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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.


The answer should be one of the followings:

  1. $(-1,1)$
  2. $(-1,1]$
  3. $[-1,1)$
  4. $[-1,1]$
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May be another way of asking the radius of convergence of the series you obtained on integrating? – user21436 Jan 31 '12 at 21:41
Let's first recognize that $x + 2x^2 + 3x^3 + \cdots$ is the derivative of the geometric series $1 + x + x^2 + \cdots$ multiplied by $x$. What's the radius of convergence for that? [It also seems like it would be better to write $f(x) = \int_0^x (t + 2t^2 + \cdots)\,dt$ but this seems like a common abuse.] – Dylan Moreland Jan 31 '12 at 21:43
@DylanMoreland: $\frac {1}{1-x}$? It's written like that in my book! – Gigili Jan 31 '12 at 22:07
I hope your book says something more than just $\frac{1}{1-x}$: it also says the series converges for $|x|<1$ and diverges otherwise. The same is true for $x + 2 x^2 + 3 x^3 + \ldots$. – Robert Israel Jan 31 '12 at 23:01
What does the book say the answer is? And why have none of those who have commented asked that question?? – Michael Hardy Feb 1 '12 at 0:52
up vote 3 down vote accepted

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


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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Thank you for your answer, but this is not correct. – Gigili Feb 1 '12 at 6:08
I think your solution is correct except the last part, when you said $|x| < 1$ so the domain should be that, am I right? – Gigili Feb 1 '12 at 6:11
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. – Gigili Feb 1 '12 at 7:13

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