19 Logarithmic Expression of Anti-Functions.

Article 19
Logarithmic Expression of Anti-Functions.

Let

\begin{array}{l} x & = cosh u, \\ then \\ \sqrt{x^{2} - 1} & = sinh u; \\ therefore \\ x + \sqrt{x^{2} - 1} & = cosh u + sinh u = e^{u}, \\ and \\ u = {cosh}^{- 1} x & = log (x + \sqrt{x^{2} - 1}) . & (38) \\ Similarly, \\ {sinh}^{- 1} x & = log (x + \sqrt{x^{2} + 1}) . & (39) \\ Also \\ {sech}^{- 1} x & = {cosh}^{- 1} \frac{1}{x} = log \frac{1 + \sqrt{1 - x^{2}}}{x}, & (40) \\ {csch}^{- 1} x & = {sinh}^{- 1} \frac{1}{x} = log \frac{1 + \sqrt{1 + x^{2}}}{x} . & (41) \\ Again, let \\ x & = tanh u = \frac{e^{u} - e^{- u}}{e^{u} + e^{- u}}, \\ therefore \\ \frac{1 + x}{1 - x} & = \frac{e^{u}}{e^{- u}} = e^{2 u}, \\ 2 u & = log \frac{1 + x}{1 - x}, {tanh}^{- 1} = \frac{1}{2} log \frac{1 + x}{1 - x}; & (42) \\ and \\ {coth}^{- 1} x & = {tanh}^{- 1} \frac{1}{x} = \frac{1}{2} log \frac{x + 1}{x - 1} . & (43) \end{array}

Prob. 46.

Show from (38), (39), that, when

x ≐ \infty

{sinh}^{- 1} x - log x ≐ log 2, {cosh}^{- 1} x - log x ≐ log 2,

and hence show that the integration-constants in (32), (33) are each equal to $log 2$ .

Prob. 47.

Derive from (42) the series for

{tanh}^{- 1} x

given in (34).

Prob. 48.

Prove the identities:

\begin{array}{l} log x = 2 {tanh}^{- 1} \frac{x - 1}{x + 1} = {tanh}^{- 1} \frac{x^{2} - 1}{x^{2} + 1} = {sinh}^{- 1} \frac{1}{2} (x - x^{- 1}) = {cosh}^{- 1} \frac{1}{2} (x + x^{- 1}); \\ log sec x = 2 {tanh}^{- 1} \frac{1}{2} x; log csc x = 2 {tanh}^{- 1} {tan}^{2} (\frac{1}{4} π + \frac{1}{2} x); \\ log tan x = - {tanh}^{- 1} cos 2 x = - {sinh}^{- 1} cot 2 x = {cosh}^{- 1} csc 2 x . \end{array}