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IV - Powers, Exponentials, Logarithms, Trigonometric Functions
if s > 1. This is the case here, and since arg cosh y and log y tend to 0 when
y tends to 1, the formula (5) shows that EUp (l) = log 2.
Exercise. Show that, for y close to 1, there is an expansion of the form
argcoshy = I>n(Y - l)n+~.
On what interval with left end point 1 is it valid?
The function y = sinh x maps JR onto JR, whence an everywhere differentiable inverse
x = arg sinh y : JR ----+ JR
with
argsinh' y = 1/ sinh' x = 1/ cosh x = (y2 + 1)-1/2.
The function y = tanh x maps JR onto] - 1,1[, whence
argtanhy :]- 1,1[----+ JR,
with
argtanh' y = l/tanh' y = 1/(1 - tanh 2 x) = (1- y2)-1
(Iyl < 1).
Finally, the function coth maps JR - [-1,1]' the (obvious) union of two
intervals, bijectively onto the same set, whence a function x = arg coth y
defined for Iyl > 1 with
argcoth' y = 1/ coth' x = 1/(1 - coth 2 x) = (1 _ y2)-1
(Iyl > 1).
These two last functions thus provide primitives of the function 1/(I- y2)
in each of three open intervals where it is defined, the two first serving to
integrate (y2 ± 1)1/2.
It remains to show how these inverse functions can be expressed in terms
of the function log. The formula
2y = 2coshx = eX + e- x = eX + l/e x
shows that e 2x - 2ye X + 1 = 0, an equation of the second degree in eX. If one
assumes x > 0 and so eX > 1, it follows that eX = y + (y2 - 1)1/2 since the
other root is < 1. In consequence,
argcoshy = log [Y + (y2 _1)1/2]
for y > 1.
An analogous argument shows that
argsinhy = log [y + (y2 + 1)1/2]
for y E JR
and that