Problem
Prove formula $\operatorname{arctanh} x = \frac{1}{2} \ln \left(\frac{1+x}{1-x}\right)$
Attempt to solve
To start off with definition of functions $\sinh(x)$ and $\cosh(x)$
$$ \cosh(x)=\frac{e^x+e^{-x}}{2} $$
$$ \sinh(x) = \frac{e^x-e^{-x}}{2} $$
Hyperbolic tangent is defined as:
$$ \tanh(x)=\frac{\sinh(x)}{\cosh(x)}=\frac{e^x-e^{-x}}{e^{x}+e^{-x}} $$
Notation $\text{arctanh}(x)$ means area tangent which is inverse of hyperbolic tangent.
$$ \operatorname{arctanh}(x)=\tanh^{-1}(x) $$
Trying to invert the $\tanh(x)$ we get:
$$\begin{align} \frac{e^x-e^{-x}}{e^x+e^{-x}} &= y &\implies\\ e^x-e^{-x}&=y(e^x+e^{-x}) &\implies\\ e^x-e^{-x}&=ye^{x}+ye^{-x} &\implies\\ e^x(1-y)&=e^{-x}(1+y) &\implies\\ \ln(e^x(1-y)) &= \ln(e^{-x}(1+y)) &\implies\\ \ln(e^x)+\ln(1-y) &= \ln(e^{-x})+\ln(1+y) &\implies\\ x + \ln(1-y) &= -x + \ln(1+y) &\implies\\ 2x &= \ln(1+y)-\ln(1-y)&\implies\\ x &= \frac{1}{2} \ln \frac{1+y}{1-y} \end{align} $$
By switching variables we get:
$$ \implies \operatorname{arctanh}(y) = \frac{1}{2} \ln \left(\frac{1+y}{1-y}\right) $$
$\endgroup$ 63 Answers
$\begingroup$Looks fine to me. Well done!
Just a small thing, a matter of taste: I'd prefer to do this:
\begin{align*} e^x(1-y) &= e^{-x}(1+y)\\ e^{2x} &=\dfrac{1+y}{1-y}\\ x &= \dfrac{1}{2}\ln\left(\dfrac{1+y}{1-y}\right). \end{align*}
$\endgroup$ $\begingroup$Shorter, using the other form of the hyperbolic tangent: $$\tanh x=\frac{\mathrm e^{2x}-1}{\mathrm e^{2x}+1},$$ we have to solve the equation in $y$ $$\frac{\mathrm e^{2y}-1}{\mathrm e^{2y}+1}=x\iff \mathrm e^{2y}- x\mathrm e^{2y}= x+1\iff \mathrm e^{2y}=\frac{1-x}{1+x}\iff \dotsm$$
$\endgroup$ $\begingroup$Just to add to the responses already here. You can also show this by solving this integral two different ways, one with trigonometric substitutions and the other with partial fraction decomposition.
$$ \int \frac {dx}{x^2+1} = \int \frac {dx}{x^2+1}$$
$\qquad x \mapsto \tan u$
$$ \int\frac{dx}{(1+ix)(1-ix)}= \int \frac {\sec^2 u}{1+ \tan^2u}du$$ $$ \int \Big(\frac{\frac 12}{1+ix}+\frac{\frac 12}{1-ix}\Big)dx\ = \int du$$ $$\frac 1{2i}\big(\ln(1+ix)- \ln(1-ix)\big)=u+C$$
$$\arctan x=\frac i2\ln \Big( \frac{1-ix}{1+ix} \Big)$$
Plug in $x=ix.$
$$\arctan(ix)=\frac i2 \ln\Big(\frac{1+x}{1-x} \Big)$$
Now use the identity$ \space\operatorname {arctanh}(x)=-i\arctan(ix).$
$\endgroup$ 2
