Decoupling a double integral

I came across this question while making some calculations.

QUESTION. Can you find some transformation to "decouple" the double integral as follows? $$\int_0^{\frac{\pi}2}\int_0^{\frac{\pi}2}\frac{d\alpha\,d\beta}{\sqrt{1-\sin^2\alpha\sin^2\beta}} =\frac14\int_0^{\frac{\pi}2}\frac{d\theta}{\sqrt{\cos\theta\,\sin\theta}}\int_0^{\frac{\pi}2}\frac{d\omega}{\sqrt{\cos\omega\,\sin\omega}}.$$

• Do you know that the equality you cite is true? – LSpice Feb 16 at 17:07
• @LSpice: that I know is true, yes. – T. Amdeberhan Feb 16 at 17:18
• for the record, this integral equals $(16/\pi) \Gamma \left(\frac{5}{4}\right)^4$. – Carlo Beenakker Feb 16 at 18:01
• The second integral is a period of $E^2 = E\times E$ where $E$ is the CM elliptic curve $Y^2 = X^3 - X$. The first is a period of the K3 surface $S: z^2 = xy(1-x)(1-y)(1-xy)$ (using Iosif Pinelis's transformation). It turns out that $S$ is the Kummer surface of $E^2$, i.e. the K3 surface birational to $E^2 / \{\pm 1\}$. One can thus find an explicit birational transformation, though it may take some work. How does this arise in your research? – Noam D. Elkies Feb 17 at 2:47
• if I input the double integral on the left-hand-side into Mathematica it evaluates it as $K(1/2)^2$, which is the integral squared on the right-hand-side, so I would think that the decoupling transformation should be a quick application of a known transformation rule for elliptic integrals; now if only I could ask Mathematica to disclose the intermediate steps... – Carlo Beenakker Feb 17 at 7:29

(Thanks go to Etanche and Jandri) \begin{align}J&=\int_0^{\frac{\pi}{2}}\int_0^{\frac{\pi}{2}} \frac{1}{\sqrt{1-\sin^2(\theta)\sin^2 \varphi}}d\varphi d\theta\\ &\overset{z\left(\varphi\right)=\arcsin\left(\sin(\theta)\sin \varphi\right)}=\int_0^{\frac{\pi}{2}} \left(\int_0^ \theta\frac{1}{\sqrt{\sin(\theta-z)\sin(\theta+ z)}}dz\right)d\theta\tag1\\ &=\frac{1}{2}\int_0^{\frac{\pi}{2}} \left(\int_{u}^{\pi-u}\frac{1}{\sqrt{\sin u\sin v}}dv\right)du \tag2\\ &=\frac{1}{2}\int_0^{\frac{\pi}{2}} \left(\int_{u}^{\frac{\pi}{2}}\frac{1}{\sqrt{\sin u\sin v}}dv\right)du+\underbrace{\frac{1}{2}\int_0^{\frac{\pi}{2}} \left(\int_{\frac{\pi}{2}}^{\pi-u}\frac{1}{\sqrt{\sin u\sin v}}dv\right)du}_{w=\pi-v}\\ &=\int_0^{\frac{\pi}{2}} \left(\int_{u}^{\frac{\pi}{2}}\frac{1}{\sqrt{\sin u\sin v}}dv\right)du\\ &\overset{u\longleftrightarrow v}=\int_0^{\frac{\pi}{2}} \left(\int_{0}^{u}\frac{1}{\sqrt{\sin u\sin v}}dv\right)du\\ &=\boxed{\frac{1}{2}\int_0^{\frac{\pi}{2}} \int_0^{\frac{\pi}{2}}\frac{1}{\sqrt{\sin u\sin v}}dudv} \end{align}

and to obtain the form in the OP, finally substitute $$u=2\theta$$, $$v=2\omega$$.

$$(1)$$: $$\displaystyle dz=\dfrac{\sqrt{\sin^2\theta-\sin^2 z}}{\sqrt{1-\sin^2 z}}d\varphi$$, $$\sin^2 a-\sin^2 b=\sin(a-b)\sin(a+b)$$

$$(2)$$: Change of variable $$u=\theta-z,v=\theta+z$$


A further comment: The main difficulty here appears to be with the integral, say $$L$$, on the left. Expanding $$\dfrac1{\sqrt{1-xy}}$$ in powers of $$xy$$: $$\dfrac1{\sqrt{1-xy}}=\sum_{m=0}^\infty a_mx^my^m$$ with $$a_m:=(-1)^m\binom{-1/2}m$$ and using the expression of the beta function in terms of the gamma function, we have
$$L=\sum_{m=0}^\infty a_m\Big(\int_0^1 dx\, x^{m-1/2}(1-x)^{-1/2}\Big)^2 =\frac{\pi^2}4\,N,$$ where $$N:=\sum_{m=0}^\infty\frac1{4^{3m}}\,\binom{2m}m^3.$$ According to Mathematica, $$N=\frac\pi{\Gamma(3/4)^4}=1.393203929\ldots\quad\text{and hence}\quad L=\frac{\pi^3}{4\Gamma(3/4)^4};$$ however, I do not know how Mathematica does this.

One may also note that $$N$$ is the expected number of times a symmetric simple random walk on $$\mathbb Z^3$$ visits its starting point. That is, $$N=\sum_{m=0}^\infty P(X_{2m}=Y_{2m}=Z_{2m}=0) =E\sum_{m=0}^\infty 1(X_{2m}=Y_{2m}=Z_{2m}=0),$$ where $$X_n:=R_1+\cdots+R_n$$, the $$R_i$$'s are iid Rademacher random variables (with $$P(R_i=\pm1)=1/2$$), and $$(Y_n)$$ and $$(Z_n)$$ are iid copies of $$(X_n)$$. It seems amusing that, whereas the dimension $$3$$ of $$\mathbb Z^3$$ is odd, $$N$$ is the square of a certain nice expression.

Closely related to $$N$$ is Pólya's random walk constant, equal $$1-1/N$$, which is the probability that the random walk will ever return to its starting point.

The equality $$N=\frac\pi{\Gamma(3/4)^4}=1.393203929\ldots$$ seems to be proved by Montroll, formulas (6.11), (6.12), (6.1).

• I have added a further comment to the answer. – Iosif Pinelis Feb 18 at 16:16
• I appreciate your transformation and the added (useful) information. – T. Amdeberhan Feb 18 at 18:19
• This equality of algebraic integrals resembles math.stackexchange.com/questions/2103054/… -- can a similar technique work here? – Matt F. Feb 19 at 10:34
• @MattF. : Thank you for this interesting reference. Unfortunately, at this point I don't see how that technique can be used here, because here $1-xy$ is under the square root. – Iosif Pinelis Feb 19 at 14:27

It might help to put the question in a bit broader perspective. Elliptic integrals provide a variety of such "decoupling" expressions, as listed in Definite Integrals of the Complete Elliptic Integral.

For example, the following identity holds for any $$z>$$1, $$\int_0^{\frac{\pi}2}\int_0^{\frac{\pi}2}\frac{d\alpha\,d\beta}{\sqrt{1-\sin^2\alpha\sin^2\beta}}\frac{\sin\alpha}{\sqrt{z^2-\sin^2\alpha}} =\left(\int_0^{\frac{\pi}2}\frac{d\theta}{\sqrt{z+\cos 2\theta}}\right)^2,\qquad(1)$$ and there is a similar expression for $$0.

There are also closed form expressions for double integrals that include the one in the OP (for $$s=1$$), $$\int_0^{\frac{\pi}{2}}\int_0^{\frac{\pi}{2}}\frac{\sin^{s-1}2\alpha\;d\alpha\,d\beta}{\sqrt{1-\sin^2\alpha\sin^2\beta}}=\left(\frac{\pi\Gamma(s/2)}{2\Gamma(3/4)\Gamma(s/2+1/4)}\right)^2,\;\;{\rm Re}\,s>0.\qquad (2)$$

For $$s=1$$ the right-hand-side equals $$K(1/2)^2$$, and it might be interesting to see if there is a more general expression in terms of the square of an elliptic integral (perhaps by evaluating $$K$$ at one of the elliptic singular values, when the complete elliptic integrals may be computed in analytic form in terms of gamma functions).

Following Iosif Pinelis's suggestion, the last integral can be worked out by substituting the series expansion of the elliptic integral (and transforming to $$u=\sin\alpha$$) $$\int_0^{\frac{\pi}{2}}\int_0^{\frac{\pi}{2}}\frac{\sin^{s-1}2\alpha\;d\alpha\,d\beta}{\sqrt{1-\sin^2\alpha\sin^2\beta}}=2^{s-1}\frac{\pi}{2}\sum_{n=0}^\infty\left(\frac{(2n)!}{2^{2n}(n!)^2}\right)^2\int_{0}^1 u^{2 n} u^{s-1} \left(1-u^2\right)^{\frac{s}{2}-1}\,du$$ $$\qquad=2^{s-3}\Gamma \left(\frac{s}{2}\right)\sum_{n=0}^\infty\frac{\Gamma \left(n+\frac{1}{2}\right)^2 \Gamma \left(n+\frac{s}{2}\right)}{\Gamma (n+1)^2 \Gamma (n+s)}.\qquad (3)$$ Mathematica evaluates the sum over $$n$$ of the gamma functions to give the expression in formula 2. I don't know how Mathematica does it, and have asked here --- with several instructive answers!

• Thank you for the relevant and instructive pointers. – T. Amdeberhan Feb 18 at 18:20
• A quick guess is that your second integral is based on formula (2) in Table I of the paper by Glasser you referred to; is this so? Glasser writes that formulas "such as (1-18) were obtained quite simply by integrating the series representing $\mathbf K(u)$ term by term." I guess, at least for $s=1$ this reduces to the evaluation of the series in the definition of $N$ in my answer. I am no specialist in these matters, but the evaluation of that series does not seem trivial to me. Is it indeed "quite simpl[e]"? – Iosif Pinelis Feb 18 at 19:59
• yes, indeed, the second integral is formula 2; the first integral is formula 11; I have worked out the series expansion, the sum over Gamma functions does not look too complicated, I might ask separately on MO. – Carlo Beenakker Feb 18 at 20:42