Integration Help [e^(-x^2)]

[sisxixon]

How would I go about in integrating

e^(-x^2)

from negative infinity to t?

I know what I'm looking for should be a linear equation, but I'm not sure how to get there. This is what I've done so far:

e^(i*x)= cos(x) + i*sin(x)

[e^(i*x)]^(i*x)
= e^(-x^2)
= cos(i*x^2) + i*sin(i*x^2), by De Moirve's theorem.

the problem is, now that I'm at this step I'm stuck. I know that integration by parts wouldn't work, and converting to either polar or spherical coordinates wouldn't be much better. What should I do next, short of using Mathematica or Maple?

 

[galactus]

This is not integrable by elementary means. In other words, it's a booger.

 

[galactus]

Knowing that:

\(\displaystyle \L\\\int_{-\infty}^{\infty}e^{-t^{2}}dt=\sqrt{\pi}\)

Let's integrate:

\(\displaystyle \L\\\int_{x}^{\infty}e^{-t^{2}}dt\) and subtract from \(\displaystyle \sqrt{\pi}\)


Expand in a series of inverse powers of x:

\(\displaystyle \L\\e^{-t^{2}}=\frac{1}{t}te^{\t^{2}}=\frac{1}{t}\frac{d}{dt}(\frac{-1}{2}e^{-t^{2}})\)

Integrate by parts:

\(\displaystyle \L\\\int_{x}^{\infty}e^{-t^{2}}dt=\int_{x}^{\infty}\frac{1}{t}\frac{d}{dt}(\frac{-1}{2}e^{-t^{2}})dt\)

=\(\displaystyle \L\\\frac{1}{t}(\frac{-1}{2}e^{-t^{2}})|_{x}^{\infty}-\int_{x}^{\infty}(\frac{-1}{2}e^{-t^{2}})(\frac{-1}{t^{2}})dt\)

=\(\displaystyle \L\\\frac{1}{2x}e^{-x^{2}}-\frac{1}{2}\int_{x}^{\infty}\frac{1}{t^{2}}e^{-t^{2}}dt\)...[1]

In the last integral, write \(\displaystyle \L\\(\frac{1}{t^{2}})e^{-t^{2}}=(\frac{1}{t^{3}}(\frac{d}{dt})(\frac{-1}{2}e^{-t^{2}})\).

Integrate by parts again, continue the process and sub [1] and the steps following it back to the beginning.. You will get a series related to the erf function

What did I tell you. It's a real sweetheart, isn't it?.

I found this in an old Mathematical Physics text. Good luck.

 

[sisxixon]

WOW, she's a beaut!

Thank you so much for taking the time to look this up. Which text did you find it in?

 

[galactus]

If you root around, you can probably find concise explanations for various versions of the Gaussian distribution. I found this in
Mathematical Methods in the Physical Sciences" 2nd Ed., Mary Boas, Wiley publisher.

The proof for \(\displaystyle \int_{-\infty}^{\infty}e^{-x^{2}}dx\) is straightforward compared to this critter. If you want it, it's easy by comparison.

You know, the standard normal table in the stats books we all use to look up z-scores?. That is derived from:

\(\displaystyle \L\\\frac{1}{\sqrt{2{\pi}}}\int_{-\infty}^{x}e^{\frac{-t^{2}}{2}}dt\)

I always wanted to know how to do that one 'by hand', but have not seen it anywhere. I never tried to figure it out myself. Too much!!!.

 

[sisxixon]

Could I have the proof for

, too, please? It's not required for this assignment, but it'd be great to see how it works, all the same. Nifty stuff, this is.

I'm also fiddling around to see how the standard normal table was found, but it looks like I'll have to keep looking.

 

[galactus]

Here's the version I like, but you have to promise if you find a techinque for deriving the solutions to the standard normal table(no calculators), you let me know.

Start out with:

\(\displaystyle \L\\\left(\int_{-\infty}^{\infty}e^{-x^{2}}dx\right)^{2}\\=\int_{-\infty}^{\infty}e^{-x^{2}}dx\int_{-\infty}^{\infty}e^{-y^{2}}dy\\=\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}e^{-x^{2}}e^{-y^{2}}dydx\\\text{remember the polar thing, x^{2}+y^{2}=r^{2}}\\=\int_{0}^{\infty}\int_{0}^{2{\pi}}e^{-r^{2}}rd{\theta}dr\\=\int_{0}^{2{\pi}}d{\theta}\int_{0}^{\infty}e^{-r^{2}}rdr\\=2{\pi}\int_{0}^{\infty}\frac{e^{-u}}{2}du={\pi}\)

\(\displaystyle \text{which gives:}\)

=\(\displaystyle \L\\\int_{-\infty}^{\infty}e^{-x^{2}}dx=\sqrt{\pi}\)


Pretty cool, huh?.

 

[sisxixon]

I promise. Thanks again!