Integration – Moments of Alternate Elliptic Integrals

conjectureselliptic integralsintegrationreference-request

Let $K_s$ denote Ramanujan's Alternate Elliptic Integrals as follows:

$$K_s:=K_s(k)=\frac{\cos(\pi s)}{2}\int_0^1\frac{t^{s-1/2}}{(1-t)^{1/2+s}(1-k^2t)^{1/2-s}}dt$$

Where in Hypergeometric Form it is:
$$K_s=\frac{\pi}{2}{}_2F_1\left[\frac{1}{2}+s,\frac{1}{2}-s,1,k^2\right]$$

The following signatures are important in this theory:
$$s=\left\{0,\frac{1}{3},\frac{1}{4},\frac{1}{6}\right\}$$

I have found a General Theorem to calculate the following Integral:

$$\int_0^1kK_uK_vdk$$

As examples I will present the corollaries that follow from the Theorem while considering the combination of $4$ signatures.

This gives us the following 10 Integrals:

Corollaries:
$$\int_0^1kK^2dk=\frac{7\zeta(3)}{4}$$

$$\int_0^1kK^2_{1/6}dk=\frac{9\sqrt{3}}{8}\operatorname{Cl_2}(\pi/3)$$

$$\int_0^1kK^2_{1/3}dk=\frac{15\sqrt{3}}{16}\operatorname{Cl_2}(\pi/3)$$

$$\int_0^1kK^2_{1/4}dk=2\beta(2)$$

$$\int_0^1kKK_{1/6}dk=\frac{27\sqrt{3}}{4}\ln 3-9\sqrt{3}\ln 2$$

$$\int_0^1kKK_{1/3}dk=\frac{27}{16}\ln 3$$

$$\int_0^1kKK_{1/4}dk=2\sqrt{2}\ln 2$$

$$\int_0^1kK_{1/3}K_{1/6}dk=\frac{3\sqrt{3}}{2}\ln 2$$

$$\int_0^1kK_{1/4}K_{1/6}dk=\frac{27\sqrt{6}}{5}\ln 2-\frac{27\sqrt{6}}{10}\ln3$$

$$\int_0^1kK_{1/4}K_{1/3}dk=\frac{27\sqrt{2}}{14}\ln 3-\frac{9\sqrt{2}}{7}\ln 2$$

The very first Integral is known in this list as $K_0$ is just the Complete Elliptic Integral of the First Kind $K$.

I have not seen the General Theorem and Other Integrals as well anywhere in Literature.

So I was wondering whether these results are new or already known?

This is only a Small List, one may produce as many Integrals as they want from the General Theorem.

Like for example after Differentiating:

$$\int_0^1kK\left[\frac{d}{dv}K_v\right]_{v=1/4}dk=2\sqrt{2}\left(8\beta(2)-\pi \ln2-8\ln 2\right)$$

This opens up new possibilities too.

The following is for example already known:

$$\int_0^1k^{2n+1}K^2dk=a_n\zeta(3)+b_n$$

where $a_n,b_n\in\mathbb{Q}$.

This is found using "FindIntegerNullVector":

Conjecture:

$$\int_0^1k^{2n+1}K^2_{1/6}dk=a_n\sqrt{3}\operatorname{Cl_2}(\pi/3)+b_n$$

$$\int_0^1k^{2n+1}K^2_{1/3}dk=a_n\sqrt{3}\operatorname{Cl_2}(\pi/3)+b_n$$

$$\int_0^1k^{2n+1}K^2_{1/4}dk=a_n\beta(2)+b_n$$

$$\int_0^1k^{2n+1}KK_{1/3}dk=a_n\ln 3+b_n$$

where $a_n,b_n\in\mathbb{Q}$.

The rest of the cases doesn't seem to continue the pattern.

I wonder if there is any greater meaning behind this:

$$\left[
\begin{array}{c}
\text{Integral}&\text{Basis}
\\\\
\displaystyle\int_0^1k^{2n+1}K_0^2dk&\zeta(3)\\\\
\displaystyle\int_0^1k^{2n+1}K_{1/6}^2dk&\sqrt{3}\operatorname{Cl_2}(\pi/3)\\\\
\displaystyle\int_0^1k^{2n+1}K_{1/3}^2dk&\sqrt{3}\operatorname{Cl_2}(\pi/3)\\\\
\displaystyle\int_0^1k^{2n+1}K_{1/4}^2dk&\beta(2)\\\\
\displaystyle\int_0^1k^{2n+1}K_0K_{1/3}dk&\ln3\\\\
\end{array}
\right]$$

Best Answer

As in this answer , these integrals can be evaluated using the decomposition of the hypergeometric functions on the Legendre polynomials (see for example here): \begin{equation} {}_2F_1\left(a,b;1;x\right)=\frac{\Gamma \left(2-a -b \right)}{\Gamma \left(a \right) \Gamma \left(b \right)}\sum_{p=0}^\infty (2 p+1)\frac{ \Gamma \left(a +p \right) \Gamma \left(b +p \right) } { \Gamma \left(2 -a +p\right) \Gamma \left(2-b +p \right)}P_p(2x-1) \end{equation} which gives \begin{align} \mathbf K_{a}\left(x \right)&=\frac\pi2{}_2F_1\left( a,1-a;1;x^2 \right)\\ &=\frac{\sin\pi a}2\sum_{p=0}^\infty \frac{(-1)^p(2p+1)}{(p+a)(p+1-a)}P_p(1-2x^2) \end{align} From the orthogonality of the Legendre polynomials, one obtains \begin{align} \int_0^1xP_p(1-2x^2)P_q(1-2x^2)\,dx&=\frac14\int_{-1}^1P_p(u)P_q(u)\,du\\ &=\frac{1}{2p+1}\delta_{p,q} \end{align} and thus \begin{equation} \int_0^1x\mathbf K_{a}\left(x \right)\mathbf K_{b}\left(x \right)\,dx=\frac{\sin\pi a\sin\pi b}{8}\sum_{p=0}^\infty \frac{2p+1}{(p+a)(p+1-a)(p+b)(p+1-b)} \end{equation} In many cases, these series with rational terms can be evaluated in closed form which are simple, as those given in the OP.

To obtain higher odd moments, \begin{equation} S_n=\int_0^1x^{2n+1}\mathbf K_{a}\left(x \right)\mathbf K_{b}\left(x \right)\,dx \end{equation} one can use the recurrence relation for the Legendre polynomials : \begin{equation} zP_p(z)=\frac{1}{2p+1}\left[(p+1)P_{p+1}(z)+pP_{p-1}(z)\right] \end{equation} with $z=1-2x^2$, one has \begin{equation} (1-2x^2)P_p(1-2x^2)=\frac{1}{2p+1}\left[(p+1)P_{p+1}(1-2x^2)+pP_{p-1}(1-2x^2)\right] \end{equation} and thus \begin{equation} x^2P_p(1-2x^2)=\frac12\left[P_p(1-2x^2)-\frac{p+1}{2p+1}P_{p+1}(1-2x^2)-\frac{p}{2p+1}P_{p-1}(1-2x^2)\right] \end{equation} For the third moments, the integrals to be evaluated read \begin{align} \int_0^1x^3P_p(1-2x^2)P_q(1-2x^2)\,dx&=\frac12I_{pq}-\frac{p+1}{2(2p+1)}I_{p+1,q}-\frac{p}{2(2p+1)}I_{p-1,q}\\ I_{n,q}&= \int_0^1xP_n(1-2x^2)P_q(1-2x^2)\,dx\\ &=\frac{1}{2q+1}\delta_{n,q} \end{align} The third moment can then be written as \begin{align} S_3&=\int_0^1x^3\mathbf K_{a}\left(x \right)\mathbf K_{b}\left(x \right)\,dx\\ &=\frac{S1}{2}-\frac{\sin\pi a\sin\pi b}{16}\sum_{p,q=0}^\infty \frac{(-1)^{p+q}(2p+1)(2q+1)}{(p+a)(p+1-a)(q+b)(q+1-b)}\\ &\hspace{3cm}\times\frac{(p+1)\delta_{p+1,q}+p\delta_{p-1,q}}{(2p+1)(2q+1)}\\ &=\frac{S1}{2}+\frac{\sin\pi a\sin\pi b}{16}\sum_{p=0}^\infty \frac{p+1}{(p+a)(p+1-a)(p+1+b)(p+2-b)}\\ &+\frac{\sin\pi a\sin\pi b}{16}\sum_{p=0}^\infty \frac{p}{(p+a)(p+1-a)(p-1+b)(p-b)} \end{align} As above, these series can often be evaluated under closed forms. For example, \begin{equation} \int_0^1x^3\mathbf K\left(x \right)\mathbf K_{1/6}\left(x \right)\,dx=\frac{1755}{2048}\log3+\frac{81}{512} \end{equation} while \begin{equation} \int_0^1x\mathbf K\left(x \right)\mathbf K_{1/6}\left(x \right)\,dx=\frac{27}{16}\log 3 \end{equation} (I think that there is a typo in the OP, $K_{1/3}$ and $K_{1/6}$ are exchanged. Present results were numerically checked).

Edit 19/01/2024

To express the moment of order $1$, one can use the summation method by the polygamma function express the series as a weighted sum of $\psi(a),\psi(1-a),\psi(b),\psi(1-b)$ with rational coefficients. Indeed, partial decomposition of the terms of the series gives \begin{equation} \frac{2p+1}{(p+a)(p+1-a)(p+b)(p+1-b)}= \frac{1}{(b-a)(1-a-b)}\left[ \frac{1}{p+a}+\frac{1}{p+1-a} -\frac{1}{p+b}-\frac{1}{p+1-b}\right] \end{equation} Then, assuming $a\ne b$, $a\ne1/2,b\ne1/2$ and non-integer, \begin{equation} \int_0^1x\mathbf K_{a}\left(x \right)\mathbf K_{b}\left(x \right)\,dx=\frac{\sin\pi a\sin\pi b}{8(a-b)(1-a-b)}\left[ \psi(a)+\psi(1-a)-\psi(b)-\psi(1-b)\right] \end{equation} With the reflection formula $\psi\left(z\right)-\psi\left(1-z\right)=-\pi/\tan\left(\pi z\right)$, \begin{align} S_1=\frac{\sin\pi a\sin\pi b}{8(a-b)(1-a-b)}\left[ 2\psi(a)-2\psi(b)+\pi\cot\pi a-\pi\cot\pi b\right] \end{align} If $a=1/2$ and/or $b=1/2$ or $a=b$ the same method gives the result in terms of polygamma functions.

To obtain higher order moments, one can apply the recurrence method described above. It can be seen that for a moment $2n+1$, Legendre polynomials of order $p$ are converted into a weighted sum of polynomials of order $p+n,p+n-2,\ldots,p-n$. In the corresponding series, the factors in the denominators $p+b$ become $p+k+b$ and the factors $p+1-b$ become $p+1-k-b$, with $k=-n,-n+2,\ldots,n$.

Assuming $a,b\ne1/2,a\ne b$ and non-integer, the corresponding poles are all different. The summation method expresses the series as weighted sums of $\psi(a),\psi(1-a),\psi(k+b),\psi(1-k-b)$ with rational coefficients.

The recurrence relation $\psi\left(z+1\right)=\psi\left(z\right)+\frac{1}{z}$ and the reflection formula $\psi\left(z\right)-\psi\left(1-z\right)=-\pi/\tan\left(\pi z\right)$ show that the resulting series are expressible with weighted terms of $\psi(b)$ and $\cot \pi b$ plus rational terms. The former terms are already present in the evaluation $S_1$. Thus, \begin{equation} S_{2n+1}=\sin\pi a\sin\pi b\left[\alpha_n\psi(b)+\beta_n\psi(b)+\gamma_n\pi\cot\pi a+\delta_n\pi\cot\pi b+a_n\right] \end{equation} where all the coefficient are rational.

Here again, the cases $a=1/2$ and/or $b=1/2$ or $a=b$ the same method can be used to express the results. These expressions could be used to validate the conjecture in the given specific cases.

Related Solutions

[Math] Relation between integral, gamma function, elliptic integral, and AGM

This one is a cakewalk. Just use the definition $$K(1/2)=\int_{0}^{\pi/2}\frac{dx}{\sqrt{1-(1/2)\sin^{2}x}}=\frac{\sqrt{2}}{2}\int_{0}^{\pi}\frac{dx}{\sqrt{2-\sin^{2}x}}$$ and then put $t=\tan (x/2)$ so that $dx=2/(1+t^2)\,dt$ and $$2-\sin^{2}x=2-\frac{4t^2}{(1+t^2)^{2}}=2\cdot\frac{1+t^{4}}{(1+t^{2})^{2}}$$ and hence we have $$K(1/2)=\frac{1}{\sqrt{2}}\int_{0}^{\infty}\frac{1+t^{2}}{\sqrt{2}\sqrt{1+t^{4}}}\cdot\frac{2}{1+t^2}\,dt=\int_{0}^{\infty}\frac{dt}{\sqrt{1+t^{4}}}$$ Similarly put $\tan t=x^2$ to get $$\frac{1}{2}\int_{0}^{\pi/2}\frac{dt}{\sqrt{\sin t\cos t}} =\int_{0}^{\infty} \frac{dx} {\sqrt{1+x^{4}}}$$ This avoids the more complicated approaches from the theory of elliptic and theta functions and AGM.

Elliptic integrals, general form

The given integral is $K(1/2)/2$ (convention!), but i will try to follow also a geometric path, the canonical one, detailed in the given particular case $x_-=1$, and $x_+=2$, so that things are easier to type. There will be also numerical check (also for intermediate steps), so that i can calmly sleep tonight.

Let us recall the Complete elliptic integral of the first kind: $$ K(k) = \int_0^{\pi/2}\frac{d\theta}{\sqrt{1-k^2\sin^2\theta}} = \int_0^1 \frac {dt}{\sqrt{(1-t^2)(1-k^2t^2)}}\ . $$ See also DLMF on Elliptic integrals, where $K(k)=F(\pi/2,k)$, and in the definition of $F$ the above convention is used.

Note that the above convention differs from the convention used by Mathematica, as also remarked in loc. cit. .

Taking in the last integral $k=\frac 12$ we obtain an integral "similar" to the given one, at least under the integral, but the limits of integration differ. What can we do?

The lower integration limit $0$ is the point of symmetry for $(1-t^2)(1-k^2t^2)$, and $1$ is one of its roots. In our case, for $(x^2-1)(x^2-4)$ the value $2$ is also a root, so the substitution $t=2/x$ is natural. We get for the given integral $I$ (well, let us have this handy notation): $$ \begin{aligned} I&= \int_{2}^\infty \frac1{\sqrt{(x^2-1)(x^2-4)}}\; dx \\ &= \int_1^0 \frac1{\sqrt{\left(\frac 4{t^2}-1\right)\left(\frac 4{t^2}-4\right)}}\; \left(-\frac 2{t^2}\right)\; dt \\ &= \int_0^1 \frac1{\sqrt{(4-t^2)(4- 4t^2)}}\; 2\; dt \\ &= \frac12 \int_0^1 \frac1{\sqrt{(1-\frac 14t^2)(1- 1t^2)}}\; dt \\ &=\frac 12K\left(\frac 12\right)\ . \end{aligned} $$

Numerical check using sage, which also suffers from the same convention as Mathematica.

sage: var('x,t');
sage: I = integral( 1/sqrt( (x^2-1)*(x^2-4) ), x, 2, oo, hold=1 )
sage: I.n()
0.8428751774069201
sage: def K(k):    return elliptic_kc(k^2)    # convention
sage: K(1/2).n() / 2
0.842875177406298

pari/gp has the wiki convention, a numerical check using pari, taking advantage of the implementation of ellK, recently made public, October 2020, is as follows:

? intnum(x=2, oo, 1/sqrt( (x^2-1)*(x^2-4) ))
%1 = 0.84287517740629802143560182889542910036

? \p 60
   realprecision = 77 significant digits (60 digits displayed)

? intnum(x=2, oo, 1/sqrt( (x^2-1)*(x^2-4) ))
%2 = 0.842875177406298021435601828899538494750400447070248008383945

? ellK(1/2)/2
%3 = 0.842875177406298021435601828899538494750400447070544522059974

It seems that Mathematica goes an other way when evaluating the given integral, thus delivering a value which reflects 19.7.3 in DLMF, Legendre's relations, explicitly: $$K(1/k) = k(K(k)\mp iK(k')\ ,\qquad k^2 + {k'}^2=1\ .$$ In our case, with $k=1/2$, we get $K(2)=\frac 12(K(1/2)\mp iK(\sqrt{3/2}))$. This is compatible with the result of Mathematica, written in its convention.

This is already an answer to the question.

$\square$

A possible digression: Please ignore, if it feels not related to the question. This is inserted for my own purpose, so that i can find it next time as an explicit example. For me there is a natural relation, the most natural way to deal with elliptic integrals is to see them explicitly in relation with elliptic curves. We will obtain (after an expensive tour) an other realization of the given integral in terms of the complete elliptic integral, $K(\sqrt{8/9})/3$.

The given integral can be written as $$ \color{blue}{ I = \int_{(2,0)}^{\infty}\frac{dx}y } \ , $$ and it can be considered on the compact manifold $Q(\Bbb R)$ or $\Bbb C$, where $Q$ is the "quartic curve" (an algebraic variety, thus analytic, thus smooth manifold) with affine equation $$ Q\ :\ y^2 =(x-1)(x+1)(x-2)(x+2)\ . $$ (The infinity point is the point $(0:1:0)$ which verifies the projective form, the homogenized form of this equation, $y^2z^2=(x-z)(x+z)(x-2z)(x+2z)$.)

The idea is to obtain a birational morphism to an elliptic curve.

We use the (birational) transformations: $$ \left\{ \begin{aligned} x &=-\frac 4Y(X-5)\ ,\\ y &=2-\frac 4{Y^2}X(X-5)^2\ , \end{aligned} \right. \qquad \left\{ \begin{aligned} X &=-\frac 4{x^2}(y-2)\ ,\\ Y &=\frac 1{x^3}(16(y-2)+20x^2)\ , \end{aligned} \right. $$ and then $$ y^2- (x^2-1)(x^2-4) = \frac{16(X - 5)^3}{Y^4}\cdot(-Y^2+ X^3 - 5X^2- 16X + 80)\ , $$ so we have a passage to the elliptic curve with the equation from the last parentesis, $$ Y^2 =(X-5)(X-4)(X+4)\ . $$ We may (or may not) want to keep it simple (and unstructural), thus involve only the substitution $$ x = -\frac{4(X-5)}{\sqrt{(X-4)(X+4)(X-5)}} \ . $$ We have already made the "natural choice" for the square root. Note that for $X=2$ we obtain $x=2$. And $X\searrow -4$ implies $x\nearrow \infty$.

The structure or brute force computations shows now that we have: $$ \color{blue}{ I= \int_2^\infty\frac {dx}{\sqrt{(x^2-1)(x^2-4)}} = \int_{-4}^2 \frac {dX}{\sqrt{(X-4)(X+4)(X-5)}}\ . } $$ This is the passage from the integral on the quartic to an integral on the elliptic curve written in Weierstraß form.

Instead of a proof, here is a numerical check using pari/gp:

? \p 60
   realprecision = 77 significant digits (60 digits displayed)
? intnum( x=2, oo, 1/sqrt( (x^2-1)*(x^2-4) ) )
%2 = 0.842875177406298021435601828899538494750400447070248008383945
? intnum( X=-4, 2, 1/sqrt( (X-4)*(X+4)*(X-5) ) )
%3 = 0.842875177406298021435601828899538494750400447070446045891423

We are thus in the position to calculate the integral of the canonical differential on the elliptic curve $E$ with equation $$ (E)\qquad Y^2=(X-e_1)(X-e_2)(X-e_3)\ ,\qquad e_1,e_2,e_3=5,4,-4,\ e_1>e_2>e_3\ . $$ The limit of integration $e_1=-4$ corresponds to one torsion point of order two, $(e_1,0)$. The other $2$-torsion points are $(e_2,0)$, and $(e_3,0)$. Can we give a structural sense to the limit of integration $2$ from the blue realtion?!

Yes, the points $(2,\pm 6)$ are ($\Bbb Q$-rational) points on $E$, and they are $4$-torsion points. For this reason we have the chain of relations: $$ \begin{aligned} I &= \int_2^\infty\frac {dx}{\sqrt{(x^2-1)(x^2-4)}} \\ &= \int_{-4}^2 \frac {dX}{\sqrt{(X+4)(X-4)(X-5)}} = \int_2^4 \frac {dX}{\sqrt{(X+4)(X-4)(X-5)}} \\ &=\frac12\int_{e_3=-4}^{e_2=4} \frac {dX}{\sqrt{(X-e_1)(X-e_2)(X-e_3)}} \\ &=\frac 12\omega_1(E)\qquad\text{(real half-period of the elliptic curve $E$)} \\ &=\frac 1{\sqrt{e_1-e_3}}K\left(\sqrt{\frac{e_2-e_3}{e_1-e_3}}\right) \\ &=\frac 1{\sqrt{5-(-4)}}K\left(\sqrt{\frac{4-(-4)}{5-(-4)}}\right) \\ &=\frac 13 K\left(\frac{2\sqrt 2}3\right)\ . \end{aligned} $$

Let us numerically check again, pari/gp.

? intnum(x=2, oo, 1/sqrt( (x^2-1)*(x^2-4) ))
%10 = 0.842875177406298021435601828899538494750400447070248008383945
? ellK(sqrt(8/9))/3
%11 = 0.842875177406298021435601828899538494750400447070544522059974

Addendum. Computer algebra support for the elliptic curve computations.

Using sage:

sage: var('x,y,X,Y');
sage: def f(x,y): return y^2 - (x^2-1)*(x^2-4)
sage: factor( f(-4*(X-5)/Y, 2-4*X*(X-5)^2/Y^2) )
16*(X^3 - 5*X^2 - Y^2 - 16*X + 80)*(X - 5)^3/Y^4
sage: E = EllipticCurve(QQ, [0, -5, 0, -16, 80])
sage: E
Elliptic Curve defined by y^2 = x^3 - 5*x^2 - 16*x + 80 over Rational Field
sage: for T in E.torsion_points():
....:     if T != E(0):
....:         print(f"Order {T.order()} for the torsion point {T.xy()}")
....: 
Order 2 for the torsion point (-4, 0)
Order 4 for the torsion point (2, -6)
Order 4 for the torsion point (2, 6)
Order 2 for the torsion point (4, 0)
Order 2 for the torsion point (5, 0)
Order 4 for the torsion point (8, -12)
Order 4 for the torsion point (8, 12)

This structural aspect explains why, when we compute the integrals of the canonical differential form $\frac{dX}Y$ of $E$ on the intervals delimited (in $\Bbb R$) by the points $$ -\infty\ ,\ -4\ ,\ 2\ ,\ 4\ ,\ 5\ ,\ 8\ ,\ +\infty $$ we obtain numerically the values for the integral of $(\ (X-4)(X+4)(X-5)\ )^{-1/2}$ on the delimited intervals... (I am taking the absolute value to not confuse sage.)

sage: f = 1/sqrt( abs((X-4)*(X+4)*(X-5)) )
sage: integral(f, X, -oo, -4, hold=True).n()
1.07825782381822
sage: integral(f, X, -4, 2, hold=True).n()
0.8428751689072135
sage: integral(f, X, 2, 4, hold=True).n()
0.8428751680336899
sage: integral(f, X, 4, 5, hold=True).n()
1.0782578072972744
sage: integral(f, X, 5, 8, hold=True).n()
0.8428751637656617
sage: integral(f, X, 8, oo, hold=True).n()
0.8428751775376374

And the above values are related to the periods of $(E)$:

sage: L = E.period_lattice()
sage: L
Period lattice associated to 
    Elliptic Curve defined by y^2 = x^3 - 5*x^2 - 16*x + 80 over Rational Field
sage: L.gens()
(1.68575035481260, 1.07825782374982*I)
sage: L.gens()[0]/2
0.842875177406298

Best Answer

Related Solutions

[Math] Relation between integral, gamma function, elliptic integral, and AGM

Elliptic integrals, general form

Related Question