# 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 $\int e^{x^2} dx$

• 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. Jun 1, 2013 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? Jun 1, 2013 at 20:14

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.
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 careful about 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) — it is the same as interchanging a limit for integration.