My Math Forum  

Go Back   My Math Forum > College Math Forum > Calculus

Calculus Calculus Math Forum

LinkBack Thread Tools Display Modes
January 4th, 2017, 07:49 AM   #1
Joined: Mar 2016
From: India

Posts: 14
Thanks: 0


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))
Prudhvi raj k is offline  
January 4th, 2017, 08:36 AM   #2
Senior Member
romsek's Avatar
Joined: Sep 2015
From: USA

Posts: 1,861
Thanks: 968

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?
romsek is online now  
January 5th, 2017, 05:00 AM   #3
Math Team
Joined: Jan 2015
From: Alabama

Posts: 3,109
Thanks: 855

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",, essentially defined as that integral.
Country Boy is offline  
January 5th, 2017, 04:32 PM   #4
Joined: Oct 2016
From: Melbourne

Posts: 77
Thanks: 35

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

$\displaystyle \begin{align*} \int_{-\infty}^{\infty}{ \mathrm{e}^{-\frac{z^2}{2}}\,\mathrm{d}z } &= \sqrt{2}\,\sqrt{\pi} \\ &= \sqrt{2\,\pi} \end{align*}$

Prove It is offline  

  My Math Forum > College Math Forum > Calculus


Thread Tools
Display Modes

Similar Threads
Thread Thread Starter Forum Replies Last Post
integration rose Calculus 8 October 25th, 2017 01:08 AM
recursive integration,integration done, how to get formula Lazar Calculus 1 December 28th, 2014 12:40 PM
integration chuackl Calculus 2 February 4th, 2014 10:53 PM
Integration charmi Calculus 1 January 2nd, 2014 08:51 PM
Integration panky Calculus 4 October 24th, 2012 06:14 AM

Copyright © 2018 My Math Forum. All rights reserved.