integral of e^(y^2)

[banana15]

Hi guys,
I have a double integral problem that involves the function e^(y^2). I don't know how to integrate this function with respect to either x or y...when I have googled it I see that it is a Gaussian integral...I am in calculus II and we haven't learned about that yet.
What do I do with this integral?

 

[ksdhart]

Well, integrating that function with respect to x is trivial, because there's no x terms involved:

$\displaystyle \displaystyle \int \:e^{y^2}dx=xe^{y^2}$

Integrating with respect to y, however, is "impossible," in the sense that it doesn't have an elementary antiderivative. If you're familiar with power series, you can get an approximation by using a large number of terms. The Taylor Series expansion is:

$\displaystyle \displaystyle e^{y^2}=\sum _{k=0}^{\infty } \frac{y^{2k}}{k!} = 1+y^2+ \frac{y^4}{2}+ \frac{y^6}{6}+...$

You can integrate the above term-by-term over your bounds of integration, and as I said you'll get a numerical approximation becomes more and more accurate as you add terms. I'd also note that this integral is, in fact, not the Gaussian Integral. For the Gaussian Integral, the exponent must be negative and it must run from negative infinity to positive infinity. Otherwise, you run into the same problem you're facing.

$\displaystyle \displaystyle \int _{-\infty }^{\infty }\:e^{-y^2}dy=\sqrt{\pi }$