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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.

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2 Answers

up vote 2 down vote accepted

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

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Do you mind putting the function in front of $dx$? Otherwise there is a potential confusion with Stieltjes integral. –  Sanchez Jan 18 '13 at 9:39
    
I don't mind, but you may notice I do this consistently with all of my posts. It comes from my background as a physicist in thinking of the integral as an operator (which seems relevant here). It is also better to express integration like this when dealing with multiple integrals. –  Ron Gordon Jan 18 '13 at 9:45
    
I beg to disagree with your reason, yet I respect your convention as long as it's consistent. –  Sanchez Jan 18 '13 at 9:47
    
Think about a double integral in which the order of integration is not obvious. In my notation, the d's stay with their respective integrals and any ambiguity is resolved. This is especially important when the integrals are not absolutely convergent and order matters. –  Ron Gordon Jan 18 '13 at 9:51
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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 $\infty$.

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Thank you, that is a cool trick, I will remember that! How may I use these expansions to show that ln(1-x)(1+x) is not L^1 but is L^2? –  Amos Joshua Jan 18 '13 at 9:35
    
@AmosJoshua: as I hinted above, the only problem is near $\infty$, and as the second series shows, $\log\,\left|\frac{1-x}{1+x}\right|=-\frac2x+O\left(\frac1{x^3}\right)$. –  robjohn Jan 18 '13 at 9:40
    
Thank you robjohn. I'm afraid I needed more hand holding than I had originally thought and I couldn't figure out the solution from just your hints, although now it all makes sense. I appreciate your help. –  Amos Joshua Jan 18 '13 at 9:54
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