Show: $4\pi=\int_0^{\frac\pi2}\int_0^{\frac\pi2}\frac{4+2\sin^2\phi\sin^2\theta}{\left(1-\frac14\sin^2\phi\sin^2\theta\right)^{\frac32}}d\theta d\phi$

Question

While working on a Ramanujan's series for $\frac1\pi$, I obtained the double integral $$\int_0^{\frac\pi2}\int_0^{\frac\pi2}\frac{4+2\sin^2\phi\sin^2\theta}{\left(1-\frac14\sin^2\phi\sin^2\theta\right)^{\frac32}}d\theta d\phi\tag1$$ which has the exact value of $4\pi$. Using the identities $\int_0^{\frac\pi2}\frac{d\phi}{\sqrt{1-k^2\sin^2\phi}}=K(k)$ and $\int_0^{\frac\pi2}\frac{d\phi}{\left(1-k^2\sin^2\phi\right)^{\frac32}}=\frac{E(k)}{1-k^2}$, $|k|<1$, I reduced the double integral to a scary integral of elliptic integrals below: $$\int_0^{\frac\pi2}\left(\frac{12E(\frac{\sin\theta}{2})}{1-\frac{\sin^2\theta}{4}}-8K(\tfrac{\sin\theta}2)\right)d\theta\tag2$$ I am stuck here. Can something be done in this way? Thanks.

Answer 1

If you do not mind, I change notations to type faster and I shall use Mathematica notation for the elliptic integrals.

$$I=\int_0^{\frac \pi 2}\int_0^{\frac \pi 2}\frac{4+2 \sin ^2(x) \sin^2(y)} {\left(1-\frac{1}{4} \sin^2(x) \sin ^2(y)\right)^{3/2}}\,dx\,dy$$

The inner antiderivative does not make much problem and, as you found, we are left with $$I=\int_0^{\frac \pi 2} \Bigg(96\frac{E\left(\frac{\sin ^2(y)}{4}\right)}{\cos (2 y)+7}-8 K\left(\frac{\sin^2(y)}{4}\right) \Bigg)\,dy$$ Let $y=\sin ^{-1}\left(2 \sqrt{t}\right)$ to make the integrand $$12\frac{ E(t)}{\sqrt{1-4 t}\, (1-t)\, \sqrt{t}}-8\frac{ K(t)}{\sqrt{1-4 t} \sqrt{t}}$$ to be integrated between $0$ and $\frac 14$.

None of the different CAS I have been able to use was able to give any result.

So, using $$E(t)=\frac \pi 2 \sum_{n=0}^\infty \frac{2^{-4 n} ((2 n)!)^2}{(n!)^4}\frac{t^n}{1-2 n}$$ $$K(t)=\frac \pi 2 \sum_{n=0}^\infty \frac{2^{-4 n} ((2 n)!)^2 }{(n!)^4}\,t^n$$

This gives $$I=\frac{\pi ^{5/2}}{2}\sum_{n=0}^\infty (-1)^n \frac{2^{-2 n}\,\, \Gamma \left(n+\frac{1}{2}\right)^2}{\Gamma \left(\frac{3}{2}-n\right)\,\, \Gamma (n+1)^3}\Bigg(2 (2 n-1)+3 \, _2F_1\left(1,n+\frac{1}{2};n+1;\frac{1}{4}\right) \Bigg)$$ what I have not been able to simplify.

But, this would converge very fast, since if $a_n$ is the summand $$\left| \frac{a_{n+1}}{a_n}\right|=\frac{1}{4}-\frac{3}{8 n}+\frac{14}{47 n^2}+O\left(\frac{1}{n^3}\right)$$

Computing the partial sums $$\left( \begin{array}{cc} p & \frac 1\pi \sum_{n=0}^p \\ 0 & 4.599610878 \\ 1 & 4.038756781 \\ 2 & 4.004926745 \\ 3 & 4.000777803 \\ 4 & 4.000137078 \\ 5 & 4.000025834 \\ 6 & 4.000005094 \\ 7 & 4.000001038 \\ 8 & 4.000000217 \\ 9 & 4.000000046 \\ 10 & 4.000000010 \\ \end{array} \right)$$

Edit

The first term is the summation can be simplified

$$\sum_{n=0}^\infty (-1)^n \frac{2^{-2 n}\,\, \Gamma \left(n+\frac{1}{2}\right)^2}{\Gamma \left(\frac{3}{2}-n\right)\,\, \Gamma (n+1)^3}\Big(2 (2 n-1) \Big)=-\frac{16}{\pi ^{3/2}}\,K\left(\frac{2-\sqrt{3}}{4}\right)^2$$ Whet is left is the summation of the Gaussian hypergeometric function (I am stuck).

But, starting from the formula given by @Martin.s. in comments, which is in fact $$\sum_{n=0}^\infty \frac{\sqrt{\pi } 2^{-2 n-3} (n+1) (2 n+1) \left(12 n^2+20 n+9\right) \Gamma \left(n+\frac{1}{2}\right)^3}{\Gamma (n+2)^3}$$

Which should converge very fast since $$ \frac{a_{n+1}}{a_n}=\frac{(2 n+1)^2 (2 n+3) \left(12 n^2+44 n+41\right)}{32 (n+1) (n+2)^2 \left(12 n^2+20 n+9\right)}=\frac{1}{4}-\frac{1}{8 n}+\frac{7}{48 n^2}+O\left(\frac{1}{n^3}\right)$$

The summation reduces to four hypergeometric functions $\,_3F_2\left(.\right)$ which simplify again to give $$\color{blue}{8\, K(u) \left(2 \sqrt{3}\,E(u)-\left(1+\sqrt{3}\right) \,K(u)\right)}\quad \text{where}\quad \color{blue}{u=\frac{2-\sqrt{3}}{4}}$$ and, numerically, this is $4\pi$.

Expanded as series $$\frac I \pi=\frac 8\pi\, K(u) \left(2 \sqrt{3}\,E(u)-\left(1+\sqrt{3}\right) \,K(u)\right)=\sum_{n=0}^\infty\pi\, b_n\,u^n$$ where the first $b_n$ are $$\left( \begin{array}{cc} n & b_n \\ 0 & 2 \left(\sqrt{3}-1\right) \\ 1 & -1-\sqrt{3} \\ 2 & \frac{1}{16} \left(-11-9 \sqrt{3}\right) \\ 3 & \frac{1}{32} \left(-17-13 \sqrt{3}\right) \\ 4 & \frac{-1787-1325 \sqrt{3}}{4096} \\ 5 & \frac{-3047-2223 \sqrt{3}}{8192} \\ 6 & \frac{-42631-30821 \sqrt{3}}{131072} \\ \end{array} \right)$$

The partial sum write $$S_p=\frac{(\alpha_p\sqrt 2-\beta_p)\pi}{\gamma_p}$$

Just a few numbers $$\left( \begin{array}{cc} p & S_p \\ 1 & 4.02465951844750554830928493188 \\ 2 & 4.00123298915036950498303291702 \\ 3 & 4.00006683083446139380248238853 \\ 4 & 4.00000378893460692756546032214 \\ 5 & 4.00000022110230785756394318954 \\ 10 & 4.00000000000018659802032155026 \\ 15 & 4.00000000000000000018698460032 \\ 20 & 4.00000000000000000000000020264 \\ \end{array} \right)$$

Using only the above coefficients $$\frac I \pi=\frac{7 \left(120602791 \sqrt{3}-111237979\right) \pi }{536870912}=4.00000001317$$

Answer 2

We will see that the integral is equivalent to the sum you have linked.

Let be I your integral: $$I=\int_{0}^{\pi/2}\left(\frac{12E(\frac{\sin{t}}{2})}{1-(\frac{\sin{t}}{2})^2}-{8K(\frac{\sin{t}}{2})} \right)dt =12\int_{0}^{\pi/2}\frac{E(\frac{\sin{t}}{2})}{1-(\frac{\sin{t}}{2})^2}dt-8\int_{0}^{\pi/2}{K(\frac{\sin{t}}{2})}dt\\=12I_{1}-8I_{2}.$$ For the first integral we proved here (Showing $\int_0^1\frac{E\left(\tfrac{x}{\sqrt{x^2+8}}\right)}{\sqrt{8-7x^2-x^4}}dx=\frac13K\left(\frac1{\sqrt2}\right)E\left(\frac1{\sqrt{2}}\right)$) that: \begin{align*} \int_{0}^{\pi/2}\frac{E\left(l\sin{x} \right)dx}{1-(l\sin{x})^2}=\frac{\pi^2}{4}\sum_{n=0}^{\infty}\frac{(2n)!^3(2n+1)}{2^{6n}n!^6}l^{2n},\tag{2} \end{align*} That is valid also for $0<l<1$. Putting $l=\frac{1}{2}$ in $(2)$ then: $$12I_{1}=\pi^2\sum_{n=0}^{\infty}\frac{(2n)!^3(6n+3)}{2^{8n}n!^6}\tag{3}.$$ For the second integral using the well known expansion: \begin{align*}K(k)=\frac{\pi}{2}\sum_{n=0}^{\infty}\frac{(2n)!^2}{2^{4n}n!^4}k^{2n},\tag{4} \end{align*} Then replacing in $(4)$ with $k=\frac{\sin{t}}{2}$ and applying Wallis's formula: $$\int_{0}^{\pi/2}\sin^{2n}x=\frac{\pi}{2}\frac{(2n)!}{2^{2n}n!^2}\tag{5}$$ we have: $$8I_{2}=\pi^2\sum_{n=0}^{\infty}\frac{2(2n)!^3}{2^{8n}n!^6}.$$ And then: $$I=12I_{1}-8I_{2}=\pi^2\sum_{n=0}^{\infty}\frac{(6n+1)(2n)!^3}{2^{8n}n!^6}=\pi^2\frac{4}{\pi}=4\pi.$$ Where we have used our answer https://math.stackexchange.com/a/5080646/1187943 to get $4/\pi$ for the sum.

Answer 3

Split the integrand by numerator. Write $I = 4I_1 + 2I_2$, where \begin{align*} I_1 &= \int_0^{\pi/2}\int_0^{\pi/2}\frac{1}{\bigl(1-\frac14\sin^2\phi\sin^2\theta\bigr)^{3/2}}\,d\theta\,d\phi, \\ I_2 &= \int_0^{\pi/2}\int_0^{\pi/2}\frac{\sin^2\phi\sin^2\theta}{\bigl(1-\frac14\sin^2\phi\sin^2\theta\bigr)^{3/2}}\,d\theta\,d\phi. \end{align*}

Apply binomial series expansion. We expand the factor $(1 - \tfrac14\sin^2\phi\sin^2\theta)^{-3/2}$ as a power series in $\sin^2\phi\sin^2\theta$ using Newton’s generalized binomial theorem. $$ (1 - z)^{-3/2} = \sum_{n=0}^\infty \binom{-3/2}{n}(-z)^n = \sum_{n=0}^\infty \frac{(2n+1)\binom{2n}{n}}{4^n}\,z^n. $$ Taking $z = \tfrac14 \sin^2\phi\sin^2\theta$ then yields a series in $(\sin^2\phi\sin^2\theta)^n$.

Interchange sum and integrals. We substitute the binomial expansion into $I_1$ and $I_2$ and integrate term-by-term. Thanks to Fubini/Tonelli justification (the series converges uniformly on $[0,\pi/2]^2$), we get double sums of integrals of the form $$ \int_0^{\pi/2}\sin^{2n}\phi \,d\phi \;\;\text{and}\;\; \int_0^{\pi/2}\sin^{2n}\theta \,d\theta. $$

Use Beta/Gamma integrals. Each one-dimensional integral $\int_0^{\pi/2}\sin^{2n}\alpha\,d\alpha$ has a known closed form in terms of Gamma functions (or a simple ratio involving factorials for integer $n$). We evaluate these exactly.

Combine results. Finally, we compute $I = 4I_1 + 2I_2$

Main Computation

Series expansion of the integrand

We set $x = \sin^2\phi\sin^2\theta$ and write the key factor as $$ \bigl(1-\tfrac14\,x\bigr)^{-3/2} \;=\;(1 - z)^{-3/2}\biggr|_{z=\frac{x}{4}}. $$ By Newton’s generalized Binomial Theorem, for any real exponent $r=-3/2$ we have $$ (1 - z)^{-3/2} = \sum_{n=0}^\infty \binom{-3/2}{n}(-z)^n, $$ where \begin{align*} \binom{-3/2}{n} &= \frac{(-3/2)(-3/2 - 1)\cdots(-3/2 - n + 1)}{n!} \\ &=\;(-1)^n\frac{(3\cdot5\cdot7\cdots (2n+1))}{2^n\,n!}. \end{align*} One can show (by simplifying factorials) that this yields the coefficient form $$ (1 - z)^{-3/2} = \sum_{n=0}^\infty \frac{(2n+1)\,\binom{2n}{n}}{4^n}\;z^n. $$ Substituting $z=\tfrac14 x$ gives \begin{align*} (1-\tfrac14 x)^{-3/2} &= \sum_{n=0}^\infty \frac{(2n+1)\,\binom{2n}{n}}{4^n}\,\Bigl(\frac{x}{4}\Bigr)^n \\ &= \sum_{n=0}^\infty \frac{(2n+1)\,\binom{2n}{n}}{4^{2n}}\;x^n. \end{align*} Thus, for our double integral region $(\phi,\theta)\in[0,\pi/2]^2$ we have uniform convergence of this series since $|\sin^2\phi\sin^2\theta|\le1$. We substitute term-by-term into $I_1$ and $I_2$:

• For $I_1$: \begin{align*} I_1 &= \int_0^{\pi/2}\int_0^{\pi/2}(1-\tfrac14\sin^2\phi\sin^2\theta)^{-3/2}\,d\theta\,d\phi \\ &= \sum_{n=0}^\infty \frac{(2n+1)\binom{2n}{n}}{4^{2n}} \int_0^{\pi/2}\int_0^{\pi/2} (\sin^2\phi\,\sin^2\theta)^n \,d\theta\,d\phi. \end{align*}

• For $I_2$: \begin{align*} I_2 &= \int_0^{\pi/2}\int_0^{\pi/2} \frac{\sin^2\phi\,\sin^2\theta}{(1-\frac14\sin^2\phi\sin^2\theta)^{3/2}}\,d\theta\,d\phi \\ &= \sum_{n=0}^\infty \frac{(2n+1)\binom{2n}{n}}{4^{2n}} \int_0^{\pi/2}\int_0^{\pi/2} \sin^2\phi\sin^2\theta \,( \sin^2\phi\sin^2\theta )^n \,d\theta\,d\phi. \end{align*} Simplifying inside gives $(\sin^2\phi\sin^2\theta)^{n+1}$, so $$ I_2 = \sum_{n=0}^\infty \frac{(2n+1)\binom{2n}{n}}{4^{2n}} \int_0^{\pi/2}\int_0^{\pi/2} (\sin^2\phi\,\sin^2\theta)^{n+1} \,d\theta\,d\phi. $$ Because the integrals over $\phi$ and $\theta$ separate, we can write each double integral as a product of one-dimensional sine-power integrals.

Evaluating one-dimensional sine-power integrals

Recall the standard Beta/Gamma integral for half-integer powers: for any integer $m\ge0$, $$ \int_0^{\pi/2} \sin^{m} x \,dx = \frac{\sqrt{\pi}\,\Gamma\!\bigl(\frac{m+1}{2}\bigr)}{2\,\Gamma\!\bigl(\frac{m}{2}+1\bigr)}. $$ In particular, for even powers $m=2k$ this simplifies (using $\Gamma(k+1/2)$ identities) to $$ \int_0^{\pi/2}\sin^{2k}x \,dx = \frac{(2k)!}{2^{2k}(k!)^2}\;\frac{\pi}{2} \;=\; \frac{\binom{2k}{k}}{4^k}\,\frac{\pi}{2}. $$ We will use this with $k=n$ or $k=n+1$ as needed.

Let us denote $$ A_n = \int_0^{\pi/2}\sin^{2n}\phi \,d\phi, \quad B_n = \int_0^{\pi/2}\sin^{2n}\theta \,d\theta. $$ Since these integrals are identical up to dummy variables, $A_n = B_n = \frac{\binom{2n}{n}}{4^n}\frac{\pi}{2}$.

Using this, each double integral becomes:

• For $I_1$, with power $(\sin^2\phi\sin^2\theta)^n = \sin^{2n}\phi\sin^{2n}\theta$: \begin{align*} \int_0^{\pi/2}\int_0^{\pi/2} (\sin^2\phi\,\sin^2\theta)^n\,d\theta\,d\phi &= \Bigl(\int_0^{\pi/2}\sin^{2n}\phi\,d\phi\Bigr)\Bigl(\int_0^{\pi/2}\sin^{2n}\theta\,d\theta\Bigr) \\ &= A_n\,B_n = \biggl(\frac{\binom{2n}{n}}{4^n}\frac{\pi}{2}\biggr)^2. \end{align*}

• For $I_2$, with power $(\sin^2\phi\sin^2\theta)^{n+1}$: \begin{align*} \int_0^{\pi/2}\int_0^{\pi/2} (\sin^2\phi\,\sin^2\theta)^{n+1}\,d\theta\,d\phi &= A_{n+1}\,B_{n+1} \\ &= \biggl(\frac{\binom{2(n+1)}{n+1}}{4^{n+1}}\frac{\pi}{2}\biggr)^2. \end{align*} Substitute these back into the series for $I_1$ and $I_2$:

• For $I_1$: \begin{align*} I_1 &= \sum_{n=0}^\infty \frac{(2n+1)\binom{2n}{n}}{4^{2n}} \Bigl(A_n B_n\Bigr) \\ &= \sum_{n=0}^\infty \frac{(2n+1)\binom{2n}{n}}{4^{2n}} \biggl(\frac{\binom{2n}{n}}{4^n}\frac{\pi}{2}\biggr)^2. \end{align*} Combine factors: $$ I_1 = \frac{\pi^2}{4} \sum_{n=0}^\infty (2n+1)\,\frac{\binom{2n}{n}^3}{4^{4n}}. $$

• For $I_2$: \begin{align*} I_2 &= \sum_{n=0}^\infty \frac{(2n+1)\binom{2n}{n}}{4^{2n}} \Bigl(A_{n+1}\,B_{n+1}\Bigr) \\ &= \sum_{n=0}^\infty \frac{(2n+1)\binom{2n}{n}}{4^{2n}} \biggl(\frac{\binom{2(n+1)}{n+1}}{4^{n+1}}\frac{\pi}{2}\biggr)^2. \end{align*} Simplify powers of 4 and binomial coefficients. Note $\binom{2(n+1)}{n+1} = \frac{(2n+2)!}{(n+1)!(n+1)!}$. Then $$ I_2 = \frac{\pi^2}{4} \sum_{n=0}^\infty (2n+1)\,\binom{2n}{n}\,\frac{\binom{2(n+1)}{n+1}^2}{4^{4n+2}}. $$ It is helpful to re-index or simplify the expression. Observe that \begin{align*} \binom{2(n+1)}{n+1} &= \frac{(2n+2)(2n+1)}{(n+1)(n+1)} \binom{2n}{n} \\ &= \frac{(2n+2)(2n+1)}{(n+1)^2}\binom{2n}{n}. \end{align*} Therefore $$ \binom{2(n+1)}{n+1}^2 = \frac{(2n+2)^2(2n+1)^2}{(n+1)^4}\binom{2n}{n}^2. $$ Substituting this into $I_2$ yields $$ I_2 = \frac{\pi^2}{4} \sum_{n=0}^\infty (2n+1)\,\binom{2n}{n} \,\frac{(2n+2)^2(2n+1)^2}{(n+1)^4} \frac{\binom{2n}{n}^2}{4^{4n+2}}. $$ Combine factors: $(2n+1)\binom{2n}{n}\cdot \binom{2n}{n}^2 = (2n+1)\binom{2n}{n}^3$ and $4^{4n+2} = 4^2 \cdot 4^{4n} = 16 \cdot 4^{4n}$. So \begin{align*} I_2 &= \frac{\pi^2}{4} \sum_{n=0}^\infty \frac{(2n+1)^3 (2n+2)^2}{(n+1)^4}\;\frac{\binom{2n}{n}^3}{16 \cdot 4^{4n}} \\ &= \frac{\pi^2}{64} \sum_{n=0}^\infty \frac{(2n+1)^3(2n+2)^2}{(n+1)^4}\,\frac{\binom{2n}{n}^3}{4^{4n}}. \end{align*}

Summing the series for $I_1$ and $I_2$

We now have $I_1$ and $I_2$ as infinite series in $n$: $$ I_1 \;=\;\frac{\pi^2}{4}\sum_{n=0}^\infty (2n+1)\,\frac{\binom{2n}{n}^3}{4^{4n}}, \quad I_2 \;=\;\frac{\pi^2}{64}\sum_{n=0}^\infty \frac{(2n+1)^3(2n+2)^2}{(n+1)^4}\,\frac{\binom{2n}{n}^3}{4^{4n}}. $$

So

\begin{aligned} I = &4 \left(\frac{\pi^2}{4}\right) \sum_{n=0}^{\infty } \frac{(2n+1) \binom{2n}{n}^3}{4^{4n}} + 2 \left(\frac{\pi^2}{64}\right) \sum_{n=0}^{\infty } \frac{(2n+1)^3 (2n+2)^2}{(n+1)^4} \cdot \frac{\binom{2n}{n}^3}{4^{4n}} \\ \end{aligned}

I don’t know how to progress further, but according to Claude Leibovici, the result of the sums is $4\pi$, which matches the proposed result.