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 '$dy$'?



double integral and Hadamard finite part
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.


