Primitive of Reciprocal of x squared minus a squared/Logarithm Form/Lemma
Lemma
Let $a \in \R_{>0}$ be a strictly positive real constant.
Then:
- $\map \ln {\dfrac {x - a} {x + a} }$ is defined if and only if $\size x > a$
- $\map \ln {\dfrac {a - x} {a + x} }$ is defined if and only if $\size x < a$
Proof
We have that the real natural logarithm is defined only on the strictly positive real numbers.
Hence:
- $\map \ln {\dfrac {x - a} {x + a} }$ is defined if and only if $\dfrac {x - a} {x + a} > 0$
- $\map \ln {\dfrac {a - x} {a + x} }$ is defined if and only if $\dfrac {a - x} {a + x} > 0$
First we note that if $\size x = a$, then either the numerator or denominator of the arguments of the logarithm functions in question are either $0$ or undefined.
Hence the expressions have no meaning unless $\size x \ne a$.
The following table indicates whether each of $x + a$, $a - x$ and $x - a$ are positive $(+)$ or negative $(-)$ on the domains in question.
$\begin {array} {c|ccc|cc} & x + a & a - x & x - a & \dfrac {x - a} {x + a} & \dfrac {a - x} {a + x} \\ \hline a < x & + & - & + & + & - \\ 0 < x < a & + & + & - & - & + \\ -a < x < 0 & + & + & - & - & + \\ x < -a & - & + & - & + & - \\ \end {array}$
Hence:
- $\map \ln {\dfrac {x - a} {x + a} }$ is defined if and only if $x > a$ or $x < -a$
- $\map \ln {\dfrac {a - x} {a + x} }$ is defined if and only if $-a < x < a$
as we were required to show.
$\blacksquare$