show $\ln \frac{1-x}{1+x}$ is $L^2$ but not $L^1$ using its taylor expansion I am trying to show that 
$$
\ln{\Big|\frac{1-x}{1+x}\Big|}
$$ 
belongs to $L^2({\Bbb{R}})$ but not to $L^1({\Bbb{R}})$ by using it's taylor expansion (this is the entire statement of the problem). More important to me than the solution to this particular problem is to understand the general technique I may use to do this.
Since there are singularities at $\pm1$ I am inclined to think that this needs to be done in three parts. Expanding around $0$ I find:
$$
ln\frac{1-x}{1+x} \approx -2x - \frac{2x^3}{3} - \frac{2x^5}{5} - \frac{2x^7}{7} ...
= \sum_0^{\infty}{\frac{-2}{2n-1}x^{2n-1}}
$$
Which is (please correct me if I am wrong) valid between $\pm1$ because of the singularities at $\pm1$.
With robjohn's suggestion an expansion for all x such that |x| > 1 is given by
$$
\ln{\Big|\frac{1-x}{1+x}\Big|} = \ln{\Big|\frac{\frac{1}{x}-1}{\frac{1}{x}+1}\Big|} = \ln{\Big|\frac{1-\frac{1}{x}}{1+\frac{1}{x}}\Big|}
$$
And since $|x|>1 => |\frac{1}{x}|<1$ we are in the domain of our previous expansion therefore:
$$
ln\frac{1-x}{1+x} \approx -\frac{2}{x} - \frac{2}{3x^3} - \frac{2}{5x^5} - \frac{2}{7x^7} ...
= \sum_0^{\infty}{\frac{-2}{2n-1x^{2n-1}}}
$$
Thank you.
 A: Let's begin with 
$$\int_{-\infty}^{\infty} \left ( \log{\Big|\frac{1-x}{1+x}\Big|} \right )^2 dx  $$
Note that the integrand is an even function.  You can then split this integral up into two pieces:
$$ 2 \int_{0}^{1} \left ( \log{\Big|\frac{1-x}{1+x}\Big|} \right )^2 dx  + 2 \int_{1}^{\infty} \left ( \log{\Big|\frac{x-1}{x+1}\Big|} \right )^2 dx $$
Taylor expand the integrand in $x$ in the first integral, and in $1/x$ in the second integral.  The series in the first integral goes as 
$$ \left ( \log{\Big|\frac{1-x}{1+x}\Big|} \right )^2 = 4 x^2 + O(x^4) $$
so we have convergence over $[0,1]$.  For the second integral:
$$ \left ( \log{\Big|\frac{x-1}{x+1}\Big|} \right )^2 = \frac{4}{x^2} + O \left ( \frac{1}{x^4} \right )$$
For 
$$\int_{-\infty}^{\infty} \log{\Big|\frac{1-x}{1+x}\Big|} dx $$
The series for the analogous second integral over $[1,\infty)$ goes as 
$$ \log{\Big|\frac{x-1}{x+1}\Big|}  = -\frac{2}{x} + O \left ( \frac{1}{x^2} \right )$$
The integral of this is divergent at $\infty$.  Thus, $\log{\Big|\frac{1-x}{1+x}\Big|}$ is $L^2(\mathbb{R})$ but not $L^1(\mathbb{R})$.
A: Hint 1:
For $|x|\lt1$, we have the standard
$$
\log\,\left|\frac{1-x}{1+x}\right|=-2\left(x+\frac{x^3}{3}+\frac{x^5}{5}+\frac{x^7}{7}+\dots\right)
$$
For $|x|>1$, we have
$$
\begin{align}
\log\,\left|\frac{1-x}{1+x}\right|
&=\log\,\left|\frac{1/x-1}{1/x+1}\right|\\
&=\log\,\left|\frac{1-1/x}{1+1/x}\right|\\
&=-2\left(\frac1x+\frac1{3x^3}+\frac1{5x^5}+\frac1{7x^7}+\dots\right)
\end{align}
$$
Hint 2:
$\log(x)$ is both $L^1$ and $L^2$ on finite intervals
$$
\int_0^1|\log(x)|\,\mathrm{d}x=\int_0^\infty t\,e^{-t}\,\mathrm{d}t=1
$$
$$
\int_0^1|\log(x)|^2\,\mathrm{d}x=\int_0^\infty t^2e^{-t}\,\mathrm{d}t=2
$$
So the difference must be near $\infty$.
