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。

反双曲函数示意图

几个反双曲函数的图形

数学符号

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

主值

下表列出基本的反双曲函数。

名称	常用符号	定义	定义域	值域	图像
反双曲正弦	${\displaystyle y=\mathrm {arsinh} x}$	${\displaystyle \ln(x+{\sqrt {x^{2}+1}})}$	${\displaystyle \mathbb {R} }$	${\displaystyle \mathbb {R} }$
反双曲余弦	${\displaystyle y=\mathrm {arcosh} x}$	${\displaystyle \ln(x\pm {\sqrt {x^{2}-1}})}$[注 1]	${\displaystyle [1,+\infty )}$	${\displaystyle [0,+\infty )}$
反双曲正切	${\displaystyle y=\mathrm {artanh} x}$	${\displaystyle {\frac {1}{2}}\ln \left({\frac {1+x}{1-x}}\right)}$	${\displaystyle (-1,1)}$	${\displaystyle \mathbb {R} }$
反双曲余切	${\displaystyle y=\mathrm {arcoth} x}$	${\displaystyle {\frac {1}{2}}\ln \left({\frac {x+1}{x-1}}\right)}$	${\displaystyle (-\infty ,-1)\cup (1,+\infty )}$	${\displaystyle (-\infty ,0)\cup (0,+\infty )}$
反双曲正割	${\displaystyle y=\mathrm {arsech} x}$	${\displaystyle \ln \left({\frac {1}{x}}+{\frac {\sqrt {1-x^{2}}}{x}}\right)}$	${\displaystyle (0,1]}$	${\displaystyle [0,+\infty )}$
反双曲余割	${\displaystyle y=\mathrm {arcsch} x}$	${\displaystyle \ln \left({\frac {1}{x}}+{\frac {\sqrt {1+x^{2}}}{\left|x\right|}}\right)}$	${\displaystyle (-\infty ,0)\cup (0,+\infty )}$	${\displaystyle (-\infty ,0)\cup (0,+\infty )}$

反双曲函数的导数

${\displaystyle {\begin{aligned}{\frac {d}{dx}}\operatorname {arsinh} \,x&{}={\frac {1}{\sqrt {1+x^{2}}}}\\{\frac {d}{dx}}\operatorname {arcosh} \,x&{}={\frac {1}{\sqrt {x^{2}-1}}},\qquad x>1\\{\frac {d}{dx}}\operatorname {artanh} \,x&{}={\frac {1}{1-x^{2}}},\qquad |x|<1\\{\frac {d}{dx}}\operatorname {arcoth} \,x&{}={\frac {1}{1-x^{2}}},\qquad |x|>1\\{\frac {d}{dx}}\operatorname {arsech} \,x&{}={\frac {-1}{x{\sqrt {1-x^{2}}}}},\qquad x\in (0,1)\\{\frac {d}{dx}}\operatorname {arcsch} \,x&{}={\frac {-1}{|x|{\sqrt {1+x^{2}}}}},\qquad x{\text{ ≠ }}0\\\end{aligned}}}$

求导范例： 设θ = arsinh x，则：

${\displaystyle {\frac {d\,\operatorname {arsinh} \,x}{dx}}={\frac {d\theta }{d\sinh \theta }}={\frac {1}{\cosh \theta }}={\frac {1}{\sqrt {1+\sinh ^{2}\theta }}}={\frac {1}{\sqrt {1+x^{2}}}}}$

幂级数展开式

${\displaystyle \operatorname {arsinh} \,x}$

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

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

${\displaystyle \operatorname {arcosh} \,x}$

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

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

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

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

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

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

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

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

${\displaystyle =\ln {\frac {2}{x}}-\sum _{n=1}^{\infty }\left({\frac {(-1)^{n}(2n)!}{2^{2n}(n!)^{2}}}\right){\frac {x^{2n}}{2n}},\qquad 0<x\leq 1}$

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

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

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

${\displaystyle \operatorname {arcosh} (2x^{2}-1)=2\operatorname {arcosh} x}$

${\displaystyle \operatorname {arcosh} (2x^{2}+1)=2\operatorname {arsinh} x}$

反双曲函数的不定积分

${\displaystyle {\begin{aligned}\int \operatorname {arsinh} \,x\,dx&{}=x\,\operatorname {arsinh} \,x-{\sqrt {x^{2}+1}}+C\\\int \operatorname {arcosh} \,x\,dx&{}=x\,\operatorname {arcosh} \,x-{\sqrt {x^{2}-1}}+C,\qquad x>1\\\int \operatorname {artanh} \,x\,dx&{}=x\,\operatorname {artanh} \,x+{\frac {1}{2}}\ln \left(1-x^{2}\right)+C,\qquad |x|<1\\\int \operatorname {arcoth} \,x\,dx&{}=x\,\operatorname {arcoth} \,x+{\frac {1}{2}}\ln \left(x^{2}-1\right)+C,\qquad |x|>1\\\int \operatorname {arsech} \,x\,dx&{}=x\,\operatorname {arsech} \,x+\arcsin \,x+C,\qquad x\in (0,1)\\\int \operatorname {arcsch} \,x\,dx&{}=x\,\operatorname {arcsch} \,x+\left|\operatorname {arsinh} \,x\right|+C,\qquad x\neq 0\end{aligned}}}$

使用分部积分法和上面的简单导数很容易得出它们。