# Why can't you find all antiderivatives by integrating a power series?

if

$f(x) = \sum\limits^{\infty}_{n=0}\frac{f^{(n)}(0)}{n!}x^n$

why can't you do the following to find a general solution

$F(x) \equiv \int f(x)dx$

$F(x) = \int (\sum\limits^{\infty}_{n=0}\frac{f^{(n)}(0)}{n!}x^n) dx = \sum\limits^{\infty}_{n=0}\frac{f^{(n)}(0)}{n!}(\int x^n dx) = \sum\limits^{\infty}_{n=0}\frac{f^{(n)}(0)}{n!}(\frac{x^{n+1}}{n+1}) = \sum\limits^{\infty}_{n=0}\frac{f^{(n)}(0)}{(n+1)!}x^{n+1}$

I was wondering about this because I tried this approach to finding the antiderivative $\int e^{x^2} dx$

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That approach will give you a series that converges to $\int e^{x^2} \, dx$, sure. You just won't be able to convert that series to an elementary function, since $e^{x^2}$ doesn't have an elementary antiderivative. – Henry T. Horton Jun 1 '13 at 20:03
You can, but most of the times the answer doesn't help you. Note for example that for any continuous function $f$ we know the anidetivative: $\int_a^x f(t)dt +C$. But this leads to the same issue, what is really this function? – N. S. Jun 1 '13 at 20:14

First of all $\int f(x) dx$ is a collections of functions, not a function. You should rather define $F(x) = \int_{t = 0}^x f(t) dt$. Then you should be carefull abou interchanging $\int$ and $\sum_{n=0}^\infty$. This is true for finite sums but not always for infinite series of integral (look e.g. Fatou-Lebesgue theorem - http://en.wikipedia.org/wiki/Fatou%E2%80%93Lebesgue_theorem) - it is the same as interchanging limit and integration. Finally you have to be able to calculate the infinite series you end up with.
If the integrand $f(x)$ can be represented as a power series (let's say with infinite radius of convergence) like $f(x)=e^{x^2}$ can, then you can use that power series representation, just like you say, to obtain a power series representation of the integral $\int f(x)dx$. But that does not mean you will be able to figure out an elementary expression for that power series. Sometimes, the integral simply does not have such an expression.