反雙曲函數

反雙曲函數是雙曲函數的反函數。與反圓函數不同之處是它的前綴是ar意即area（面積），而不是arc（弧）。因為雙曲角是以雙曲線、通過原點直線以及其對x軸的映射三者之間所夾面積定義的，而圓角是以弧長與半徑的比值定義。

## 數學符號

符號$\mathrm{sinh}^{-1}, \mathrm{cosh}^{-1}$等常用於$\mathrm{arsinh}, \mathrm{arcosh}$等。但是這種符號有時在$\mathrm{sinh}^{-1} x$和$\frac{1}{\mathrm{sinh}x}$之間易造成混淆。

## 主值

下表列出基本的反雙曲函數。

名稱｜常用符號｜定義｜定義域｜值域｜圖像

反雙曲正弦｜$y=\mathrm{arsinh} x$｜$\ln(x + \sqrt{x^2 + 1})$｜$\mathbb{R}$｜$\mathbb{R}$｜

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反雙曲餘弦｜$y=\mathrm{arcosh} x$｜$\ln(x \pm \sqrt{x^2 - 1})$｜$[1,+\infty)$｜$[0,+\infty)$｜

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反雙曲正切｜$y=\mathrm{artanh} x$｜$\frac{1}{2} \ln\left(\frac{1+x}{1-x}\right)$｜$(-1,1)$｜$\mathbb{R}$｜

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反雙曲餘切｜$y=\mathrm{arcoth} x$｜$\frac{1}{2} \ln\left(\frac{x+1}{x-1}\right)$｜$(-\infty,-1)\cup(1,+\infty)$｜$(-\infty,0)\cup(0,+\infty)$｜

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反雙曲正割｜$y=\mathrm{arsech} x$｜$\ln\left(\frac{1}{x}+ \frac{\sqrt{1 - x^2}}{x}\right)$｜$(0,1]$｜$[0,+\infty)$｜

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反雙曲餘割｜$y=\mathrm{arcsch} x$｜$\ln\left(\frac{1}{x}+ \frac{\sqrt{1 + x^2}}{\left|x\right|}\right)$｜$(-\infty,0)\cup(0,+\infty)$｜$(-\infty,0)\cup(0,+\infty)$｜

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## 加法公式

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

## 其他恆等式

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

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

## 反雙曲函數的導數

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

求導範例：

設θ = arsinh x，則：

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

## 冪級數展開式

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

## 反雙曲函數的不定積分

$$\begin{align} \int \operatorname{arsinh}\,x\,dx &{}= x\,\operatorname{arsinh}\,x - \sqrt{x^2+1} + C\\ \int \operatorname{arcosh}\,x\,dx &{}= x\,\operatorname{arcosh}\,x - \sqrt{x^2-1} + C,\qquad x >1\\ \int \operatorname{artanh}\,x\,dx &{}= x\,\operatorname{artanh}\,x + \frac{1}{2}\ln\left(1-x^2\right) + C,\qquad |x| <1\\ \int \operatorname{arcoth}\,x\,dx &{}= x\,\operatorname{arcoth}\,x + \frac{1}{2}\ln\left(x^2-1\right) + C,\qquad |x| >1\\ \int \operatorname{arsech}\,x\,dx &{}= x\,\operatorname{arsech}\,x + \arcsin\,x + C,\qquad x \in (0,1)\\ \int \operatorname{arcsch}\,x\,dx &{}= x\,\operatorname{arcsch}\,x + \left|\operatorname{arsinh}\,x\right| + C,\qquad x\ne0 \end{align}$$

使用分部積分法和上面的簡單導數很容易得出它們。

## 外部連結

- Inverse trigonometric functions at MathWorld