反双曲函数

反双曲函数是双曲函数的反函数。与反圆函数不同之处是它的前缀是ar意即area（面积），而不是arc（弧）。因为双曲角是以双曲线、通过原点直线以及其对x轴的映射三者之间所夹面积定义的，而圆角是以弧长与半径的比值定义。

數學符號
符号\mathrm{sinh}^{-1}, \mathrm{cosh}^{-1}等常用于\mathrm{arsinh}, \mathrm{arcosh}等。但是这种符号有时在\mathrm{sinh}^{-1} x和\frac{1}{\mathrm{sinh}x}之间易造成混淆。

主值
下表列出基本的反双曲函数。
\right)|| (-\infty,0)\cup(0,+\infty) || (-\infty,0)\cup(0,+\infty) ||
|-
|}

加法公式
\operatorname{arsinh} u \pm \operatorname{arsinh} v = \operatorname{arsinh} \left(u \sqrt{1 + v^2} \pm v \sqrt{1 + u^2}\right)
\operatorname{arcosh} u \pm \operatorname{arcosh} v = \operatorname{arcosh} \left(u v \pm \sqrt{(u^2 - 1) (v^2 - 1)}\right)
\operatorname{artanh} u \pm \operatorname{artanh} v = \operatorname{artanh} \left( \frac{u \pm v}{1 \pm uv} \right)
\operatorname{arcoth} u \pm \operatorname{arcoth} v = \operatorname{arcoth} \left( \frac{1 \pm uv}{u \pm v} \right)
\begin{align}\operatorname{arsinh} u + \operatorname{arcosh} v & = \operatorname{arsinh} \left(u v + \sqrt{(1 + u^2) (v^2 - 1)}\right) \\
& = \operatorname{arcosh} \left(v \sqrt{1 + u^2} + u \sqrt{v^2 - 1}\right) \end{align}

其他恒等式
\begin{align}
2\operatorname{arcosh}x&=\operatorname{arcosh}(2x^2-1) &\quad \hbox{ for }x\geq 1 \\
4\operatorname{arcosh}x&=\operatorname{arcosh}(8x^4-8x^2+1) &\quad \hbox{ for }x\geq 1 \\
2\operatorname{arsinh}x&=\pm\operatorname{arcosh}(2x^2+1) \\
4\operatorname{arsinh}x&=\operatorname{arcosh}(8x^4+8x^2+1) &\quad \hbox{ for }x\geq 0
\end{align}

\ln(x) = \operatorname{arcosh} \left( \frac{x^2 + 1}{2x}\right) = \operatorname{arsinh} \left( \frac{x^2 - 1}{2x}\right)
= \operatorname{artanh} \left( \frac{x^2 - 1}{x^2 + 1}\right)

反双曲函数的导数
:
\begin{align}
\frac{d}{dx} \operatorname{arsinh}\, x & {}= \frac{1}{\sqrt{1+x^2}}\\
\frac{d}{dx} \operatorname{arcosh}\, x & {}= \frac{1}{\sqrt{x^2-1}}, \qquad x>1\\
\frac{d}{dx} \operatorname{artanh}\, x & {}= \frac{1}{1-x^2}, \qquad |x| 1\\
\frac{d}{dx} \operatorname{arsech}\, x & {}= \frac{-1}{x\sqrt{1-x^2}}, \qquad x \in (0,1)\\
\frac{d}{dx} \operatorname{arcsch}\, x & {}= \frac{-1}{|x|\sqrt{1+x^2}}, \qquad x \text{ ≠ }0\\
\end{align}

求导范例：
设θ = arsinh x，则：
:\frac{d\,\operatorname{arsinh}\, x}{dx} = \frac{d \theta}{d \sinh \theta} = \frac{1} {\cosh \theta} = \frac{1} {\sqrt{1+\sinh^2 \theta}} = \frac{1}{\sqrt{1+x^2}}

幂级数展开式
:\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|

:\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|

:\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|

:\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

:\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

:\operatorname{arcosh}(2x^2-1) = 2\operatorname{arcosh} x
:\operatorname{arcosh}(2x^2+1) = 2\operatorname{arsinh} x

反双曲函数的不定积分
:
\begin{align}
\int \operatorname{arsinh}\,x\,dx &{}= x\,\operatorname{arsinh}\,x - \sqrt{x^2+1} + C\\
\int \operatorname{arcosh}\,x\,dx &{}= x\,\operatorname{arcosh}\,x - \sqrt{x^2-1} + C,\qquad x >1\\
\int \operatorname{artanh}\,x\,dx &{}= x\,\operatorname{artanh}\,x + \frac{1}{2}\ln\left(1-x^2\right) + C,\qquad |x| 1\\
\int \operatorname{arsech}\,x\,dx &{}= x\,\operatorname{arsech}\,x + \arcsin\,x + C,\qquad x \in (0,1)\\
\int \operatorname{arcsch}\,x\,dx &{}= x\,\operatorname{arcsch}\,x + \left|\operatorname{arsinh}\,x\right| + C,\qquad x\ne0
\end{align}

使用分部积分法和上面的简单导数很容易得出它们。

註釋
外部链接

• [http://mathworld.wolfram.com/InverseHyperbolicFunctions.html Inverse trigonometric functions] at MathWorld

参见
*双曲函数