Is there an easy way to find the value of e^(-x^2) using numerical methods?

[C0nfused]

Hi everybody,
Do you have any idea how this is solved?
\int_{-\infty}^{+\infty} e^{-x^2} dx =?
Thanks

 

[matt grime]

yes. first note the integral exists, let its value be T.

By some theorem

T^2 = \int_{-\infty}^{+\infty} e^{-x^2} dx\int_{-\infty}^{+\infty} e^{-y^2} dy

which equals

\int_{-\infty}^{+\infty} e^{-x^2-y^2} dxdy

now put it into polars

 

[Oggy]

I=\int_{0}^{+\infty} e^{-x^2} dx, is solved using a trick, and integrating in polar coordinates, in which the Jacobian helps solving it. It is equal to \frac{\sqrt{\pi}}{2}. Your integral is just 2I, I think.

 

[dextercioby]

Oggy,the "trick" applies for his integral.Yours,if u apply Fubini's theorem,will not lead to an integral over \mathbb{R}^{2},but over the semiplane x\geq 0 and it wouldn't be the same...

To evaluate that integral (and to get the value u wrote),u need another "trick":the integrand is even,therefore

\int_{0}^{+\infty} e^{-x^{2}} \ dx =\frac{1}{2}\int_{\mathbb{R}} e^{-x^{2}} \ dx



Daniel.

 

[Theelectricchild]

elaborating with what dexter had stated:

\int_{0}^{\infty} e^{-x^2} dx

Convert to DI problem:

{(\int_{0}^{\infty} e^{-x^2} dx)}^2

=(\int_{0}^{\infty} e^{-x^2} dx) (\int_{0}^{\infty} e^{-y^2} dy)

=\int_{0}^{\infty}\int_{0}^{\infty} e^{-x^2}e^{-y^2} dxdy

=\int_{0}^{\infty}\int_{0}^{\infty} e^{-{(x^2+y^2)}} dxdy

From here using a Change of Variables step or noticing the region we are integrating is simply a semicircular arc, we can use polar coordinates:

=\int_{0}^{\frac{\pi}{2}}\int_{0}^{R} e^{-r^2}r drd\theta

Where we take the limit of this as R --> infinity and the extra r in the integrand is from the Jacobian when switching to polar coordinates (proven in the change of variables theorem)

And NOW we can use a U-substitution to easily solve this iterated double integral:

{(\int_{0}^{\infty} e^{-x^2} dx)}^2 = \frac{\pi}{4}

Thus:

\int_{0}^{\infty} e^{-x^2} dx = \frac{\sqrt \pi}{2}

It seems like we can solve quite a few "definite" or "improper" integrals of non elementary functions using multivar. calculus. Does anyone know any other methods?

 

[dextercioby]

The theorem of residues of A.Cauchy never fails...

Daniel.

 

[Oggy]

That's exactly what I referred to.

 

[Theelectricchild]

Well... you didn't explicitly state residue calculus now did you

 

[C0nfused]

Thanks for your help. I am not very familiar with integration theorems but i wanted to calculate this integral because the normal distibution function has a term like this
(e^(-x^2)) and the integral is 1. I checked it in Mathematica and the indefinite integral of the function g(x)=e^(-x^2) is (1/2)*[(pi)^(1/2)]*erf(x). Do u know what this erf(x) is? (a simple explanation as I am not very advanced in analysis)

 

[dextercioby]

The way i know it (it coincides with the definition swallowed by Mathematica and my Maple)

\mbox{erf} (x)=:\frac{2}{\sqrt{\pi}}\int_{0}^{x} e^{-t^{2}} \ dt



Daniel.

 

[Gokul43201]

Numerical values of erf(x) or the "error function" can be readily looked up in any book of math tables.

http://www.uni-mainz.de/FB/Geo/Geologie/Geophysik/Lithosphere/erfTable.htm