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The definition of an analytic function is: A function $f$ is (real) analytic on an open set $D$ in the real line if for any $x_0\in D$ one can write
$$f(x) = \sum_{n=0}^\infty a_n(x-x_0)^n,$$ in which the coefficients $a_n$ are real numbers and the series is convergent to $f(x)$ for $x$ in a neighborhood of $x_0$.

If we integrate this we get $$\int f(x)\,\mathrm{d}x = \int\sum_{n=0}^\infty a_n(x-x_0)^n\,\mathrm{d}x.$$ Fubinis theorem states that we can interchange an integral with a series if $$\int\sum\lvert f_n\rvert < \infty.$$ Let's assume this is true for our integral and series. [I think we cannot proof it in general.] Now we get $$\int f(x)\,\mathrm{d}x = \sum_{n=1}^\infty\cfrac{a_n}{n+1}(x-x_0)^{n+1}+C = C + (x-x_0)\sum_{n=1}^\infty b_n(x-x_0)^{n},$$ where $b_n = {a_n}/(n+1)$. Since $C\in\mathbb{R}$, $x_0\in D$ and the new sum is per defintion analytic too we can say that:

If the function $f$ satisfies (3), the integral of an analytical function is also analytic.

Would this be a valid proof to show that the integral of an analytic function is analytic too?

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    $\begingroup$ The main problem is: how do you justify that you are integrate term by term? Then, you still need to prove that this new series has the same radius of convergence (which is quite straightforward). But then, yes. $\endgroup$
    – Mushu Nrek
    Aug 8, 2020 at 8:09

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Assuming we already know that an analytic function is infinitely differentiable, this is not too difficult to see.

First (rather than integrating term by term), define the two series $f = \sum_{n=0}^\infty a_n(x-a)^n$ and $F = \sum_{n=0}^\infty \frac{a_n}{n+1}(x-a)^{n+1}$. Then by our differentiation property, $F$ has the same radius of convergence as $f$. From this property again, we also know that $F$ is differentiable, and that its derivative is $f$. Hence $F$ is a primitive of $f$.

Once you have followed this logic you can justify the integration term by term, but as Mushu says you must be able to justify this in the first place. This is one way to do it.

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