Formelsammlung Mathematik: Areafunktionen

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Definition der Areafunktionen durch den Logarithmus

$\operatorname {arsinh} (z)=\ln \left(z+{\sqrt {z^{2}+1}}\right)$

$\operatorname {arcosh} (z)=\ln \left(z+{\sqrt {z-1}}\,{\sqrt {z+1}}\right)$

$\operatorname {artanh} (z)={\frac {1}{2}}{\Big (}\ln(1+z)-\ln(1-z){\Big )}$ für $z\neq \pm 1\,$

$\operatorname {arcoth} (z)=\left\{{\begin{matrix}\operatorname {artanh} \left({\frac {1}{z}}\right)&&z\neq 0\\{\frac {\mathrm {i} \pi }{2}}&&z=0\end{matrix}}\right.$

$\operatorname {arsech} (z)=\ln \left({\frac {1+{\sqrt {1-z^{2}}}}{z}}\right)$

$\operatorname {arcsch} (z)=\ln \left({\frac {1+{\sqrt {1+z^{2}}}}{z}}\right)$

Argument iz

${\begin{matrix}\arcsin(\mathrm {i} z)&=&\;\;\,\mathrm {i} \;\operatorname {arsinh} \,z&\qquad &\operatorname {arsinh} (\mathrm {i} z)&=&\;\;\mathrm {i} \;\arcsin z\\\arctan(\mathrm {i} z)&=&\;\;\,\,\mathrm {i} \;\operatorname {artanh} \,z&\qquad &\operatorname {artanh} (\mathrm {i} z)&=&\;\;\;\mathrm {i} \;\arctan z\\\operatorname {arccot}(\mathrm {i} z)&=&-\mathrm {i} \;\operatorname {arcoth} \,z&\qquad &\operatorname {arcoth} (\mathrm {i} z)&=&-\mathrm {i} \;\operatorname {arccot} z\\\operatorname {arccsc}(\mathrm {i} z)&=&-\mathrm {i} \;\operatorname {arcsch} \,z&\qquad &\operatorname {arcsch} (\mathrm {i} z)&=&-\mathrm {i} \;\operatorname {arccsc} z\\\end{matrix}}$

Verkettung einer Hyperbelfunktion mit einer Areafunktion

	$\operatorname {arsinh} \!$	$\operatorname {arcosh} \!$	$\operatorname {artanh} \!$	$\operatorname {arcoth} \!$	$\operatorname {arsech} \!$	$\operatorname {arcsch} \!$
$\sinh \!$	$z\!$	${\sqrt {z-1}}\;{\sqrt {z+1}}$	${\frac {z}{\sqrt {1-z^{2}}}}$	${\frac {1}{z\,{\sqrt {1-{\frac {1}{z^{2}}}}}}}$	${\sqrt {{\frac {1}{z}}-1}}\;{\sqrt {{\frac {1}{z}}+1}}$	${\frac {1}{z}}$
$\cosh \!$	${\sqrt {1+z^{2}}}$	$z\!$	${\frac {1}{\sqrt {1-z^{2}}}}$	${\frac {1}{\sqrt {1-{\frac {1}{z^{2}}}}}}$	${\frac {1}{z}}$	${\sqrt {1+{\frac {1}{z^{2}}}}}$
$\tanh \!$	${\frac {z}{\sqrt {1+z^{2}}}}$	${\frac {{\sqrt {z-1}}\;{\sqrt {z+1}}}{z}}$	$z\!$	${\frac {1}{z}}$	$z\,{\sqrt {{\frac {1}{z}}-1}}\;{\sqrt {{\frac {1}{z}}+1}}$	${\frac {1}{z\,{\sqrt {1+{\frac {1}{z^{2}}}}}}}$
$\coth \!$	${\frac {\sqrt {1+z^{2}}}{z}}$	${\frac {z}{{\sqrt {z-1}}\;{\sqrt {z+1}}}}$	${\frac {1}{z}}$	$z\!$	${\frac {1}{z}}{\frac {1}{{\sqrt {{\frac {1}{z}}-1}}\;{\sqrt {{\frac {1}{z}}+1}}}}$	$z\,{\sqrt {1+{\frac {1}{z^{2}}}}}$
$\operatorname {sech} \!$	${\frac {1}{\sqrt {1+z^{2}}}}$	${\frac {1}{z}}$	${\sqrt {1-z^{2}}}$	${\sqrt {1-{\frac {1}{z^{2}}}}}$	$z\!$	${\frac {1}{\sqrt {1+{\frac {1}{z^{2}}}}}}$
$\operatorname {csch} \!$	${\frac {1}{z}}$	${\frac {1}{{\sqrt {z-1}}\;{\sqrt {z+1}}}}$	${\frac {\sqrt {1-z^{2}}}{z}}$	$z\,{\sqrt {1-{\frac {1}{z^{2}}}}}$	${\frac {1}{{\sqrt {{\frac {1}{z}}-1}}\;{\sqrt {{\frac {1}{z}}+1}}}}$	$z\!$