How to treat absolute value bars in $\log\left|\frac{1+x}{1-x}\right|$ to show is an even function?

When dealing with strictly numbers, one defines the absolute value of $x$ as: $$|x| =\begin{cases} x, & x\geq 0, \\ -x, & x < 0. \\ \end{cases}$$

But we can't do that in this situation. I ask because I am working through a solution for the expression $$\int_{-\infty}^{\infty}\log\left|\frac{1+x}{1-x}\right| \frac{1}{x}dx = \pi^{2}$$

One of the steps is to establish that the integrand is even which I say I "somewhat" did:

$$f(-x) = \log\left|\frac{1-x}{1+x}\right|\frac{1}{-x} = (\log(1-x) - \log(1+x))\frac{1}{-x} = (\log(1+x) - \log(1-x))\frac{1}{x} = f(x) $$

But I am extremely wary of how I just "casually" dropped the absolute value signs. Because what I want to do is rewrite:

$$\int_{-\infty}^{\infty}\log\left|\frac{1+x}{1-x}\right| \frac{1}{x}dx = \int_{0}^{\infty}\log\left|\frac{1+x}{1-x}\right| \frac{1}{x}dx + \int_{-\infty}^{0}\log\left|\frac{1+x}{1-x}\right| \frac{1}{x}dx$$

And then flip the integral sign in the second term to $$\int_{0}^{\infty}\log\left|\frac{1+x}{1-x}\right| \frac{1}{x}dx$$ as well. But the step doesn't feel right to me in the same sense as if I was doing this to a number with absolute value bars which would have the following progression:

$$\int_{-\infty}^{\infty} |x| dx = \int_{0}^{\infty}x dx + \int_{-\infty}^{0}-x dx = \int_{0}^{\infty}x dx + \int_{0}^{\infty}x dx$$

So after all this the question remains: How do I "formally" handle this absolute value bars in the log to extract the same relationship?


This one is rather straight forward actually. When we say that the function is even, we mean


and this can be seen clearly. Here we have


and thus


one of the Laws of Logarithms is that


so if we let $A=-1$ here, we can write

$$f(-x)=\frac{1}{x}\log\left|\frac{1-x}{1+x}\right|^{-1}$$ $$=\frac{1}{x}\log\left|\frac{1+x}{1-x}\right|$$ $$=f(x)$$

  • $\begingroup$ Makes sense. Thanks for this part.....Any idea how to deal with the integration part? $\endgroup$ – dc3rd Aug 12 '19 at 22:59

Just pull down the exponent from the log:

$$\frac{1}{-x} \cdot\log \left|\frac{1+(-x)}{1-(-x)} \right| = \frac{1}{-x} \log \left|\frac{1+x}{1-x}\right|^{-1} = \frac{1}{x}\log \left|\frac{1+x}{1-x} \right|.$$


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