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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?

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

$$f(x)=f(-x)$$

and this can be seen clearly. Here we have

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

and thus

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

one of the Laws of Logarithms is that

$$A\log(x)=\log(x^A)$$

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

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

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