show $\ln \frac{1-x}{1+x}$ is $L^2$ but not $L^1$ using its taylor expansion

Question

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.

Answer 1

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 $L^1(\mathbb{R})$.

Answer 2

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 $\infty$.