Given the divergent integral
$$ \int _{0}^{\infty}dx \int_{0}^{\infty}dy \frac{x^{2}y+1}{1+x+y} $$
how can I apply Hadamard's finite part to give a finite meaning to it ?
It is just made by applying the itetrated finite part first to '$dx$' and then '$dy$'?
The Hadamard finite part description is strictly defined only for $1$-dimensional integrals. Also, it is usually defined for singularities within the domain of integration or at a finite boundary. On the other hand, your integrals is divergent at infinity. It is of course possible to try to do something similar in this case, but there is no guarantee that one can come up with a unique prescription.
Take the integration intervals to be $[0,X]$ and $[0,Y]$ instead of semi-infinite. The idea is then to evaluate the integral and try to subtract off a polynomial in $X$, $Y$, $\log X$ and $\log Y$ (or something like that) such that the remainder is finite. Doing the integral I get the following: \begin{gather} (X^4/4+X^3/3-X-11/12)\log(X+1) \\ - (Y^4/4+2Y^3/3+Y^2/2+Y+11/12)\log(Y+1) \\ + (Y^4/4-X^4/4+2Y^3/3-X^3/3+Y^2/2+Y+X+11/12)\log(X+Y+1) - XY^3/4+X^2Y^2/8+X^3Y/4-5XY^2/12+X^2Y/12-XY/12 \end{gather} Unfortunately, the singularity structure of $\log(X+Y+1)$ makes it difficult to subtract a unique polynomial in $X$, $Y$, $\log(X)$ and $\log(Y)$ and get a finite result in a way that is independent of the way $X$ and $Y$ go to infinity. So, from this point, you'll need more information about how you want to take these limits to get any further.
Rather than a simple numerical value, one might consider a more meaningful interpretation of the integral as an attempt to evaluate the Fourier transform of the integrand (restricted to the positive quadrant) as tempered distribution, at $(0,0)$. Any "regularization scheme" compatible with that is a-priori guaranteed to produce the same outcome. (The question of whether a tempered distribution is locally given by a continuous function, hence having canonical point-wise values, is answerable in an unambiguous way by looking at local Sobolev spaces and thinking of Sobolev imbedding.)
Replacing the denominator by $(1+x+y)^s$ for complex $s$ and taking $Re(s)$ large allows a typical computation, producing, I think, ${2\over s(s-1)(s-2)(s-3)(s-4)} + {1\over s(s-1)}$. Unfortunately, the constants do not conspire to cancel the pole at $s=1$, but we do see the behavior in $s$. E.g., as in Hadamard's finite-part, for $s$ a non-integer we can assign a finite regularized value.
