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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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    $\begingroup$ 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. $\endgroup$ Jun 1, 2013 at 20:03
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    $\begingroup$ 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? $\endgroup$
    – N. S.
    Jun 1, 2013 at 20:14

2 Answers 2

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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.

However, certainly for purposes of numerical integration but also for purposes of studying the solution, this method is very useful. Power series are great!

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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.

Finally, you have to be able to calculate the infinite series you end up with.

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    $\begingroup$ The nice thing with power series is that they converge absolutely in their range of convergence and can be integrated and differentiated term by term. So, whenever a function admits a power series representation, it can be integrated there term wise. $\endgroup$ Jun 1, 2013 at 21:49

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