# 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.

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:

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

• Wow, many thanks I am going to study your answer. – user243301 Jul 3 '17 at 17:41

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 $$\frac{-x^2-x y^2+x+1}{-x^2y-y^2+y+1}$$

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