# How to integrate e^(-x^2)?

Alright, I just got out of a calculus final and this was one of the questions on it? As far as I knew this integration was really difficult to do and beyond the scope of a second-year vector calculus course. But since my prof put it on the final, there has to be some way to do it. I mean, he wouldn’t just put a question on the final that was unanswerable, in order to shatter our already small calculus egos. So, anyone out there got any idea how to do it?

And I already typed it in to google and absolutely nothing of any value; except for erf.

Thanks.

No, it can’t be done analytically.

There is, however, a trick that gets you the definite improper integral of exp(-x[sup]2[/sup]) on the interval -infinity…infinity. Is that what was asked?

I should add that the range of integration was between 0 and infinity.

Or 0 to infinity. Otherwise, you look it up in a table or Maple or Mathematica, probably under the name erf (for error function–it is the classical bell-shaped curve). The trick for integrating it from 0 to infinity would be expected from you only if it was done in class. What, you didn’t go to class that day, too bad. Essentially, you write the square of the integral from 0 to infinity as a double integral of e^{-x^2 + y^2}dx dy which is the double integral over the first quadrant of e^{-r^2} r dr d theta, which is easy and turns out, IIRC, to be pi/4 and then taking square root, you get sqrt(pi)/2. The integral from -infinity to infinity is twice that.

Should the exponent on the e in the integral be -x^2 - y^2, since that would be equal to -r^2, because -x^2 + y^2 wouldn’t equal that?

That’s right. It comes from the integral of exp(-x[sup]2[/sup])dx times the integral of exp(-y[sup]2[/sup])dy.

The integral ranges for theta is 0 and pi/2 and for r is 0 and infinity? Am I right to be thinking that?

And I understand the reason for multiplying by e^(-y^2)dy; to get -r^2. But what I don’t get is the theory behind the multiplication. By multiplying the original integrand with something else will this not cause the final answer to be skewed by it. As far as I can tell at no point is the multiplication of e^(-y^2)dy balanced out by something else in the integral.

Maybe it’s just one of those calculus things you do without understanding the meaning behind it.

Kid_A

Int e^-(x^2+y^2) dx dy = Int (e^-x^2)(e^-y^2) dx dy = (Int e^-x^2 dx)(Int e^-y^2 dy)

Perhaps he meant for you to do it using power series?
It’s pretty easy (assuming you know the basics of power series) to come up with a power series for e^(-x^2) and then integrate it term-by-term.

My roommate was thinking about doing it as a power series, but we didn’t learn that this semester, we learned in our previous calculus class and there was no mention of it this semester. So I don’t think that it was the intended way of solving the problem.

RN , I understand what you’re showing but I still don’t understand why the multiplication of (Int e^-y^2 dy) doesn’t have an effect on the final answer.

Let J = Int exp(-x[sup]2[/sup]) dx = Int exp(-y[sup]2[/sup]) dy. Then when you compute the double integral, you’re computing J[sup]2[/sup]. And take the square root when you’re done to get J itself.