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January 4th, 2017, 08:49 AM   #1
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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))
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January 4th, 2017, 09:36 AM   #2
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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?
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January 5th, 2017, 06:00 AM   #3
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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 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.
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January 5th, 2017, 05:32 PM   #4
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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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