions. Its integral is given in terms of the "error function", https://en.wikipedia.org/wiki/Error_function, essentially defined as that integral.
 January 5th, 2017, 05:32 PM #4 Member   Joined: Oct 2016 From: Melbourne Posts: 37 Thanks: 18 I'll assume that you want the value of the DEFINITE integral \displaystyle \begin{align*} \int_{-\infty}^{\infty}{ \mathrm{e}^{-\frac{z^2}{2}} \,\mathrm{d}z } \end{align*}, because as has been stated before, this does not have an indefinite integral. Start by considering the double integral over the entire real plane \displaystyle \begin{align*} \int{ \int_{\mathbf{R}^2}{ \mathrm{e}^{-\left( x^2 + y^2 \right) } \,\mathrm{d}x }\,\mathrm{d}y } \end{align*}. This can be evaluated by converting to polars: \displaystyle \begin{align*} \int{ \int_{\mathbf{R}^2}{ \mathrm{e}^{-\left( x^2 + y^2 \right) } \,\mathrm{d}x } \,\mathrm{d}y } &= \int_0^{2\,\pi}{ \int_0^{\infty}{ \mathrm{e}^{-r^2} \,r\,\mathrm{d}r } \,\mathrm{d}\theta } \\ &= \int_0^{2\,\pi}{ -\frac{1}{2}\int_0^{\infty}{ -2\,r\,\mathrm{e}^{-r^2} \,\mathrm{d}r } \,\mathrm{d}\theta } \end{align*} Let \displaystyle \begin{align*} u = -r^2 \implies \mathrm{d}u = -2\,r\,\mathrm{d}r \end{align*} and note that \displaystyle \begin{align*} u\left( 0 \right) = 0 \end{align*} and as \displaystyle \begin{align*} r \to \infty , \, u \to -\infty \end{align*}, giving \displaystyle \begin{align*} \int_0^{2\,\pi}{ -\frac{1}{2} \int_0^{\infty}{ -2\,r\,\mathrm{e}^{-r^2}\,\mathrm{d}r } \,\mathrm{d}\theta } &= \int_0^{2\,\pi}{-\frac{1}{2} \int_0^{-\infty}{ \mathrm{e}^u\,\mathrm{d}u } \,\mathrm{d}\theta } \\ &= \int_0^{2\,\pi}{ \frac{1}{2} \int_{-\infty}^0{ \mathrm{e}^u\,\mathrm{d}u } \,\mathrm{d}\theta } \\ &= \int_0^{2\,\pi}{ \frac{1}{2} \lim_{b \to -\infty} \int_b^0{ \mathrm{e}^u\,\mathrm{d}u } \,\mathrm{d}\theta } \\ &= \int_0^{2\,\pi}{\lim_{b \to -\infty} \left[ \mathrm{e}^u \right] _b^0 \,\mathrm{d}\theta } \\ &= \int_0^{2\,\pi}{ \frac{1}{2}\lim_{b \to -\infty} \left( \mathrm{e}^0 - \mathrm{e}^b \right) \,\mathrm{d}\theta } \\ &= \int_0^{2\,\pi}{ \frac{1}{2} \left( 1 - 0 \right) \,\mathrm{d}\theta } \\ &= \int_0^{2\,\pi}{ \frac{1}{2}\,\mathrm{d}\theta } \\ &= \left[ \frac{1}{2}\,\theta \right] _0^{2\,\pi} \\ &= \frac{1}{2} \left( 2\,\pi \right) - \frac{1}{2} \left( 0 \right) \\ &= \pi \end{align*} Now notice that the original double integral can be rearranged: \displaystyle \begin{align*} \int{\int_{\mathbf{R}^2}{ \mathrm{e}^{-\left( x^2 + y^2 \right) }\,\mathrm{d}x }\,\mathrm{d}y} &= \int_{-\infty}^{\infty}{ \int_{-\infty}^{\infty}{ \mathrm{e}^{-x^2 - y^2} \,\mathrm{d}x } \,\mathrm{d}y } \\ &= \int_{-\infty}^{\infty}{ \int_{-\infty}^{\infty}{ \mathrm{e}^{-x^2} \,\mathrm{e}^{-y^2}\,\mathrm{d}x } \,\mathrm{d}y } \\ &= \int_{-\infty}^{\infty}{ \mathrm{e}^{-x^2}\,\mathrm{d}x } \int_{-\infty}^{\infty}{ \mathrm{e}^{-y^2}\,\mathrm{d}y } \end{align*} Now notice that \displaystyle \begin{align*} \int_{-\infty}^{\infty}{\mathrm{e}^{-x^2}\,\mathrm{d}x} \end{align*} is exactly the same integral as \displaystyle \begin{align*} \int_{-\infty}^{\infty}{ \mathrm{e}^{-y^2}\,\mathrm{d}y } \end{align*}, one is just taken over the x axis and the other over the y. But the areas underneath the two integrands will be exactly the same. Thus \displaystyle \begin{align*} \int_{-\infty}^{\infty}{ \mathrm{e}^{-x^2}\,\mathrm{d}x } \int_{-\infty}^{\infty}{ \mathrm{e}^{-y^2}\,\mathrm{d}y } &= \left( \int_{-\infty}^{\infty}{ \mathrm{e}^{-x^2}\,\mathrm{d}x } \right) ^2 \end{align*} and don't forget we have already evalutated the value of this double integral at the beginning, so \displaystyle \begin{align*} \left( \int_{-\infty}^{\infty}{ \mathrm{e}^{-x^2}\,\mathrm{d}x } \right) ^2 &= \pi \\ \int_{-\infty}^{\infty}{ \mathrm{e}^{-x^2}\,\mathrm{d}x } &= \sqrt{\pi} \end{align*} Now, as for your original question, \displaystyle \begin{align*} \int_{-\infty}^{\infty}{ \mathrm{e}^{-\frac{z^2}{2} }\,\mathrm{d}z} \end{align*}, notice that the transformation from \displaystyle \begin{align*} \mathrm{e}^{-x^2} \end{align*} to \displaystyle \begin{align*} \mathrm{e}^{-\frac{x^2}{2}} = \mathrm{e}^{-\left( \frac{x}{\sqrt{2}} \right) ^2} \end{align*} is a dilation of factor \displaystyle \begin{align*} \sqrt{2} \end{align*} in the x direction. When a shape is magnified in ONE dimension, then the area of the shape is also magnified by the same factor. Thus \displaystyle \begin{align*} \int_{-\infty}^{\infty}{ \mathrm{e}^{-\frac{z^2}{2}}\,\mathrm{d}z } &= \sqrt{2}\,\sqrt{\pi} \\ &= \sqrt{2\,\pi} \end{align*} Q.E.D.

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