1. ## Integral hint

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

2. It's well known that such integral has no elementary primitive.

Makes sense if you're lookin' for $\displaystyle \int_{ - \infty }^{ + \infty } {e^{ - y^2 } \,dy} .$

3. Originally Posted by Pinsky How to solve $\displaystyle I=\int e^{y^2} dy$
You could expand this out as a power series, though, if you want.

-Dan

4. Originally Posted by Krizalid It's well known that such integral has no elementary primitive.

Makes sense if you're lookin' for $\displaystyle \int_{ - \infty }^{ + \infty } {e^{ - y^2 } \,dy} .$
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.

5. Originally Posted by mr 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.
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.

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