<s>[INST] Get an approximation of $\int_0^1\int_0^1\frac{1+x+y^2+x^3+\ldots}{1+y+x^2+y^3+\ldots}dxdy$ I am interested in to know how get an approximation of the double integral defined as limit of these $$\int_0^1\int_0^1\frac{1+x}{1+y}dxdy\,,$$
$$\int_0^1\int_0^1\frac{1+x+y^2}{1+y+x^2}dxdy\,,$$
$$\int_0^1\int_0^1\frac{1+x+y^2+x^3}{1+y+x^2+y^3}dxdy\,,\ldots$$

Question. Provide me an approximation of the definite integral defined as limit of previous sequence. Thanks in advance.

Using the formula to get the sum of a geometric series I know how get in closed-form the integrand. Addtitionally I presume that a it's possible to get a closed-form of it, since Wolfram Alpha can deduce the antiderivatives. Any case even the evaluation of the integral limits seem to me now complicated.
 [/INST] The numerator can be written as
$$\sum _{n=0}^{\infty } y^{2 n}=\frac{1}{1-y^2};\quad \sum _{n=1}^{\infty } x^{2 n-1}=-\frac{x}{x^2-1}$$
that is
$$\frac{1}{1-y^2}-\frac{x}{x^2-1}=\frac{-x^2-x y^2+x+1}{\left(x^2-1\right) \left(y^2-1\right)}$$
The denominator is pretty the same work swapping $x$ and $y$
$$\frac{1}{1-x^2}-\frac{y}{y^2-1}=\frac{-x^2y-y^2+y+1}{\left(x^2-1\right) \left(y^2-1\right)}$$
The fraction becomes simply
$$\frac{-x^2-x y^2+x+1}{-x^2y-y^2+y+1}$$
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