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:

$$ \implies \operatorname{arctanh}(y) = \frac{1}{2} \ln \left(\frac{1+y}{1-y}\right) $$

  • $\begingroup$ Isn't that fine? $\endgroup$ – Henry Lee Sep 19 '18 at 14:36
  • $\begingroup$ Should be $\mathrm{artanh}(y)$. in the last line, since there are no $x$ in the RHS. $\endgroup$ – user10354138 Sep 19 '18 at 14:38
  • $\begingroup$ @HenryLee Looks fine to me but i wanted to post this so i can get confirmation from other people that this looks fine or if it doesn't then i can fix it. $\endgroup$ – Tuki Sep 19 '18 at 14:38
  • $\begingroup$ @user10354138 yes there is typo. $\endgroup$ – Tuki Sep 19 '18 at 14:39
  • $\begingroup$ @user10354138 fixed it. $\endgroup$ – Tuki Sep 19 '18 at 14:39

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*}


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 $y$ $$\frac{\mathrm e^{2y}-1}{\mathrm e^{2y}+1}=x\iff \mathrm e^{2y}- x\mathrm e^{2y}= x+1\iff \mathrm e^{2y}=\frac{1-x}{1+x}\iff \dotsm$$


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).$

  • $\begingroup$ I think you want arctanh in that last identity. $\endgroup$ – Thomas Andrews Sep 20 '18 at 0:05
  • $\begingroup$ @ThomasAndrews thanks, fixed. $\endgroup$ – guy600 Sep 20 '18 at 0:15

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