Inverse hyperbolic function

From Wikipedia, the free encyclopedia

The function artanh.

The inverses of the hyperbolic functions are the area hyperbolic functions. The names hint at the fact that they compute the area of a sector of the unit hyperbola $x 2 - y 2 = 1$ in the same way the inverse trigonometric functions compute the arclength of a sector on the unit circle $x 2 + y 2 = 1.$ The usual abbreviations for them in mathematics are arsinh, arcsinh (US) or asinh in computer science. The notation sinh^-1 (x), cosh^-1(x) etc. are also used, despite the fact that care must be taken to avoid misinterpretations of the superscript -1 as a power as opposed to a shorthand for inverse. The acronyms arcsinh, arccosh etc. are commonly used in the US, even though they are – technically speaking – misnomers. The prefix arc is the abbreviation for arcus in contrast to the prefix ar which means area.

$\operatorname{arsinh}\, x = \ln(x + \sqrt{x^2 + 1})$

$\operatorname{acosh}\, x = \ln(x \pm \sqrt{x^2 - 1})$

$\operatorname{artanh}\, x = \ln\left(\frac{\sqrt{1 - x^2}}{1-x}\right) = \frac{1}{2} \ln\left(\frac{1+x}{1-x}\right)$

$\operatorname{arcoth}\, x = \ln\frac{\sqrt{x^2 - 1}}{x-1} = \frac{1}{2} \ln\frac{x+1}{x-1}$

$\operatorname{arsech}\, x = \pm \ln\frac{1 + \sqrt{1 - x^2}}{x}$

$\operatorname{arcsch}\, x = \begin{cases} \ln\frac{1 - \sqrt{1 + x^2}}{x}, & \mbox{for }x < 0\!\, \\ \ln\frac{1 + \sqrt{1 + x^2}}{x}, & \mbox{for }x > 0\!\, \end{cases}$

Expansion series can be obtained for the above functions:

$\operatorname{arsinh}\, x$

$= x - \left( \frac {1} {2} \right) \frac {x^3} {3} + \left( \frac {1 \cdot 3} {2 \cdot 4} \right) \frac {x^5} {5} - \left( \frac {1 \cdot 3 \cdot 5} {2 \cdot 4 \cdot 6} \right) \frac {x^7} {7} +\cdots$

$= \sum_{n=0}^\infty \left( \frac {(-1)^n(2n)!} {2^{2n}(n!)^2} \right) \frac {x^{2n+1}} {(2n+1)} , \qquad \left| x \right| < 1$

$\operatorname{arcosh}\, x$

$= \ln 2x - \left( \left( \frac {1} {2} \right) \frac {x^{-2}} {2} + \left( \frac {1 \cdot 3} {2 \cdot 4} \right) \frac {x^{-4}} {4} + \left( \frac {1 \cdot 3 \cdot 5} {2 \cdot 4 \cdot 6} \right) \frac {x^{-6}} {6} +\cdots \right)$

$= \ln 2x - \sum_{n=1}^\infty \left( \frac {(-1)^n(2n)!} {2^{2n}(n!)^2} \right) \frac {x^{-2n}} {(2n)} , \qquad x > 1$

$\operatorname{artanh}\, x = x + \frac {x^3} {3} + \frac {x^5} {5} + \frac {x^7} {7} +\cdots = \sum_{n=0}^\infty \frac {x^{2n+1}} {(2n+1)} , \qquad \left| x \right| < 1$

$\operatorname{arcsch}\, x = \operatorname{arsinh}\, x^{-1}$

$= x^{-1} - \left( \frac {1} {2} \right) \frac {x^{-3}} {3} + \left( \frac {1 \cdot 3} {2 \cdot 4} \right) \frac {x^{-5}} {5} - \left( \frac {1 \cdot 3 \cdot 5} {2 \cdot 4 \cdot 6} \right) \frac {x^{-7}} {7} +\cdots$

$= \sum_{n=0}^\infty \left( \frac {(-1)^n(2n)!} {2^{2n}(n!)^2} \right) \frac {x^{-(2n+1)}} {(2n+1)} , \qquad \left| x \right| < 1$

$\operatorname{arsech}\, x = \operatorname{arcosh}\, x^{-1}$

$= \ln \frac{2}{x} - \left( \left( \frac {1} {2} \right) \frac {x^{2}} {2} + \left( \frac {1 \cdot 3} {2 \cdot 4} \right) \frac {x^{4}} {4} + \left( \frac {1 \cdot 3 \cdot 5} {2 \cdot 4 \cdot 6} \right) \frac {x^{6}} {6} +\cdots \right)$

$= \ln \frac{2}{x} - \sum_{n=1}^\infty \left( \frac {(-1)^n(2n)!} {2^{2n}(n!)^2} \right) \frac {x^{2n}} {2n} , \qquad 0 < x \le 1$

$\operatorname{arcoth}\, x = \operatorname{artanh}\, x^{-1}$

$= x^{-1} + \frac {x^{-3}} {3} + \frac {x^{-5}} {5} + \frac {x^{-7}} {7} +\cdots$

$= \sum_{n=0}^\infty \frac {x^{-(2n+1)}} {(2n+1)} , \qquad \left| x \right| > 1$

Asymptotic expansion for the arsinh x is given by

$\operatorname{arsinh}\, x = \ln 2x + \sum\limits_{n = 1}^\infty {\left( { - 1} \right)^{n - 1} \frac{{\left( {2n - 1} \right)!!}}{{2n\left( {2n} \right)!!}}} \frac{1}{{x^{2n} }}$

[edit] Applications of inverse trigonometric functions and inverse hyperbolic functions to integrals

$\int \frac {dx} {\sqrt{1 - x^2}} = \operatorname{arcsin}\, x + {C} = - \operatorname{arccos}\, x + \frac {\pi}{2}+{C}$

$\int \frac {dx} {\sqrt{x^2 + 1}} = \operatorname{arsinh}\, x + {C} = \ln (x + \sqrt{x^2 + 1}) + {C}$

$\int \frac {dx} {\sqrt{x^2 - 1}} = \operatorname{arcosh}\, x + {C} = \ln (x + \sqrt{x^2 - 1}) + {C}$

$\int \sqrt{1 - x^2}\; dx = \frac{\operatorname{arcsin}\, x + x\sqrt{1 - x^2}}{2} + {C}$

$\int \sqrt{x^2 + 1}\; dx = \frac{\operatorname{arsinh}\, x + x\sqrt{x^2 + 1}}{2} + {C} = \frac{\ln(x + \sqrt{x^2 + 1}) + x\sqrt{x^2 + 1}}{2} + {C}$

$\int \sqrt{x^2 - 1}\; dx = \frac{- \operatorname{arcosh}\, x + x\sqrt{x^2 - 1}}{2} + {C} = \frac{- \ln(x + \sqrt{x^2 - 1}) + x\sqrt{x^2 - 1}}{2} + {C}$