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I came across this question:

What's the domain of $f(x)=\int (x+2x^2+3x^3+\dots) dx$?

I think it'd be a polynomial, something like $\frac{x^2}{2}+\frac{2x^3}{3}+\dots$, which domain is all real numbers ($\mathbb R$) but it's not correct according to the book. I'm stuck.

EDIT:

The answer should be one of the followings:

  1. $(-1,1)$
  2. $(-1,1]$
  3. $[-1,1)$
  4. $[-1,1]$
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    @MichaelHardy: I've added it to my question.2012-02-01

2 Answers 2

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Given the answer choices, it's probably intended for you to integrate term-wise and then investigate the convergence of the resulting power series. (Also: polynomials have finite degree, but these don't and hence are not polynomials.) Thus

$\int(x+2x^2+3x^3+\cdots)dx=\frac{1}{2}x^2+\frac{2}{3}x^3+\frac{3}{4}x^4+\cdots.$

The ratio test tells us that this does converge when $|x|<1$ but not $|x|>1$. Similarly, plugging in either of $x=\pm1$ results in a divergent series, so the domain is $(-1,1)$.

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Lets write your function this way

$x+2x^2+3x^3+\cdots = x \frac{d}{dx} (x+x^2+x^3+x^4+\cdots)$

It is elementary that $x+x^2+x^3+x^4 + \cdots = \frac{x}{1-x} $

Thus we have $x+2x^2+3x^3+\cdots = x \frac{d}{dx} \left(\frac{x}{1-x}\right)$

This produces $x+2x^2+3x^3+\cdots = \frac{x}{(1-x)^2}$

Remember we always have to consider $ |x| < 1 $ to make sure the sum converges.

Returning to you problem, we can integrate to get:

$ \int {\frac{x}{(1-x)^2} dx} = \frac{1}{1-x} + \log(1-x) + C $

EDIT: Rigorously, the series will only converge for $|x|<1$ and you'll need $x \neq 1 $ for the $\log$ to be defined. So $\mathbb{D} = (-1,1)$.

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    I accepted the other answer because it was correct according to my book, but I guess you missed a part of your own explanation, otherwise yours is correct as well. I'll accept your answer when you edit it. Thank you again.2012-02-01