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Recently I have been working on the following question:

Let \begin{equation*} I_n:= \int\int_{[0,1]^2}\frac{(3x)^n(1-x^3)^n(3y)^n(1-y^3)^n}{(1+xy+x^2 y^2)^{2n+1}}dxdy. \end{equation*} Define \begin{equation*} L(\chi_{-3}, 2) := \sum_{n=1}^{\infty}\frac{\chi_{-3}(n)}{n^2} = \frac{1}{1^2}-\frac{1}{2^2}+\frac{1}{4^2}-\frac{1}{5^2}+... \end{equation*} I want to show that \begin{equation*} I_0=L(\chi_{-3}, 2), \qquad I_1=3L(\chi_{-3}, 2)-2. \end{equation*}

The reason I am working on this is because these $I_n$ satisfy an Apéry-like recursion: \begin{equation*} (n+1)^2 I_{n+1}-(10n^2+10n+3)I_n+9n^2 I_{n-1}=0, \end{equation*} from which it would follow that $I_n \in \mathbb{Q}+\mathbb{Q}L(\chi_{-3}, 2)$ for all $n$. However, I cannot find an original source on this, which is why I tried showing these initial values on my own. I can only show the case where $n=0$, which is as follows:

Let us recall the identity \begin{equation*} 1+z+z^2 = \frac{1-z^3}{1-z} \implies \frac{1}{1+z+z^2}=\frac{1-z}{1-z^3}. \end{equation*} Let $z=xy$, which results in \begin{equation*} I_0 = \int\int_{[0,1]^2}\frac{1}{1+xy+x^2 y^2}dxdy = \int\int_{[0,1]^2}\frac{1-xy}{1-(xy)^3}dxdy = \int\int_{[0,1]^2}\frac{1}{1-(xy)^3}-\frac{xy}{1-(xy)^3}dxdy. \end{equation*} Let us note that on the open unit square we have $|xy| < 1$, so we can write \begin{equation*} \frac{1}{1-(xy)^3} = \sum_{n=0}^{\infty}(xy)^{3n}, \qquad \frac{xy}{1-(xy)^3}= \sum_{n=0}^{\infty}(xy)^{3n+1}. \end{equation*} Plugging this back in our integral gives \begin{equation*} I_0 = \int\int_{[0,1]^2}\sum_{n=0}^{\infty}(xy)^{3n}-\sum_{n=0}^{\infty}(xy)^{3n+1}dxdy = \int_0^1\int_0^1\sum_{n=0}^{\infty}\left((xy)^{3n}-(xy)^{3n+1}\right)dxdy. \end{equation*} Now, using Fubini-Tonelli, we get that \begin{equation*} I_0 = \sum_{n=0}^{\infty} \int_0^1\int_0^1\left((xy)^{3n}-(xy)^{3n+1}\right)dxdy = \sum_{n=0}^{\infty} \frac{1}{(3n+1)^2}-\frac{1}{(3n+2)^2} = \sum_{n=1}^{\infty}\frac{\chi_{-3}(n)}{n^2}=L(\chi_{-3}, 2). \end{equation*}

Does have anyone have an idea for how to handle $n=1$? The same methods as in $n=0$ do not seem to work out if we expand it all (or maybe I do not see how all these terms combined to yield the right result?). Does anyone have a source which talks about this integral and what it is called? I could not find it online either.

Tipeg
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1 Answers1

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Consider $J=I(1)-3I(0)$ so that,

$$J = \int_0^1\int_0^1 \frac{9xy(x^3-1)(y^3-1)}{\left(x^{2} y^{2}+x y+1\right)^{3}} - \frac{3}{x^{2} y^{2}+x y+1} dxdy.$$

This integral can be done with elementary techniques.

First, the integral over $x$ can be done with partial fractions. The result is,

$$J = \int_0^1 2\sqrt{3}\left(y^{2}-\frac{2}{y^{4}}\right)\arctan\! \left(\frac{1+2 y}{\sqrt{3}}\right) -\left(\frac{\pi \sqrt{3}}{3}+\frac{9}{2}\right) y^{2} +3y-\frac{3}{2}-\frac{3}{y^{2}}+\frac{6}{y^{3}}+\frac{2 \pi \sqrt{3}}{3 y^{4}}dy.$$

To avoid worrying about improper integrals, let's find the antiderivative of the integrand. The rational terms are straightforward and so it suffices to compute,

$$H = \int \left(y^{2}-\frac{2}{y^{4}}\right)\arctan\! \left(\frac{1+2 y}{\sqrt{3}}\right)dy.$$

This can be done using integration by parts and partial fractions to get

$$H = \left(\frac{y^{3}}{3}+\frac{2}{3 y^{3}}\right)\arctan\! \left(\frac{1+2 y}{\sqrt{3}}\right)-\frac{1}{2\sqrt{3}}\int \frac{\, y^{6}+2}{ y^{3} \left(y^{2}+y+1\right)}dx,$$ $$\implies H =\left(\frac{y^{3}}{3}+\frac{2}{3 y^{3}}-1\right)\arctan\! \left(\frac{1+2 y}{\sqrt{3}}\right)-\frac{\sqrt{3}\left(y^{4}-2 y^{3}+4 y-2\right)}{12 y^{2}} +C. $$ Therefore, $$J = 2\sqrt{3}\left.\left(\frac{y^{3}}{3}+\frac{2}{3 y^{3}}-1\right)\arctan\! \left(\frac{1+2 y}{\sqrt{3}}\right)-\left(\frac{\pi \sqrt{3}}{9}+\frac{3}{2}\right) y^{3}+y^{2}-\frac{y}{2}+\frac{1}{y}-\frac{2}{y^{2}}-\frac{2 \sqrt{3}\, \pi}{9 y^{3}}\right|_0^1.$$ Evaluating the limits gives $J = \left(-\frac{25 \pi \sqrt{3}}{72}-\frac{43}{16}\right)-\left(-\frac{25 \pi \sqrt{3}}{72}-\frac{11}{16}\right)=-2$.

So, $I(1) = 3I(0)-2 = 3 L(\chi_{-3}, 2)-2$.

Andrew
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