# Integration of e^(-x^2)

1. Apr 13, 2005

### C0nfused

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

2. Apr 13, 2005

### 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

3. Apr 13, 2005

### 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.

4. Apr 13, 2005

### 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.

5. Apr 13, 2005

### 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?

6. Apr 13, 2005

### dextercioby

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

Daniel.

7. Apr 14, 2005

### Oggy

That's exactly what I refered to.

8. Apr 14, 2005

### Theelectricchild

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

9. Apr 14, 2005

### 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)

10. Apr 14, 2005

### 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.

11. Apr 14, 2005

### Gokul43201

Staff Emeritus
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 [Broken]

Last edited by a moderator: May 2, 2017