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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.

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    $\begingroup$ Expand it into series. Since $\alpha < 1$, the denominator of the integrand is always positive, so the integrand is continuous, thus interchanging $\sum$ and $\int$ is permitted by Fubini. $\endgroup$ – GNUSupporter 8964民主女神 地下教會 Feb 9 '18 at 13:57
  • $\begingroup$ @GNUSupporter Thank you for your suggestion. Would you elaborate, please? $\endgroup$ – user519686 Feb 9 '18 at 15:49
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    $\begingroup$ It has closed-form ,solution by hypergeometric function that CAS says. $\endgroup$ – Mariusz Iwaniuk Feb 9 '18 at 17:06
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    $\begingroup$ I mistakenly thought that $\beta,\gamma \ge 0$. If not, this may not be calculable. $\endgroup$ – GNUSupporter 8964民主女神 地下教會 Feb 9 '18 at 17:09
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    $\begingroup$ @Lolita It would be helpful to know the conditions on $\beta$ and $\gamma$. Are they real or complex numbers? If real, are they positive? $\endgroup$ – David H Feb 9 '18 at 19:50
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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 $$I = \ _2F_1 \left (2, \frac{1}{\beta}; 1 + \frac{1}{\beta}; \alpha \right ) - \frac{2 \alpha \beta}{(1 + \beta)^2} \ _3F_2 \left (3,1 + \frac{1}{\beta}, 1 + \frac{1}{\beta};2 + \frac{1}{\beta}, 2 + \frac{1}{\beta}; \alpha \right ).$$

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    $\begingroup$ This last integral can be evaluated by derivation under the integral: $$ I= -\frac{1}{\beta^2}\left. \frac{d}{ds} \int_0^1 \frac{z^{s+\frac{1}{\beta} - 1} \ln z}{(1 - \alpha z)^2} \, dz\right|_{s=0}$$ which can be calculated as the derivative of an hypergeometric function. The result is $$ I=\ _2 F_1 \left (2, \frac{1}{\beta}; 1 + \frac{1}{\beta}; \alpha \right )-\frac{2\alpha\beta}{\left( 1+\beta \right)^2}\ _3 F_2 \left (3, \frac{1}{\beta}, \frac{1}{\beta}; 2 + \frac{1}{\beta}, 2+\frac{1}{\beta}; \alpha \right ) $$ $\endgroup$ – Paul Enta Feb 16 '18 at 18:28
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    $\begingroup$ Alternatively, from the solution obtained for $\beta\neq\gamma$, it can also be expressed as $$ \lim_{\beta\to\gamma}\frac{\beta \ _2 F_1 \left (2, \frac{1}{\beta}; 1 + \frac{1}{\beta}; \alpha \right )-\gamma \ _2 F_1 \left (2, \frac{1}{\gamma}; 1 + \frac{1}{\gamma}; \alpha \right )}{\beta-\gamma}=\frac{d}{d\beta}\left[\beta \ _2 F_1 \left (2, \frac{1}{\beta}; 1 + \frac{1}{\beta}; \alpha \right )\right]$$ $\endgroup$ – Paul Enta Feb 16 '18 at 18:28
  • $\begingroup$ @Paul Enta - Thanks for your suggestion on how to evaluate the integral when $\beta = \gamma$. It takes quite an effort to evaluate, so thanks again. Just one remark. I have updated my solution where you will find you have made a small error in the argument for the $_3F_2$ term. $\endgroup$ – omegadot Feb 17 '18 at 3:54
  • $\begingroup$ You're welcome. Sorry for the typo. To construct the hypergeometric function, you may find it easier to work with Gamma functions from the beginning, to build the ratio of two successive terms of the series and to identify the constants just like here. $\endgroup$ – Paul Enta Feb 17 '18 at 9:35
  • $\begingroup$ Beautiful answer! Unfortunately, I have reached my daily vote limit and I have to wait a couple of hours before I can vote $(+1)$. $\endgroup$ – Mr Pie Feb 19 '18 at 14:59
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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 $$\frac1{(1-{\alpha}x^{\beta}y^{\gamma})^2}=\sum_{n=0}^\infty(n+1)\alpha^nx^{n\beta}y^{n\gamma}$$

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    $\begingroup$ Sure, but as the OP explicitly asked for a closed-form solution, and by closed-form solution I assumed a solution in terms of any of the known functions, special or otherwise, of mathematical physics, this is what I attempted to provide. I usually don't think of infinite sums as closed in form but of course all functions of mathematical physics can be represented by infinite sums. $\endgroup$ – omegadot Jan 5 at 12:42
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    $\begingroup$ @omegadot The difference is specious. As hinted at in my answer, one could easily express the sum of the series above using one or several from the vast list of special functions with a name, but this brings zero extra understanding. (In addition, and since the goal of your comment here seems actually to justify your answer, I might mention that I fail to understand why all the complications and separate cases in it are needed at all.) $\endgroup$ – Did Jan 5 at 14:37

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