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.
 A: Well, first of all we can write:
$$\mathscr{F}\left(\text{x},\text{y}\right):=\frac{1+\text{x}+\text{y}^2+\text{x}^3+\dots}{1+\text{y}+\text{x}^2+\text{y}^3+\dots}=\frac{\sum\limits_{\text{n}=0}^\infty\text{x}^{2\text{n}+1}+\sum\limits_{\text{n}=0}^\infty\text{y}^{2\text{n}}}{\sum\limits_{\text{n}=0}^\infty\text{x}^{2\text{n}}+\sum\limits_{\text{n}=0}^\infty\text{y}^{2\text{n}+1}}\tag1$$
Now, we can use:


*

*When $\left|\text{x}\right|<1$:
$$\sum_{\text{n}=0}^\infty\text{x}^{2\text{n}+1}=\frac{\text{x}}{1-\text{x}^2}\tag2$$

*When $\left|\text{x}\right|<1$:
$$\sum_{\text{n}=0}^\infty\text{x}^{2\text{n}}=\frac{1}{1-\text{x}^2}\tag3$$


So, we get:
$$\mathscr{F}\left(\text{x},\text{y}\right)=\frac{\frac{\text{x}}{1-\text{x}^2}+\frac{1}{1-\text{y}^2}}{\frac{1}{1-\text{x}^2}+\frac{\text{y}}{1-\text{y}^2}}=\frac{\text{x}\cdot\left(\text{x}+\text{y}^2-1\right)-1}{\text{y}\cdot\left(\text{x}^2+\text{y}-1\right)-1}\tag4$$
When $\left|\text{x}\right|<1\space\wedge\space\left|\text{y}\right|<1$
For the $\text{x}$ integral we get:
$$\mathscr{I}_{\space\text{x}}\left(\text{y}\right):=\int_0^1\mathscr{F}\left(\text{x},\text{y}\right)\space\text{d}\text{x}=\int_0^1\frac{\text{x}\cdot\left(\text{x}+\text{y}^2-1\right)-1}{\text{y}\cdot\left(\text{x}^2+\text{y}-1\right)-1}\space\text{d}\text{x}=$$
$$\left(\text{y}^2-1\right)\int_0^1\frac{\text{x}-1}{\text{y}\cdot\left(\text{x}^2-1\right)+\text{y}^2-1}\space\text{d}\text{x}+\frac{1}{\text{y}}\int_0^11\space\text{d}\text{x}\tag5$$
So, we get for the indefinite integrals:


*

*$$\int1\space\text{d}\text{x}=\text{x}+\text{C}_1\tag6$$

*$$\int\frac{\text{x}-1}{\text{y}\cdot\left(\text{x}^2-1\right)+\text{y}^2-1}\space\text{d}\text{x}=$$
$$\int\frac{\text{x}}{\text{y}\cdot\left(\text{x}^2-1\right)+\text{y}^2-1}\space\text{d}\text{x}-\int\frac{1}{\text{y}\cdot\left(\text{x}^2-1\right)+\text{y}^2-1}\space\text{d}\text{x}\tag7$$

*Substitute $\text{u}:=\text{y}\cdot\left(\text{x}^2-1\right)+\text{y}^2-1$:
$$\int\frac{\text{x}}{\text{y}\cdot\left(\text{x}^2-1\right)+\text{y}^2-1}\space\text{d}\text{x}=\frac{1}{2\cdot\text{y}}\int\frac{1}{\text{u}}\space\text{d}\text{u}=\frac{\ln\left|\text{u}\right|}{2\cdot\text{y}}+\text{C}_2\tag8$$

*Substitute $\text{p}:=\frac{\text{x}\cdot\sqrt{\text{y}}}{\sqrt{\text{y}^2-\text{y}-1}}$:
$$\int\frac{1}{\text{y}\cdot\left(\text{x}^2-1\right)+\text{y}^2-1}\space\text{d}\text{x}=\frac{1}{\sqrt{\text{y}}\cdot\sqrt{\text{y}^2-\text{y}-1}}\int\frac{1}{1+\text{p}^2}\space\text{d}\text{p}=$$
$$\frac{\arctan\left(\text{p}\right)}{\sqrt{\text{y}}\cdot\sqrt{\text{y}^2-\text{y}-1}}+\text{C}_3\tag9$$


So, for the $\text{x}$ integral we get:
$$\mathscr{I}_{\space\text{x}}\left(\text{y}\right)=\left(\text{y}^2-1\right)\cdot\left[\frac{\ln\left|\text{y}\cdot\left(\text{x}^2-1\right)+\text{y}^2-1\right|}{2\cdot\text{y}}-\frac{\arctan\left(\frac{\text{x}\cdot\sqrt{\text{y}}}{\sqrt{\text{y}^2-\text{y}-1}}\right)}{\sqrt{\text{y}}\cdot\sqrt{\text{y}^2-\text{y}-1}}\right]_0^1+\frac{1}{\text{y}}=$$
$$\left(\text{y}^2-1\right)\cdot\left\{\frac{\ln\left|\frac{\text{y}^2-1}{\text{y}^2-\text{y}-1}\right|}{2\cdot\text{y}}-\frac{\arctan\left(\frac{\sqrt{\text{y}}}{\sqrt{\text{y}^2-\text{y}-1}}\right)}{\sqrt{\text{y}}\cdot\sqrt{\text{y}^2-\text{y}-1}}\right\}+\frac{1}{\text{y}}\tag{10}$$
A: 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
$$\frac{-x^2-x y^2+x+1}{-x^2y-y^2+y+1}$$
