I am reading an article and came across this expression and would appreciate some explanation.


"We have a function
$$u(F)=2\Big[\frac{F-L}{L}-ln\frac{F}{L}\Big]$$ 


Since $u''(F)>0$ a Taylor series expansion with second-order remainder of $u(F\_n)$ about $F\_n=L$ implies
$$u(F\_n)=\int\_0^L\frac{2}{K^2}(K-F\_n)^+dK+\int\_L^\infty\frac{2}{K^2}(F\_n-K)^+dK$$"


Now, with what I know I would have the Taylor series to be 
$$\frac{2}{L^2}(L-F\_n)$$ I do not see how the integrals play a part in the series expansion. If someone could shed some light over this it would be extremely helpful. Thanks!



Taylor Series Expansion
you can find this worked out in Appendix 1 of [Towards a Theory of Volatility Trading](http://pricing.online.fr/docs/TradingVolatilityStrat.pdf)


the general identity (for $L>0$) is


$$u(F)=u(L)+u'(L)[(F-L)^+-(L-F)^+]$$
$$\qquad+\int\_0^L u''(K)(K-F)^+\,dK+\int\_L^\infty u''(K)(F-K)^+ dK,$$


where $(K-F)^+$ means $K-F$ if $F<K$ and zero if $F>K$.


