# A tricky double integral

What is

$$\int_0^1 \int_0^1 \frac{ dx \; dy}{1+xy+x^2y^2} ?$$

Can you do one of the integrals and turn it into a single integral? I get lost in a sea of inverse tangents.

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Can you provide us your approach or at least your last step? – Inquest Feb 20 '12 at 16:55
+1, and I wonder why I'm the only one who voted for this. I see questions where someone put a lot of work into a very good answer, that have zero votes in their favor. Should that happen? I never post an answer without up-voting the question. – Michael Hardy Feb 20 '12 at 18:19

The first step would be one $\arctan$ only. The denominator is $$1+xy+x^2y^2 =\left(\frac{1}{2}+xy\right)^2 + \frac{3}{4}$$ Hence, with $a=\frac{2}{\sqrt{3}}$ we have $$\int\frac{dy}{1+xy+x^2y^2} =\int\frac{dy}{\left(\frac{1}{2}+xy\right)^2 + 1/a^2}=\qquad\qquad\qquad\text{ }\\ a^2\int\frac{dy}{a^2\left(\frac{1}{2}+xy\right)^2 + 1}= a^2\frac{\arctan(\frac{a}{2}+axy)}{ax} + C=\\a\frac{\arctan\left(\frac{1}{\sqrt{3}}+\frac{2xy}{\sqrt{3}}\right)}{x} + C$$ so that $$\int_0^1\frac{dy}{1+xy+x^2y^2} = a\frac{\arctan(\frac{1}{\sqrt{3}}+\frac{2x}{\sqrt{3}})}{x} -\frac{\pi}{3\sqrt{3}}\frac{1}{x}$$ The integral of $\arctan(a+bx)/x$ is not that easy, it ends up in dilogarithms...

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Integrating first in $y$ gives $$-\int_0^1\frac{\pi-6\,\arctan\Bigl(\dfrac{2\,x+1}{\sqrt{3}}\Bigr)}{3\,\sqrt{3} x}\,dx,$$ which you probably already know. Mathematica does not evaluate it.

The fact that the integrand depends on $x\,y$ suggests the change of variables $\xi=x\,y$, $\eta=y$. The domain $\{(x,y):0\le x\le1,\ 0\le y\le x\}$ is transformed into $\{(\xi,\eta):0\le\xi\le1,\ \xi\le\eta\le\sqrt{\xi}\,\}$, and $dx\,dy=d\xi\,d\eta/\eta$. Then \begin{align*} \int_{0}^{1} \int_{0}^{1} \frac{ dx\,dy}{1+x\,y+x^2\,y^2}&=2\int_{0}^{1} \int_{0}^{x} \frac{dx\,dy}{1+x\,y+x^2\,y^2}\\ &=2\int_0^1\int_{\xi}^{\sqrt\xi}\frac{1}{1+\xi+\xi^2}\frac{d\xi\,d\eta}{\eta}\\ &=-\int_0^1\frac{\log\xi}{1+\xi+\xi^2}\,d\xi. \end{align*} This last integral is not elementary, but at least Mathematica gives a value interms of the PolyGamma function: $$\frac{1}{9}\Bigl(\operatorname{PolyGamma}[1, \frac23]-\operatorname{PolyGamma}[1,\frac13]\Bigr).$$ The numerical value is $0.781302$.

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Could you explain why $\int_{0}^{1} \int_{0}^{1} \frac{ dx\,dy}{1+x\,y+x^2\,y^2}=2\int_{0}^{1} \int_{0}^{x} \frac{dx\,dy}{1+x\,y+x^2\,y^2}$? (It is not true "pointwise": $\int_{0}^{1} \frac{dy}{1+x\,y+x^2\,y^2} \neq 2 \int_{0}^{x} \frac{dy}{1+x\,y+x^2\,y^2}$ for $x = \frac{1}{3}$ [wolfram].) – savick01 Feb 20 '12 at 18:36
The integrand is symmetric in the variables $x$ and $y$. It is the same if you change $x$ by $y$ and $y$ by $x$. The domain of integration is also symmetric. Then the integral on half the square under the main diagonal ($x=y$) is equal to the integral over it. – Julián Aguirre Feb 20 '12 at 19:06
Of course! Thank you. – savick01 Feb 20 '12 at 19:12

You can rewrite the integral as $$\int_0^1 \int_0^1 {1 - xy \over 1 - x^3y^3}\,dx\,dy$$ Expanding ${\displaystyle{1 \over 1 - x^3y^3}}$ as a geometric series $\sum_n x^{3n}y^{3n}$ this becomes $$\int_0^1 \int_0^1 \sum_{n=0}^{\infty} x^{3n}y^{3n} - \sum_{n=0}^{\infty} x^{3n+1}y^{3n+1}\,dx\,dy$$

Integrating this termwise this becomes $$\sum_{n = 0}^{\infty} {1 \over (3n + 1)^2 } - \sum_{n = 0}^{\infty} {1 \over (3n + 2)^2 }$$

I don't think these sums have a closed form in terms of well-known functions; in effect these are the polygamma functions that Julian Aguirre put in his answer.

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+1 Nice and simple! – AD. Feb 20 '12 at 19:29