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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.

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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:

$$\mathscr{I}_{\space\text{x}}\left(\text{y}\right)=\left(\text{y}^2-1\right)\cdot\left[\frac{\ln\left|\text{y}\cdot\left(\text{x}^2-1\right)+\text{y}^2-1\right|}{2\cdot\text{y}}-\frac{\arctan\left(\frac{\text{x}\cdot\sqrt{\text{y}}}{\sqrt{\text{y}^2-\text{y}-1}}\right)}{\sqrt{\text{y}}\cdot\sqrt{\text{y}^2-\text{y}-1}}\right]_0^1+\frac{1}{\text{y}}=$$ $$\left(\text{y}^2-1\right)\cdot\left\{\frac{\ln\left|\frac{\text{y}^2-1}{\text{y}^2-\text{y}-1}\right|}{2\cdot\text{y}}-\frac{\arctan\left(\frac{\sqrt{\text{y}}}{\sqrt{\text{y}^2-\text{y}-1}}\right)}{\sqrt{\text{y}}\cdot\sqrt{\text{y}^2-\text{y}-1}}\right\}+\frac{1}{\text{y}}\tag{10}$$

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  • $\begingroup$ Wow, many thanks I am going to study your answer. $\endgroup$ – user243301 Jul 3 '17 at 17:41
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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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  • $\begingroup$ IMHO the integral diverges $\endgroup$ – Raffaele Jul 1 '17 at 10:47
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    $\begingroup$ imo the fraction stays between $4/5$ and $5/4$ on the unit square so the integral converges perfectly nicely. $\endgroup$ – mercio Jul 1 '17 at 12:35
  • $\begingroup$ Many thanks for your contribution Raffaele, and @mercio many thanks helping here. $\endgroup$ – user243301 Jul 1 '17 at 14:49

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