Integrating $\int_0^1 \frac{\ln x \ln (1+x) \ln(1-x)}{x^{\frac12}} \,dx$

Question

$$\int_0^1 \frac{\ln x \ln (1+x) \ln(1-x)}{x^{\frac12}} \,dx$$

Now I have seen a similar integral here,

I am unable to deal with the $x^{\frac12}$ term in the denominator.

Here's one idea,

$$\int_0^1x^{n-1}\ln(x)\ln(1-x)\ dx=\frac{H_n}{n^2}+\frac{H_n^{(2)}}{n}-\frac{\zeta(2)}{n}$$

How do we take these forward? Do they involve the substitution of $$x^{\frac{3}{2}}=t, x^{\frac{5}{2}}=t, x^{\frac{7}{2}}=t, x^{\frac{9}{2}}=t$$

$$\int_0^1 \frac{\ln x \ln (1+x) \ln(1-x)}{x^{\frac32}} \,dx$$

$$\int_0^1 \frac{\ln x \ln (1+x) \ln(1-x)}{x^{\frac52}} \,dx$$

$$\int_0^1 \frac{\ln x \ln (1+x) \ln(1-x)}{x^{\frac72}} \,dx$$

$$\int_0^1 \frac{\ln x \ln (1+x) \ln(1-x)}{x^{\frac92}} \,dx$$

Answer 1

Let $t=\sqrt x $ to convert the integral to $4I$, with $$I= \int_0^1 {\ln t \ln (1+t^2) \ln(1-t^2)}dt $$ which can be broken up via integration-by-parts into a number of simpler integrals, i.e. $$I= \int_0^1 \bigg[ {\ln (1+t^2) \ln(1-t^2)+ 2\ln t\ln(1-t^2)} +2\ln t\ln(1+t^2)\\ -\frac{2\ln t\ln(1-t^2)}{1+t^2} -\frac{2\ln t\ln(1+t^2)}{1-t^2}\bigg] dt\\ $$ The reduced integrals above, one-rank lower than the original one, are relatively easy to evaluate. However, they do not offset much among them, adding up to a lengthy closed-form below $$I=-4\Im\text{Li}_3(\frac{1+i}2)+\frac73\zeta(3)+G(6-4\ln2-\pi)+ \frac{\pi^2}{8}(5-2\ln2)\\ +\frac{5\pi^3}{32} +\pi(2-\frac12\ln2+\frac18\ln^22)-\frac{15}4\ln^22+12\ln2-24 $$

Answer 2

May be, we could use $$\log(1+x)\,\log(1-x)=-\sum_{n=1}^\infty \frac{a_n}{b_n}\,x^{2n}$$ where the $a_n$ and $b_n$ correspond to sequences $A049281$ and $A069685$ in $OEIS$ and use $$\int x^{2 n-\frac{1}{2}} \log (x)\,dx=\frac{2 x^{2 n+\frac{1}{2}} ((4 n+1)\log (x)-2)}{(4 n+1)^2}$$ $$\int_0^1 x^{2 n-\frac{1}{2}} \log (x)\,dx=-\frac{4}{(4 n+1)^2}$$ which would lead to $$\int_0^1\frac{\log (1-x) \log (x) \log (1+x)}{\sqrt{x}}\,dx=\sum_{n=1}^\infty \frac{a_n}{b_n}\,\frac{4}{(4 n+1)^2}$$

If $c_n$ is the summand $$\frac {c_{n+1}}{c_n} \sim 1-\frac 3n$$

Computing the partial sums $$S_p=\sum_{n=1}^p \frac{a_n}{b_n}\,\frac{4}{(4 n+1)^2}$$ and converting to decimals, the converegnce is quite slow $$\left( \begin{array}{cc} p & S_p\\ 10 & 0.1927571 \\ 20 & 0.1933295 \\ 30 & 0.1934418 \\ 40 & 0.1934820 \\ 50 & 0.1935008 \\ 60 & 0.1935110 \\ 70 & 0.1935173 \\ 80 & 0.1935213 \\ 90 & 0.1935241 \\ 100 & 0.1935261 \\ 200 & 0.1935325 \\ 300 & 0.1935337 \\ 400 & 0.1935341 \\ 500 & 0.1935343 \\ \end{array} \right)$$

The various inverse symbolic calculators I tried do not seem to enjoy $$0.193534656289778253633161982923\cdots$$

Edit

We can also write (using the Lerch transcendent function) $$\log(1+x)\,\log(1-x)=-\sum_{n=1}^\infty \frac{\Phi (-1,1,2 k)+\log (2)}{k}\,x^{2n}$$ which leads to $$I=\log (2) \left(32-8 C-2 \pi -\pi ^2-12 \log (2)\right)+4\sum_{k=1}^\infty \frac{\Phi (-1,1,2 k)}{k (4 k+1)^2}$$

If $c_n$ is the summand

$$\frac {c_{n+1}}{c_n} \sim 1-\frac 4n$$

Summing the first $100$ terms leads to $0.1935346$ which is much better than the previous and next solutions.

Comment

Using Feynman's trick for $$I(a)=\int_0^1\frac{\log (1-x) \log (x) \log (1+ax)}{\sqrt{x}}\,dx$$ $$I'(a)=\int_0^1\frac{\sqrt{x} \log (1-x) \log (x)}{1+a x}\,dx$$ leads to a monster involving the Lerch transcendent function and a few (even multiple and cross) derivatives of the generalized Gaussian hypergeometric function with respect to its second and third arguments.

It is a pity since $$\,_2\tilde{F}_1\left(1,\frac{5}{2};\frac{7}{2};-a\right)=\frac{8 \left(\sqrt{a} (a-3)+3 \tan ^{-1}\left(\sqrt{a}\right)\right)}{9 \sqrt{\pi} a^{5/2}}$$

Edit

Using your idea of expanding only $\log(1+x)$ we should have $$I=\sum_{n=1}^\infty \frac{(-1)^{n+1}}n \int_0^1 x^{n-\frac{1}{2}} \log (1-x) \log (x)\,dx$$ and $$J_n=\int_0^1 x^{n-\frac{1}{2}} \log (1-x) \log (x)\,dx$$ $$J_n=\frac{8}{(2 n+1)^3}+\frac{4 \left(\psi ^{(0)}\left(n+\frac{1}{2}\right)+\gamma \right)}{(2 n+1)^2}-\frac{2 \psi ^{(1)}\left(n+\frac{3}{2}\right)}{2 n+1}$$ which would give $$I=K+2\sum_{n=1}^\infty\frac{ (-1)^{n+1}}{n(2n+1) }\Bigg(\frac{2 \psi ^{(0)}\left(\frac{2n+1}{2}\right)}{2 n+1}-\psi ^{(1)}\left(\frac{2n+3}{2}\right) \Bigg)$$ with $$K=8 (2+\gamma ) C+\frac{\pi ^3}{2}-4 (12-\pi -2\log (2))-2 \gamma (8-\pi -2\log (2))$$

The convergence of the summation will not be very fast since, if $a_n$ is the summand $$\left|\frac{a_{n+1}}{a_n}\right|=1-\frac{3 \log (n)-4}{n (\log (n)-1)}+O\left(\frac{1}{n^2 \log ^2(n)}\right)$$

For $p=100$ the result would be $0.1935329$

Answer 3

Let's use

$$\ln (1+x) \ln(1-x)=- \sum_{k=1}^{\infty}\frac{H_{2k-1}}{k}x^{2k}$$

where

$$H_{k}=\sum_{m=1}^{k}\frac{(-1)^{m-1}}{m}$$

and consider $$I(a)=\int_{0}^{1}x^a\ln (1+x) \ln(1-x)dx=$$

$$=-\sum_{k=1}^{\infty}\frac{H_{2k-1}}{k(2k+1+a)}$$ Now we differentiate with respect to a and take $a=-\frac{1}{2}$ and we get the result we are looking for:

$$I=4\sum_{k=1}^{\infty}\frac{H_{2k-1}}{k(4k+1)^2}$$

Answer 4

We are looking for a closed expression for the integral

$$i = \int_{0}^{1}\frac{1}{\sqrt{x}}\log(x) \log(1-x) \log(1+x)\;dx\tag {1}$$

The simplifying idea is now to use the relation

$$\log (1-x) \log (x+1)=\frac{1}{2} \left(\log ^2\left(1-x^2\right)-\log ^2(1-x)-\log ^2(x+1)\right)\tag{2}$$

Then the integral splits up into the following three simpler integrals

$$i1 = \int_0^1 \frac{\log (x) \log ^2\left(1-x^2\right)}{2 \sqrt{x}} \, dx\tag{3a}$$ $$i2= -\int_0^1 \frac{\log (x) \log ^2(1-x)}{2 \sqrt{x}} \, dx\tag{3b}$$ $$i3=-\int_0^1 \frac{\log (x) \log ^2(1+x)}{2 \sqrt{x}} \, dx\tag{3c}$$

The integrals are easily solved in Mathematica (for simplicity) and the total integral can be written as

$$\begin{align}& i=i1+i2+i3 \\ = & -4 C (\pi -6+\log (16))+\Re\left(16 i \text{Li}_3\left(\frac{1}{2}+\frac{i}{2}\right)\right)+14 \zeta (3)+8 \pi +96+\frac{1}{24} \left(15 \pi ^2 (4+\pi )+12 (\pi -16) \log ^2(2)+2 \pi (-24-12 \pi ) \log (2)\right)+8 \log (64)\end{align}\tag{4}$$

Here $C$ is Catalan's constant.

We can simplify the term with the Polylog function using its defining sum

$$\text{Li}_3(z)=\sum _{k=1}^{\infty } \frac{z^k}{k^3}$$

and splitting it into sums with $k=m + 4 j$,$m=0..3$ to obtain

$$\begin {align}&\Re\left(16 i \text{Li}_3\left(\frac{1}{2}+\frac{i}{2}\right)\right)\ \\= &\frac{1}{16} \left(-2 \left(\Phi \left(-\frac{1}{4},3,\frac{1}{2}\right)+\Phi \left(-\frac{1}{4},3,\frac{1}{4}\right)\right)-\Phi \left(-\frac{1}{4},3,\frac{3}{4}\right)\right)\end {align}\tag{5}$$

here $\Phi(a,b,c)$ is the Lerch transcendent.

The numeric value of our integral and its solution is

$$N(i) = 0.193535...$$