## integrate exp(-x^2), dx

hi all,

i've tried to solve this thing with Derive, but it gave me some vague erf(x) function (error function??). Is there some gosu-mathematician who can help me solve the integral?

$$\int exp(-x^2) dx$$

tnx
 Recognitions: Gold Member Science Advisor Staff Emeritus There is no analytic form to the gaussian integral. You need to look up values in the erf table for definite integrals.
 ah, that clarifies a lot, thank you.

## integrate exp(-x^2), dx

Tom did this using double integration here: http://www.physicsforums.com/showthr...t=25798&page=2

Have fun!

 Quote by Tom Mattson This integral can be done the same way that the integral of exp(-x2) can be done. First, write the integral of x2exp(-x2) from zero to infinity. Then write the integral of y2exp(-y2) from zero to infinity (they're both exactly the same as your integral). Now multiply the integrands together double integrate over x and y. When you convert to polar coordinates, you will get an integral that can be done by parts. Just don't forget to take the square root at the end. Note: x2y2=r4sin2(θ)cos2(θ) x2+y2=r2 dx dy=rdr dθ

 Quote by NSX Tom did this using double integration here: http://www.physicsforums.com/showthr...t=25798&page=2 Have fun!
Remember, that that only works for solving certain definite integrals.

 Quote by master_coda Remember, that that only works for solving certain definite integrals.
Why's that?
 It's the limits of integration that count here. Say you're trying to integrate a hard function, but there's a neat little trick for working out the integral from zero to infinity. That trick probably won't help you if you're integrating from, say, 1 to 5.729.
 When I first did that integral (the trick way with the nice limits) I thought it was the neatest thing.
 Hello to you all, i've tried hard to solve this problem related with the wind resources, but so far like Tom Mattson said in is post, i solved the problem to k=2, but i can't solve it to any k>0!!! integrate exp(-(x/C)^k), dx with k>0 and C>0 Is there anyone willing to help me ???? Perinhas.