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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?

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JMJ
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    seriously, downvoting without even leaving a comment? What is so stupid about this question? – JMJ May 30 '16 at 19:02
  • Don't you need all $a_n$ to be positive too? –  May 30 '16 at 19:04
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    Many many problems. Taylor's theorem does not show that $f$ is equal to its Talylor series! You should learn what Taylor's theorem actually says. To apply Fubini that way you need $a_n\ge0$, as Rahul pointed out. Now if $f$ equals its Talor series and if $a_n\ge0$, and if $f$ is not identically $0$ then $I=\infty$, making the whole thing more or less vacuous. – David C. Ullrich May 30 '16 at 19:08
  • Thank you Rahul...I appear to have missed that! Thanks also David. I see you are a math professor; perhaps you can point me and anyone else who sees this with the same confusion to a resource to learn what Taylor's theorem actually says, because it really isn't discussed in intro books and the expansion of functions like I wrote (at least formally) is done all the time in physics and engineering. Regarding your tone, I don't see the point of being so cross, but thank you nonetheless for giving me the opportunity to learn more about the fundamentals. – JMJ May 30 '16 at 19:22
  • https://en.wikipedia.org/wiki/Taylor%27s_theorem – David C. Ullrich May 30 '16 at 19:30

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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: $$ \int_{0}^{2}\arctan(x)\,dx =\\= \sum_{n\geq 0}\frac{(-1)^n}{(2n+1)(2n+2)}+\sum_{n\geq 0}\frac{(-1)^n}{4^{n+1}}\left(\frac{4}{(4n+1)(4n+2)(4n+3)}+\frac{1}{(4n+3)(4n+4)}\right).$$

Jack D'Aurizio
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