How does one take the 'Hadamard finite part' of an integral? I am following a paper and the result stated is that
$$\int_{0}^{\infty}U_{B}^{-2}(Y)-U_{B}^{2}(Y)\,\textrm{d}Y=-2.7950.$$
The function $U_{B}(Y)$ is calculated numerically (in fact it's the classical Blasius flat plate boundary-layer velocity profile - http://jhuang79.weebly.com/uploads/7/2/1/5/7215285/4475122_orig.jpg from this image read 'df1' as $U_{B}$ and 'eta' as $Y$). I know that
$$U_{B}(0)=0,\quad U_{B}(\infty)=1,\quad U_{B}'(0)=0.3321,\quad U_{B}'(\infty)=0.$$
Plotting the function $F=U_{B}^{-2}-U_{B}^{2}$ against $Y$ one observes that $F(0)=\infty$ and $F(\infty)=0$ therefore the integral is undefined? How does taking the 'Hadamard finite part' return the above stated result?
Any help would be greatly appreciated! Very confused here.