Understanding an apparent contradiction from naively applying Taylor's theorem and Fubini's theorem together

Question

Suppose $f(x)$ is a bounded, $\mathcal{C}^{\infty}$ function on $\mathbb{R}$ for which the integral $$I = \int_{0}^{\infty}f(x)\ dx,$$ exists. Taylor's theorem implies $f$ admits a MacLaurin expansion $$f(x) = \sum_{n=0}^{\infty}a_nx^n$$ and therefore $$I = \int_{0}^{\infty}\sum_{n=0}^{\infty}a_nx^n\ dx.$$ When can a sum and integral be interchanged? features an answer which argues the sum and integral can be exchanged (so long as $x^n >0$ for all $x$ and $n$, which is obvious) via Fubini's theorem since the sum can be viewed as an integral with respect to the counting measure on $\mathbb{N}$ and the integral with respect to the Lebesgue measure on $\mathbb{R}$. Thus $$I = \sum_{n=0}^{\infty}a_n\int_{0}^{\infty}x^n\ dx.$$ Now we have constructed a sequence $$s = \left\{a_n \int_{0}^{\infty}x^n\ dx\right\}_{n\in\mathbb{N}}$$ which diverges at every $n$ but whose series must converge by assumption! What went wrong? How is the series convergence justified here?

Answer 1

A real function $f(x)$ may be analytic at $x=0$ with a finite radius of convergence, due to the existence of singularities in the complex plane: take, for instance, $f(x)=\frac{1}{x^2+1}$ or its primitive $\arctan x$. In such a case the Taylor series actually converges to $f(x)$ only if $x$ is sufficiently close to the origin, so, even if $$\forall x\in(-1,1),\qquad \arctan(x) = \sum_{n\geq 0}\frac{(-1)^n}{2n+1}\,x^{2n+1}$$ we are not allowed to state that: $$ \int_{0}^{2}\arctan(x)\,dx = \sum_{n\geq 0}\frac{(-1)^n 2^{2n+2}}{(2n+2)(2n+1)},$$ also because the RHS is not convergent. In such a case, however, the radius of convergence at $x=0$ is $1$ and the radius of convergence at $x=1$ is $\sqrt{2}$, so we may get past the previous issue through: $$ \forall x\in(-\sqrt{2},\sqrt{2}),\qquad \arctan(1+x) = \frac{\pi}{4}+\sum_{n\geq 0}\left(\frac{(-1)^n x^{4n+1}}{2^{2n+1}(4n+1)}-\frac{(-1)^n x^{4n+2}}{2^{2n+1}(4n+2)}+\frac{(-1)^n x^{4n+3}}{2^{2n+2}(4n+3)}\right) $$ and correctly state that: $$ \int_{0}^{2}\arctan(x)\,dx =\\= \sum_{n\geq 0}\frac{(-1)^n}{(2n+1)(2n+2)}+\sum_{n\geq 0}\frac{(-1)^n}{4^{n+1}}\left(\frac{4}{(4n+1)(4n+2)(4n+3)}+\frac{1}{(4n+3)(4n+4)}\right).$$