Solving the integral $\int_0^1\int_0^1\frac{1}{(1-\alpha x^\beta y^\gamma)^2}dxdy$

How to solve $$\int_0^1\int_0^1\frac{1}{(1-{\alpha}x^{\beta}y^{\gamma})^2}dxdy$$ where $0\leq\alpha<1$, $\beta\geq0$, and $\gamma\geq0$? Does it have any closed-form?

I'm not getting any method to solve this integral. Any hint is appreciated.

Solution 1:

Case 1: $\beta \neq \gamma$ and $\beta, \gamma > 0$

As has been noted in the comments, the integral can be expressed in closed form in terms of the hypergeometric function $_2 F_1(a,b;c;z)$ provided $\beta \neq \gamma$. To do this we will make use of the following integral representation for the hypergeometric function of $$_2 F_1 (a,b;c;z) = \frac{\Gamma (c)}{\Gamma (b) \Gamma (c - b)} \int_0^1 \frac{t^{b - 1} (1 - t)^{c - b - 1}}{(1 - zt)^a} \, dt. \tag1$$

Now, let $$I = \int_0^1 \int_0^1 \frac{1}{(1 - \alpha x^\beta y^\gamma)^2} \, dx dy,$$ where $0 \leqslant \alpha < 1, \beta > 0, \gamma > 0$. Setting $z = x^\beta y^\gamma$, then $$dx = \frac{y^{-\gamma/\beta}}{\beta z^{1 - 1/\beta}} \, dz,$$ and the integral becomes $$I = \frac{1}{\beta} \int_0^1 \int_0^{y^\gamma} \frac{1}{(1 - \alpha z)^2} \frac{y^{-\gamma/\beta}}{z^{1 - 1/\beta}} \, dz dy.$$ Changing the order of integration gives $$I = \frac{1}{\beta} \int_0^1 \int_{z^{1/\gamma}}^1 \frac{1}{(1 - \alpha z)^2} \frac{y^{-\gamma/\beta}}{z^{1 - 1/\beta}} \, dy dz.$$ The $y$-integration can now be readily performed. Provided $\beta \neq \gamma$ the result is \begin{align*} I &= \frac{1}{\beta - \gamma} \int_0^1 \frac{z^{\frac{1}{\beta} - 1} - z^{\frac{1}{\gamma} - 1}}{(1 - \alpha z)^2} \, dz \tag2 \end{align*}

We now rewrite (2) in the form of (1) as follows \begin{align*} I &= \frac{1}{\beta - \gamma} \int_0^1 \frac{z^{\frac{1}{\beta} - 1} (1 - z)^{(\frac{1}{\beta} + 1) - \frac{1}{\beta} - 1}}{(1 - \alpha z)^2} \, dz - \frac{1}{\beta - \gamma} \int_0^1 \frac{z^{\frac{1}{\gamma} - 1} (1 - z)^{(\frac{1}{\gamma} + 1) - \frac{1}{\gamma} - 1}}{(1 - \alpha z)^2} \, dz\\ &= \frac{1}{\beta - \gamma} \frac{\Gamma \left (\frac{1}{\beta} \right ) \Gamma (1)}{\Gamma \left (1 + \frac{1}{\beta} \right )}\ _2 F_1 \left (2, \frac{1}{\beta}; 1 + \frac{1}{\beta}; \alpha \right ) - \frac{1}{\beta - \gamma} \frac{\Gamma \left (\frac{1}{\gamma} \right ) \Gamma (1)}{\Gamma \left (1 + \frac{1}{\gamma} \right )}\ _2 F_1 \left (2, \frac{1}{\gamma}; 1 + \frac{1}{\gamma}; \alpha \right ). \end{align*}

From the following property for the Gamma function, namely $\Gamma (1 + z) = z\Gamma (z)$, the result can be simplified and leads to $$\int_0^1 \int_0^1 \frac{1}{(1 - \alpha x^\beta y^\gamma)^2} \, dx dy = \frac{\beta}{\beta - \gamma}\ _2 F_1 \left (2, \frac{1}{\beta}; 1 + \frac{1}{\beta}; \alpha \right ) - \frac{\gamma}{\beta - \gamma}\ _2 F_1 \left (2, \frac{1}{\gamma}; 1 + \frac{1}{\gamma}; \alpha \right ),$$ provided $\beta \neq \gamma$.

Case 2: $\beta = 0, \gamma > 0$

The integral reduces to $$I = \int_0^1 \int_0^1 \frac{1}{(1 - \alpha y^\gamma)^2} \, dx dy = \int_0^1 \frac{dy}{(1 - \alpha y^\gamma)^2} \, dy.$$ Enforcing a substitution of $y \mapsto y^{1/\gamma}$ gives \begin{align*} I &= \frac{1}{\gamma} \int_0^1 \frac{y^{\frac{1}{\gamma} - 1}}{(1 - \alpha y)^2} \, dy\\ &= \frac{1}{\gamma} \int_0^1 \frac{y^{\frac{1}{\gamma} - 1}(1 - y)^{(\frac{1}{\gamma} + 1) - \frac{1}{\gamma} - 1}}{(1 - \alpha y)^2} \, dy\\ &= \frac{1}{\gamma} \frac{\Gamma (1/\gamma) \Gamma (1)}{\Gamma (1 + 1/\gamma)}\ _2 F_1 \left (2, \frac{1}{\gamma}; 1 + \frac{1}{\gamma}; \alpha \right )\\ &= \ _2 F_1 \left (2, \frac{1}{\gamma}; 1 + \frac{1}{\gamma}; \alpha \right ). \end{align*}

Case 3: $\gamma = 0, \beta > 0$

Similarly, by symmetry we have $$I = \ _2 F_1 \left (2, \frac{1}{\beta}; 1 + \frac{1}{\beta}; \alpha \right ).$$

Case 4: $\beta = \gamma = 0$

Here we trivially have $$I = \frac{1}{(1 - \alpha)^2}.$$

Case 5: $\beta = \gamma$ and $\beta, \gamma > 0$

Thanks to @Paul Enta the solution to the integral for this particular case can now be given.

If $\beta = \gamma$, after performing the $y$-integration rather than the result given in (2) one will instead have $$I = -\frac{1}{\beta^2} \int_0^1 \frac{z^{\frac{1}{\beta} - 1} \ln z}{(1 - \alpha z)^2} \, dz. \tag3$$

Now consider the integral $$J(s) = -\frac{1}{\beta^2} \int_0^1 \frac{z^{s + \frac{1}{\beta} - 1}}{(1 - \alpha z)^2} \, dz.$$ Now observe that $J'(0) = I$ so in order to find $I$ we just need to find $J$ first before taking its derivative with respect to the parameter $s$ before setting it equal to zero.

Now \begin{align*} J(s) &= -\frac{1}{\beta^2} \int_0^1 \frac{z^{s + \frac{1}{\beta} - 1} (1 - z)^{(s + \frac{1}{\beta} + 1) - s - \frac{1}{\beta} - 1}}{(1 - \alpha z)^2} \, dz\\ &= -\frac{1}{\beta^2} \frac{\Gamma (s + 1/\beta) \Gamma (1)}{\Gamma (s + 1/\beta + 1)} \ _2F_1 \left (2, s + \frac{1}{\beta}; s + 1 + \frac{1}{\beta}; \alpha \right )\\ &= -\frac{1}{\beta^2} \frac{1}{(s + \frac{1}{\beta})} \ _2F_1 \left (2, s + \frac{1}{\beta}; s + 1 + \frac{1}{\beta}; \alpha \right ). \end{align*} Thus \begin{align*} J'(s) &= \frac{1}{\beta^2 (s + \frac{1}{\beta})^2} \ _2F_1 \left (2,s + \frac{1}{s}; s + 1 + \frac{1}{s}; \alpha \right ) - \frac{1}{\beta^2 (s + \frac{1}{\beta})} \frac{\partial}{\partial s} \ _2F_1 \left (2,s + \frac{1}{\beta}; s + 1 + \frac{1}{\beta}; \alpha \right ).\tag4 \end{align*} Now \begin{align*} \frac{\partial}{\partial s} \ _2F_1 \left (2,s + \frac{1}{\beta}; s + 1 + \frac{1}{\beta}; \alpha \right ) &= \frac{\partial}{\partial s} \sum_{n = 0}^\infty \frac{(2)_n (s + \frac{1}{\beta})_n}{(s + 1 + \frac{1}{\beta})_n} \frac{\alpha^n}{n!}\\ &= \sum_{n = 0}^\infty (2)_n \frac{\alpha^n}{n!} \frac{\partial}{\partial s} \left [\frac{(s + \frac{1}{\beta})_n}{(s + 1 + \frac{1}{\beta})_n} \right ]\\ &= \sum_{n = 0}^\infty (2)_n \frac{n}{(s + n + \frac{1}{\beta})^2} \frac{\alpha^n}{n!}. \end{align*}

We now make use of the following property for the Pochhammer symbol $(x)_n$ (the raising factorial) of $$(x)_n = \frac{\Gamma (x + n)}{\Gamma (x)}.$$

So \begin{align*} \frac{\partial}{\partial s} \ _2F_1 \left (2,s + \frac{1}{\beta}; s + 1 + \frac{1}{\beta}; \alpha \right ) &= \sum_{n = 0}^\infty \frac{n(n +1) \alpha^n}{(s + n + \frac{1}{\beta})^2} = \sum_{n = 1}^\infty \frac{n(n +1) \alpha^n}{(s + n + \frac{1}{\beta})^2}, \end{align*} and we have $$J'(0) = \ _2F_1 \left (2,\frac{1}{\beta};1 + \frac{1}{\beta}; \alpha \right ) - \frac{1}{\beta} \sum_{n = 1}^\infty \frac{n (n + 1) \alpha^n}{(n + \frac{1}{\beta})^2}. \tag5$$

It now remains to write the infinite sum appearing in (5) in closed form in terms of a hypergeometric function. Setting $$S = \sum_{n = 1}^\infty \frac{n(n + 1) \alpha^n}{(n + \frac{1}{\beta})^2}.$$ Shifting the index $n \mapsto n + 1$ gives $$S = \alpha \sum_{n = 0}^\infty \frac{(n + 1)(n + 2) \alpha^n}{(n + \frac{1}{\beta} + 1)^2}.\tag6 $$ Now, noting that $$\left (1 + \frac{1}{\beta} \right )_n = \frac{\Gamma \left (n + \frac{1}{\beta} + 1 \right )}{\Gamma \left (1 + \frac{1}{\beta} \right )} = \frac{\left (\frac{1}{\beta} + n \right ) \Gamma \left (\frac{1}{\beta} + n \right )}{\frac{1}{\beta} \Gamma \left (\frac{1}{\beta} \right )}.$$ and $$\left (2 + \frac{1}{\beta} \right )_n = \frac{\Gamma \left (n + \frac{1}{\beta} + 2 \right )}{\Gamma \left (2 + \frac{1}{\beta} \right )} = \frac{\left (\frac{1}{\beta} + n + 1 \right ) \left (\frac{1}{\beta} + n \right ) \Gamma \left (\frac{1}{\beta} + n \right )}{\left (1 + \frac{1}{\beta} \right ) \frac{1}{\beta} \Gamma \left (\frac{1}{\beta} \right )},$$ we see that $$\frac{\left (1 + \frac{1}{\beta} \right )_n}{\left (2 + \frac{1}{\beta} \right )_n} = \frac{1 + \frac{1}{\beta}}{1 + \frac{1}{\beta} + n}.\tag7$$ Also $$(3)_n = \frac{\Gamma (3 + n)}{\Gamma (3)} = \frac{(2 + n)!}{2!} = \frac{(n + 1)(n + 2) n!}{2}.\tag8$$

Using (7) and (8), the sum in (6) can be written as \begin{align*} S &= \frac{2 \alpha}{(1 + \frac{1}{\beta})^2} \sum_{n = 0}^\infty \frac{(3)_n \left (1 + \frac{1}{\beta} \right )_n \left (1 + \frac{1}{\beta} \right )_n}{\left (2 + \frac{1}{\beta} \right )_n \left (2 + \frac{1}{\beta} \right )_n} \frac{\alpha^n}{n!}\\ &= \frac{2 \alpha \beta^2}{(1 + \beta)^2} \ _3F_2 \left (3, 1 + \frac{1}{\beta}, 1 + \frac{1}{\beta}; 2 + \frac{1}{\beta}, 2 + \frac{1}{\beta}; \alpha \right ). \end{align*}

So finally, for the case $\beta = \gamma$ such that $\beta, \gamma > 0$ we have $$I = \ _2F_1 \left (2, \frac{1}{\beta}; 1 + \frac{1}{\beta}; \alpha \right ) - \frac{2 \alpha \beta}{(1 + \beta)^2} \ _3F_2 \left (3,1 + \frac{1}{\beta}, 1 + \frac{1}{\beta};2 + \frac{1}{\beta}, 2 + \frac{1}{\beta}; \alpha \right ).$$

Solution 2:

One feels slightly sorry to post this in view of the truly Herculean task that a previously posted answer represents, but it happens that, at least for every $\alpha$ in $[0,1)$ and every nonnegative $\beta$ and $\gamma$,

$$\int_0^1\!\!\int_0^1\frac{dxdy}{(1-{\alpha}x^{\beta}y^{\gamma})^2}=\sum_{n=0}^\infty\frac{n+1}{(n\beta+1)(n\gamma+1)}\alpha^n$$

Two remarks:

• This identity actually holds for every complex parameters such that $|\alpha|<1$, $\Re\beta>0$ and $\Re\gamma>0$.

• Various equivalent presentations of the RHS exist, as series and/or as special functions evaluated at $\alpha$, but these alternative formulas/denominations seem to bring no further mathematical understanding of the situation hence we omit them.

The proof of the identity above is a one-liner: integrate term by term the series expansion $$\frac1{(1-{\alpha}x^{\beta}y^{\gamma})^2}=\sum_{n=0}^\infty(n+1)\alpha^nx^{n\beta}y^{n\gamma}$$