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Consider the below power series:

$\sum\limits_{n=1}^\infty \dfrac{x^{n}}{n^{2}}$

I know that it converges for $x\in [-1,1]$ and the sum $s(x)$ of the series is given by:

$s(x) = - \int\limits_{0}^{x} \frac{\ln(1-t)}{t} dt$ for all $x\in ]-1,1[$

I now have to explain why the integrand can be extended uniquely to be a contionuos function on $]-1,1[$, even though it is only defined in $]-1,0[\cup ]0,1[$.

I'm completely lost on this question. What do they want me to do? I can see from the series that $s(0) = 0$ but what is the statement "extended uniquely" supposes to mean?

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It looks to me as if the question has nothing to do with power series or integration, but just the function $g: ]-1,0[ \cup ]0,1[ \to \mathbb{R}$ defined by $g(t) = \ln(1-t)/t$. I think you are supposed to show that there is a unique continuous function $G: ]-1,1[ \to \mathbb{R}$ satisfying $G(t) = g(t)$ for all $t$ in the domain of $g$. This is equivalent to showing that $\lim_{t \to 0} g(t)$ exists (defining $G(0)$ to be the value of this limit provides the required function $G$). –  leslie townes May 24 '12 at 21:34
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They want you to reason that the integrand $f(t)=\log(1-t)/t$, while technically only defined on $(-1,0)\cup(0,1)$, has a removable discontinuity at $t=0$ and extends to a unique continuous function on the whole interval $(-1,1)$ which equals $f(t)$ when restricted to $t\ne0$. –  anon May 24 '12 at 21:35
    
Thank you. Still trying to understand. Can I then extend f(t) to be a piecewise function: $f(t) = log(1-t)/t$ when $t\neq 0$ and $f(t) = 1$ when $t=0$. And then argue that f(t) is continuos for $t\in ]-1,1[$ because f(t) is continuos in $]-1,0[$ and $]0,1[$ and $f(t)\rightarrow 1$ when $t\rightarrow 0+$ or $t\rightarrow 0-$? –  characters May 24 '12 at 23:15
    
$\lim_{t\to0}f(t)=-1$, not $+1$, no? –  Gerry Myerson May 25 '12 at 1:04
    
Sorry I forgot to multiply it with -1 –  characters May 25 '12 at 8:11
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