Evaluation of $\int_0^1\frac{\log x\,dx}{\sqrt{x(1-x)(1-cx)}}$

Question

Assume $c$ is a small real number.

QUESTION. What is the value of this integral in terms of the complete elliptic function $K(k)$? $$\int_0^1\frac{\log x}{\sqrt{x(1-x)(1-cx)}}\,dx.$$

I got as far as (give or take some silly errors) expressing the integral as $$\int_{-\omega_1}^{\omega_1}\log\left(\wp(z)-\wp(\omega_2)\right)dz$$ where $\wp(\omega_1)=e_1, \wp(\omega_2)=e_3$ and $\wp(\omega_3)=e_2$ while the $e_j$'s are the (real) roots of the cubic equation associated to the Weierstrass elliptic function of the current problem.

UPDATE. I have found a solution to this problem using the Weierstrass functions. However, I welcome any sort of alternative approach.

Answer 1

---

Define the function $\mathcal{I}:(-\infty,1]\rightarrow\mathbb{R}$ via the definite integral

$$\mathcal{I}{\left(c\right)}:=-\int_{0}^{1}\mathrm{d}t\,\frac{\ln{\left(t\right)}}{\sqrt{t\left(1-t\right)\left(1-ct\right)}}.$$

We seek a closed form expression for $\mathcal{I}$ in terms of the complete elliptic integral of the first kind $K{\left(k\right)}$, where $k$ is the elliptic modulus. Recall

$$F{\left(\varphi,k\right)}:=\int_{0}^{\sin{\varphi}}\mathrm{d}x\,\frac{1}{\sqrt{\left(1-x^{2}\right)\left(1-k^{2}x^{2}\right)}};~~~\small{\varphi\in\mathbb{R}\land k\in(-1,1)},$$

$$K{\left(k\right)}:=F{\left(\frac{\pi}{2},k\right)};~~~\small{k\in(-1,1)}.$$

Below is an evaluation of $\mathcal{I}{\left(c\right)}$ for $0<c<1$ that invokes the following formula from Gradshteyn's Table of Integrals, Series, and Products:

$$\mathbf{(6.111)}~~~~~\int_{0}^{\frac{\pi}{2}}\mathrm{d}\varphi\,\cot{\varphi}\,F{\left(\varphi,k\right)}=\frac{\pi}{4}K{\left(\sqrt{1-k^{2}}\right)}+\frac12\ln{\left(k\right)}K{\left(k\right)};~~~\small{0<k<1}.$$

I do not currently know how to prove this integration formula, nor do I know how to extend the final result to negative $c$, but hopefully this answer will be useful to you as is.

---

Suppose $c\in(0,1)$, and set $k:=\sqrt{c}$.

$$\begin{align} \mathcal{I}{\left(c\right)} &=\mathcal{I}{\left(k^{2}\right)}\\ &=-\int_{0}^{1}\mathrm{d}t\,\frac{\ln{\left(t\right)}}{\sqrt{t\left(1-t\right)\left(1-k^{2}t\right)}}\\ &=\int_{0}^{1}\mathrm{d}t\,\frac{\left[\ln{\left(1\right)}-\ln{\left(t\right)}\right]}{\sqrt{t\left(1-t\right)\left(1-k^{2}t\right)}}\\ &=\int_{0}^{1}\mathrm{d}t\,\frac{1}{\sqrt{t\left(1-t\right)\left(1-k^{2}t\right)}}\int_{t}^{1}\mathrm{d}u\,\frac{1}{u}\\ &=\int_{0}^{1}\mathrm{d}t\int_{t}^{1}\mathrm{d}u\,\frac{1}{u\sqrt{t\left(1-t\right)\left(1-k^{2}t\right)}}\\ &=\int_{0}^{1}\mathrm{d}u\int_{0}^{u}\mathrm{d}t\,\frac{1}{u\sqrt{t\left(1-t\right)\left(1-k^{2}t\right)}}\\ &=\int_{0}^{1}\mathrm{d}u\int_{0}^{\sqrt{u}}\mathrm{d}x\,\frac{2}{u\sqrt{\left(1-x^{2}\right)\left(1-k^{2}x^{2}\right)}};~~~\small{\left[\sqrt{t}=x\right]}\\ &=\int_{0}^{1}\mathrm{d}y\,2y\int_{0}^{y}\mathrm{d}x\,\frac{2}{y^{2}\sqrt{\left(1-x^{2}\right)\left(1-k^{2}x^{2}\right)}};~~~\small{\left[\sqrt{u}=y\right]}\\ &=4\int_{0}^{1}\mathrm{d}y\,\frac{1}{y}\int_{0}^{y}\mathrm{d}x\,\frac{1}{\sqrt{\left(1-x^{2}\right)\left(1-k^{2}x^{2}\right)}}\\ &=4\int_{0}^{\frac{\pi}{2}}\mathrm{d}\varphi\,\cot{\varphi}\int_{0}^{\sin{\varphi}}\mathrm{d}x\,\frac{1}{\sqrt{\left(1-x^{2}\right)\left(1-k^{2}x^{2}\right)}};~~~\small{\left[y=\sin{\varphi}\right]}\\ &=4\int_{0}^{\frac{\pi}{2}}\mathrm{d}\varphi\,\cot{\varphi}\,F{\left(\varphi,k\right)}\\ &=4\left[\frac{\pi}{4}K{\left(\sqrt{1-k^{2}}\right)}+\frac12\ln{\left(k\right)}K{\left(k\right)}\right]\\ &=\pi\,K{\left(\sqrt{1-c}\right)}+\ln{\left(c\right)}K{\left(\sqrt{c}\right)}.\\ \end{align}$$

---

Answer 2

$\def\F{\operatorname F}$ Expand $\ln(x)$ as a limit:

$$\int_0^1\frac{\ln(x)}{\sqrt{x(1-x)(1-kx)}}dx=\lim_{s\to0}\frac1s \int_0^1 x^{s-\frac12}(1-x)^{-\frac12}(1-cx)^{-\frac12}dx-\frac1s\int_0^1 dx x^{-\frac12}(1-x)^{-\frac12}(1-cx)^{-\frac12}$$

Now recall the regularized Gauss hypergeometric function:

$$\Gamma(b)\Gamma(c-b)\,_2\tilde\F_1(a,b;c;z)= \frac{\Gamma(b)\Gamma(c-b)}{\Gamma(c)}\,_2\F_1(a,b;c;z) =\int_0^1 x^{b-1}(1-x)^{c-b-1}(1-xz)^{-a}dx$$

Therefore, we take the derivative using its limit definition:

$$\int_0^1\frac{\ln(x)}{\sqrt{x(1-x)(1-kx)}}dx=\lim_{s\to0}\frac1s\left(\sqrt\pi\Gamma\left(s+\frac12\right)\,_2\tilde\F_1\left(\frac12,s+\frac12;s+1;k\right)-2\text K(k)\right)=\pi\left.\frac{d\,_2\F_1\left(\frac12,b;1;k\right)}{db}\right|_{b=\frac12}+ \pi\left.\frac{d\,_2\F_1\left(\frac12,\frac12;c;k\right)}{dc}\right|_{c=1}-2\ln(4)\text K(k) $$

Answer 3

This is not an answer.

I did not see any way to introduce elliptic functions in this problem.

For the time being, I wrote $$\frac{\log (x)}{\sqrt{x(1-x)(1-cx)}}=\sum_{n=0}^\infty (-1)^n \,\binom{-\frac{1}{2}}{n} \,c^n \,\,\frac{x^{n-\frac{1}{2}}\log (x)}{\sqrt{1-x}}$$ The antiderivative of the summand express in terms of hypergeometric function and if $$I_n=\int_0^1 \frac{x^{n-\frac{1}{2}}\log (x)}{\sqrt{1-x}}\,dx$$ $$I_n=\sqrt{\pi }\,\,\frac{ \Gamma \left(n+\frac{1}{2}\right)}{\Gamma (n+1)}\left(\psi ^{(0)}\,\,\left(n+\frac{1}{2}\right)-\psi ^{(0)}(n+1)\right)$$ If $$a_n=(-1)^n\,\binom{-\frac{1}{2}}{n} \,c^n\,I_n \quad \implies \quad \frac {a_{n+1}}{a_n}=c\left(1-\frac{2}{n}+O\left(\frac{1}{n^2}\right)\right)$$ giving probably a quite fast convergence.

Computing the partial sums for $c=\frac 12$ $$\left( \begin{array}{cc} p & \sum_{n=0}^p a_n \\ 0 & -4.35517 \\ 1 & -4.50687 \\ 2 & -4.53113 \\ 3 & -4.53699 \\ 4 & -4.53871 \\ 5 & -4.53928 \\ 6 & -4.53948 \\ 7 & -4.53955 \\ 8 & -4.53958 \\ 9 & -4.53959 \\ 10 & -4.53960 \\ \end{array} \right)$$

Answer 4

In his thesis, Anil evaluated tons of integrals like these (without $\log$-term), even those having fourth degree polynomials in the square-root, by $x=x_0+\tan^2\theta$ substitutions.

$$I=\int_0^1\frac{\log x}{\sqrt{x(1-x)(1-cx)}}\,dx\overset{x\rightarrow 1/x}{=}-\int_1^{\infty}\frac{\log x}{\sqrt{x(x-1)(x-c)}}\,dx\overset{x\rightarrow 1+\tan^2\theta}=4\int_0^{\pi/2}\frac{\ln\cos\theta\,d\theta}{\sqrt{1-c\cos^2\theta}}$$ and now with $\ln\cos\theta=\frac12\ln(1-\sin^2\theta)=-\frac12\sum_{n=1}^{\infty}\frac1n\sin^{2n}\theta$ we get $$I=-2\sum_{n=1}^{\infty}\frac1n\color{blue}{\int_0^{\pi/2}\frac{\sin^{2n}\theta\,d\theta}{\sqrt{1-c\cos^2\theta}}}=-2\sum_{n=1}^{\infty}\frac1n \color{blue}{K_{2n}(\sqrt c)}$$ where $K_{2n}(\sqrt c)$ can be calculated recursively in terms of $K(\sqrt c)$ and $E(\sqrt c)$ not just $K(\sqrt c)$ alone, as OP requested. Also, the convergence is slow. My numerical experience so far is that: the sum is approaching to $-4$ at $c=\frac12$.

Answer 5

Firstly, $$I(c)=\int\limits_0^1 \dfrac{\ln x\,\text dx}{\sqrt{x(1-x)(1-cx)\large\mathstrut}} =4\int\limits_0^1 \dfrac{\ln\sqrt x\,\text d\sqrt x}{\sqrt{(1-x)(1-cx)\large\mathstrut}} =4\int\limits_0^1 \dfrac{\ln y\,\text dy}{\sqrt{(1-y^2)(1-cy^2)\large\mathstrut}}.$$

If $\mathbf{c=0},$ then $$I(0) = 4\int\limits_0^1\dfrac{\ln y}{1-y^2}\,\text dy =-\dfrac{\pi^2}2 \approx-4.9348.$$

If $\mathbf{c=1},$ then $$I(1) = 4\int\limits_0^1\dfrac{\ln y}{\sqrt{1-y^2}}\,\text dy =-2\pi\ln2\approx-4.35517.$$

Assume $\mathbf{c\in(0,1).}$

Using elliptic antiderivative $$\int \dfrac{\text dy}{\sqrt{(1-y^2)(1-c y^2)}} = \operatorname{F}\big(\arcsin y\ |\,c\big)\tag1$$ and integration by parts, one can get: $$I(c)=4\int\limits_0^1 \ln y\,\text d\operatorname{F}\big(\arcsin(y)\,|\,c\big) =4\ln y\operatorname{F}\big(\arcsin(y)\,|\,c\big)\bigg|_0^1 -4\int\limits_0^1 \operatorname{F}\big(\arcsin(y)\,|\,c\big)\,\dfrac{\text dy}{y},$$ $$\color{brown}{\mathbf{I(c)=-4\int\limits_0^1 \operatorname{F}\big(\arcsin(y)\,|\,c\big)\,\dfrac{\text dy}{y}.}}\tag1$$ Applying Maclaurin series in the form of $$\operatorname{F}\big(\arcsin(y)\,|\,c\big)= \dfrac{c^k \left(1/2\right)_{(k)}}{(1 + 2 k) k!} \operatorname{_2F_1}\left(\dfrac12, -k, \dfrac12 - k, \dfrac1c\right) y^{2k+1}, \qquad (|y|<1)$$ easily tu get

$$I(c)=-4\int_0^1\dfrac{c^k \left(1/2\right)_{(k)}}{(1 + 2 k)^2 k!} \operatorname{_2F_1}\left(\dfrac12, -k, \dfrac12 - k, \dfrac1c\right) y^{2k+1}, \qquad (|y|<1)\tag3.$$

Sadly, obtained series converges badly (5 digs by 200 terms).

Alternative way is based on the orthogonal polynomials technic.

Let $c=p^2,\; z=py^2,\; t=\dfrac{1+p^2}{2p},$ then $$(1-y^2)(1-p^2y^2) = 1-2 t z+z^2,$$ and from the generating function representation it follows $$\dfrac1{\sqrt{1-2tz+z^2}} = \sum\limits_{k=0}^\infty \operatorname P_k(t) z^k =\sum\limits_{k=0}^\infty c^{k}\operatorname P_k\left(\dfrac{1+c}{2\sqrt c}\right) y^{2k},\tag4$$ where $$\operatorname P_k(t)= \sum_{m=0}^n \dbinom nm \dbinom{-n-1}m \left(\dfrac{1-x}2\right)^m\tag5$$ are the Legendre orthogonal polynomials. Taking in account definite integration of $$\int\limits_0^1 x^k\ln x dx = -\dfrac1{(k+1)^2},\qquad (k>-1),\tag6$$ finally we can obtain $$\color{brown}{\mathrm{I(c) = -4\,\sum_{k=0}^\infty \binom{k-\frac12}k \dfrac1{(k+1)^2} \operatorname{_1F_2}\left(\dfrac12, -k, -\dfrac12-k, \dfrac1c\right)}}.\tag7$$

However, this series converges similarly bad, as the previous one.