Is the integral of a real analytic function analytic? If it is so, how do you go about the proof? In particular, what guarantees that the Taylor series expansion of this integral converges to itself?
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1Are you familiar with the definition of analyticity? – Vim Oct 15 '17 at 17:34
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@Vim Yes. I am. – Oct 15 '17 at 17:40
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if so, you should have mentioned power series instead of Taylor series. Note the fact that power series converges uniformly within a radius smaller than its radius of convergence, then integrate term by term. – Vim Oct 15 '17 at 17:43
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@Vim https://www.encyclopediaofmath.org/index.php/Real_analytic_function#Taylor_series This page says that the power series expansion of an analytic function is the same as its Taylor series expansion. – Oct 15 '17 at 17:50
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usually I would take this as meaning their coefficients are identical. But in strict terminology, the Taylor series is a "pointwise" expression without any specification of "radius of convergence" (in fact in the notorious example $f(x)=e^{-1/x^2}$ it has zero Taylor series about $0$, but has no power series expansion around $0$ at all), whereas the power series is defined on a non-degenerate interval with strictly positive radius of convergence. The intrinsic difference between them is huge. – Vim Oct 15 '17 at 17:59
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@Vim Thanks. Got it! – Oct 15 '17 at 18:05