Originally Posted by

**james** Thanks for the quick reply, think I've got somethingwhich looks alright. What do you think?

$\displaystyle

x-a \ln\left(\cosh \left(\frac{x}{a}\right)\right)

$

over b and zero gives

$\displaystyle

b-a \ln\left(\cosh \left(\frac{b}{a}\right)\right) - 0 + a \ln\left(\cosh \left(0\right)\right)$

$\displaystyle

= b - a \ln \left(\frac{e^{b/a}+e^{-b/a}}{2} \right)

$

$\displaystyle

= b - a \ln \left(e^{b/a}+e^{-b/a} \right) + a \ln 2

$

then as b -> infinity the negative exponential vanishes so

$\displaystyle

= b - a \ln \left(e^{b/a}\right) + a \ln 2

$

$\displaystyle

= b - a (b/a) + a \ln 2

$

$\displaystyle

= a \ln 2

$