How to solve $\displaystyle I=\int e^{y^2} dy$

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- Jan 23rd 2008, 06:19 AMPinskyIntegral hint
How to solve $\displaystyle I=\int e^{y^2} dy$

- Jan 23rd 2008, 06:47 AMKrizalid
It's well known that such integral has no elementary primitive.

Makes sense if you're lookin' for $\displaystyle \int_{ - \infty }^{ + \infty } {e^{ - y^2 } \,dy} .$ - Jan 23rd 2008, 09:12 AMtopsquark
- Jan 23rd 2008, 11:46 AMmr fantastic
Liouville first proved in 1835 that if f (x) and g(x) are rational functions (where g(x) is not a constant), then $\displaystyle \int f(x) e^{g(x)} \, dx$ is elementary if and only if there exists a rational function R(x) such that $\displaystyle f (x) = \frac{dR}{dx} + R(x)g(x)$.

There are several other theorems that can prove that integrals of various sorts have no elementary primitive. Here is the briefest of brief reviews that barely scratches the surface. - Jan 23rd 2008, 12:42 PMThePerfectHacker
There is even an entire branch to math on this. Part of Galois theory answers which polynomials can be solved and which cannot. Similarly, differencial Galois theory address the question which primitive can be solved and which cannot. It looks interesting to me, because algebra is what I study the most, maybe one day I learn more about it.