Evaluating $\int_0^{\frac{1}{2}} \frac{1}{x} \cdot \ln(1+x) \cdot \ln(\frac{1}{x} -1)dx$

Question

I was trying to evaluate the definite integral

$$\int_0^{\frac{1}{2}} \frac{1}{x} \cdot \ln(1+x) \cdot \ln\left(\frac{1}{x} -1\right)\,\mathrm{d}x.$$

On WolframAlpha, I found out that this converges to 0.651114. This seems to be pretty close (and possibly equal) to $\frac{13}{24}\cdot\zeta(3)$, where $\zeta(3)$ is the value of the Riemann Zeta function $\zeta(n)$ at $\ n = 3$. Is this true? How can we prove that they are equal?

I tried substituting $\ x = e^{-t}$ and the integral became

$$\ I = \int_0^{\ln2}\ln(1+e^{-t}) \cdot \ln\left(\frac{e^{-t}}{1-e^{-t}}\right)\,\mathrm{d}t.$$

How do I proceed further? Any help is appreciated. Thanks for reading.

Answer 1

I'm outsourcing most of the work from another answer. Perform integration by parts with $f = \ln(\frac{1}{x}-1) =\ln(1-x)-\ln(x) $ and $g' = \frac{\ln(1+x)}{x}$ to observe

$$\begin{align}\int_0^{1/2} \frac{1}{x}\ln(1+x)\ln\left(\frac{1}{x}-1\right)dx &= -\ln\left(\frac{1}{x}-1\right)\text{Li}_2(-x)\bigg\vert_{0}^{1/2} -\int_0^{1/2}\frac{\text{Li}_2(-x)}{x}+\frac{\text{Li}_2(-x)}{1-x} dx\\ &=-\int_0^{1/2}\frac{\text{Li}_2(-x)}{x}dx-\int_0^{1/2}\frac{\text{Li}_2(-x)}{1-x} dx\\ \end{align}$$

We have $$-\int_0^{1/2}\frac{\text{Li}_2(-x)}{x}dx =-\text{Li}_3\left(-\frac{1}{2}\right)$$ more or less by definition and the second is seen here to be given by

$$-\int_0^{1/2}\frac{\text{Li}_2(-x)}{1-x}dx = \text{Li}_3\left(-\frac{1}{2}\right) + \frac{13}{24}\zeta(3) $$

Edit: Looking at Bob's answer, it is really the calculation of his integral that is most of the work in the separate question I linked.

Answer 2

This integral is not so bad if we use Feynman's trick.

$$I(a)=\int_{0}^{\frac 12} \frac{1}{x} \, \ln(1+ax) \, \log\left(\frac{1}{x} -1\right)\,dx$$ $$I'(a)=\int_{0}^{\frac 12} \frac 1 {1+ax}\, \log\left(\frac{1}{x} -1\right)\,dx$$ $$I'(a)=\frac{12 \text{Li}_2\left(\frac{a+1}{a+2}\right)+6 \log ^2(a+2)-\pi ^2}{12 a}$$ $$I(a)=\int_0^1 \frac{12 \text{Li}_2\left(\frac{a+1}{a+2}\right)+6 \log ^2(a+2)-\pi ^2}{12 a}\,da$$ The corresponding antiderivative is not bad (in terms of logarithms and polylogarithms) and $$\color{blue}{I(1)=\frac{\log ^3(3)}{3}-\frac{13}{8} \zeta (3)+\left(2 \text{Li}_2\left(\frac{1}{3}\right)-\text{Li}_2\left(-\frac{1} {3}\right)\right) \log (3)+}$$ $$\color{blue}{\left(2 \text{Li}_3\left(\frac{1}{3}\right)-\text{Li}_3\left(-\frac{1} {3}\right)\right)}$$

Edit

I do not know how to reduce the result to $\frac{13 }{24}\zeta (3)$ which is exact for at least $10,000$ significant figures.

Answer 3

Not an answer... Too long for a comment.

We can split the given integral as $I_1-I_2.$

The second piece is easy. By integration by parts, we have

$$\int\frac{\ln(1+x)\ln x}x dx\\=-\text{Li}_2(-x)\ln x+\int\frac{\text{Li}_2(-x)}xdx\\=-\text{Li}_2(-x)\ln x+\text{Li}_3(-x)+C$$ as in WolframAlpha's output. Hence, $I_2=\text{Li}_2(-\frac12)\ln 2+\text{Li}_3(-\frac12)\approx-0.783415.$

For the first piece WA gave a complicated antiderivative output and I questioned the purpose of my life. Why should İ try to find it? Anyway, we can do series expansion $$\int_0^{\frac12}\frac{\ln(1+x)\ln(1-x)}x dx\\=-\sum_{n=1}^\infty \frac1{n}\int_0^{\frac12}x^{n-1}\ln(1+x)dx$$ The integrals in the sum are doable by integration by parts for numerical approximation. WA gives $I_1\approx-0.132301$.

Unfortunately, I cannot see any connection with the value of Riemann Zeta Function at 3.

Answer 4

\begin{align} I&=\int_0^\frac{1}{2}\frac{\ln(1+x)}{x}\ln\left(\frac{1}{x}-1\right)\mathrm{d}x\\ &=\frac{1}{2}\int_0^\frac{1}{2}\frac{2\ln(1-x)\ln(1+x)}{x}\mathrm{~d}x\\ &\quad-\int_0^\frac{1}{2}\frac{\ln x\ln(1+x)}{x}\mathrm{~d}x\\ &=\frac{1}{2}\int_0^\frac{1}{2}\frac{\ln^2\left(1-x^2\right)-\ln^2(1-x)-\ln^2(1+x)}{x}\mathrm{~d}x\\ &\quad-\int_0^\frac{1}{2}\frac{\ln x\ln(1+x)}{x}\mathrm{~d}x\\ &=\frac{1}{2}\int_0^\frac{1}{2}\frac{\ln^2\left(1-x^2\right)}{x}\mathrm{~d}x-\frac{1}{2}\int_0^\frac{1}{2}\frac{\ln^2(1-x)}{x}\mathrm{~d}x\\ &\quad-\frac{1}{2}\int_0^\frac{1}{2}\frac{\ln^2(1+x)}{x}\mathrm{~d}x-\int_0^\frac{1}{2}\frac{\ln x\ln(1+x)}{x}\mathrm{~d}x\\ &=\frac{1}{2}I_1-\frac{1}{2}I_2-\frac{1}{2}I_3-I_4. \end{align} Next, we introduce the $\text{Polylog}$ function defined as:

$$\mathbf{Li}_n(x)=\sum_{k=1}^\infty\frac{x^k}{k^n},\quad|x|\leqslant1.$$ It is easy to get

$$\frac{\mathrm{d}\mathrm{Li}_{n+1}(x)}{\mathrm{d}x}=\frac{\mathbf{Li}_n(x)}{x}.$$ Next, we have

\begin{align} I_1&=\int_0^\frac{1}{2}\frac{\ln^2\left(1-x^2\right)}{x}\mathrm{~d}x\\ &\xlongequal{x^2\to x}\frac{1}{2}\int_0^\frac{1}{4}\frac{\ln^2\left(1-x\right)}{x}\mathrm{~d}x\\ &=\frac{1}{2}\int_0^\frac{1}{4}\ln^2(1-x)\mathrm{~d}\ln x\\ &=-\ln2\ln^2\left(\frac{3}{4}\right)+\underbrace{\int_0^\frac{1}{2}\frac{\ln(1-x)\ln x}{1-x}\mathrm{~d}x}_{1-x\to x}\\ &=-\ln2\ln^2\left(\frac{3}{4}\right)+\int_\frac{1}{2}^1\frac{\ln(1-x)\ln x}{x}\mathrm{~d}x\\ &=-\ln2\ln^2\left(\frac{3}{4}\right)-\int_\frac{1}{2}^1\ln x\mathrm{~d}\mathbf{Li}_2(x)\\ &=-\ln2\left(\ln^23+4\ln^22-4\ln2\ln3\right)\\ &\quad-\ln2\mathbf{Li}_2\left(\frac{1}{2}\right)+\int_\frac{1}{2}^1\frac{\mathbf{Li}_2(x)}{x}\mathrm{~d}x\\ &=-\ln2\left(\ln^23+4\ln^22-4\ln2\ln3\right)\\ &\quad-\ln2\mathbf{Li}_2\left(\frac{1}{2}\right)+\mathbf{Li}_3\left(1\right)-\mathbf{Li}_3\left(\frac{1}{2}\right), \end{align}

we have

\begin{align} \mathbf{Li}_2\left(\frac{1}{2}\right)&=\frac{\pi^2}{12}-\frac{\ln^22}{2};\\ \mathbf{Li}_3\left(1\right)&=\zeta(3);\\ \mathbf{Li}_3\left(\frac{1}{2}\right)&=\frac{\ln^32}{6}-\frac{\pi^2\ln2}{12}+\frac{7\zeta(3)}{8}. \end{align} Hence, we get

$$I_1=4\ln^22\ln3-\ln2\ln^23-\frac{11\ln^32}{3}+\frac{\zeta(3)}{8};$$ similarly, we have

\begin{align} I_2&=\int_0^\frac{1}{2}\frac{\ln^2\left(1-x\right)}{x}\mathrm{~d}x\\ &=-\ln^32-2\ln2\mathbf{Li}_2\left(\frac{1}{2}\right)\\ &\quad+2\mathbf{Li}_3\left(1\right)-2\mathbf{Li}_3\left(\frac{1}{2}\right)\\ &=\frac{\zeta(3)}{4}-\frac{\ln^32}{6}; \end{align} and

\begin{align} I_3&=\int_0^\frac{1}{2}\frac{\ln^2\left(1+x\right)}{x}\mathrm{~d}x\\ &=\int_0^\frac{1}{2}\ln^2(1+x)\mathrm{~d}\ln x\\ &=\ln^2\left(\frac{3}{2}\right)-2\underbrace{\int_0^\frac{1}{2}\frac{\ln x\ln(1+x)}{1+x}\mathrm{~d}x}_{1+x\to x}\\ &=\ln^23+\ln^22-2\ln2\ln3\\ &\quad-2\int_1^\frac{3}{2}\frac{\ln x\ln(x-1)}{x}\mathrm{~d}x\\ &=\ln^23+\ln^22-2\ln2\ln3\\ &\quad-2\int_1^\frac{3}{2}\frac{\ln x}{x}\left[\ln x-\ln\left(1-\frac{1}{x}\right)\right]\mathrm{d}x\\ &=\ln^23+\ln^22-2\ln2\ln3-2\int_1^\frac{3}{2}\ln^2 x\mathrm{~d}\ln x\\ &\quad-2\int_1^\frac{3}{2}x\ln\left(1-\frac{1}{x}\right)\ln x\mathrm{~d}\left(\frac{1}{x}\right)\\ &=\ln^23+\ln^22-2\ln2\ln3-\frac{2}{3}\ln^3\left(\frac{3}{2}\right)\\ &\quad-2\int_1^\frac{3}{2}\ln x\mathrm{~d}\mathbf{Li}_2\left(\frac{1}{x}\right)\\ &=\ln^22+\ln^23-2\ln2\ln3\\ &\quad-\frac{2\ln^32}{3}-\frac{2\ln^33}{3}-2\ln^22\ln3+2\ln2\ln^23\\ &\quad-2(\ln3-\ln2)\mathbf{Li}_2\left(\frac{2}{3}\right)-2\int_1^\frac{3}{2}x\mathrm{Li}_2\left(\frac{1}{x}\right)\mathrm{d}\left(\frac{1}{x}\right)\\ &=\ln^22+\ln^23-2\ln2\ln3\\ &\quad-\frac{2\ln^32}{3}-\frac{2\ln^33}{3}-2\ln^22\ln3+2\ln2\ln^23\\ &\quad-2(\ln3-\ln2)\mathbf{Li}_2\left(\frac{2}{3}\right)-2\mathbf{Li}_3\left(\frac{2}{3}\right)+2\mathbf{Li}_3(1)\\ &=\ln^22+\ln^23-2\ln2\ln3-\frac{\pi^2\ln2}{6}+\frac{7\zeta(3)}{4}\\ &\quad-\frac{\ln^32}{3}-\frac{2\ln^33}{3}-2\ln^22\ln3+2\ln2\ln^23\\ &\quad-2(\ln3-\ln2)\mathbf{Li}_2\left(\frac{2}{3}\right)-2\mathbf{Li}_3\left(\frac{2}{3}\right); \end{align}

\begin{align} I_4&=\int_0^\frac{1}{2}\frac{\ln x\ln\left(1+x\right)}{x}\mathrm{~d}x\\ &=\int_0^\frac{1}{2}\frac{\ln x\ln\left[1-(x)\right]}{x}\mathrm{~d}x\\ &=\int_0^\frac{1}{2}\ln x\mathrm{~d}\mathbf{Li}_2(-x)\\ &=-\ln2\mathbf{Li}_2\left(-\frac{1}{2}\right)-\int_0^\frac{1}{2}\frac{\mathbf{Li}_2(-x)}{x}\mathrm{~d}x\\ &=-\ln2\mathbf{Li}_2\left(-\frac{1}{2}\right)-\mathbf{Li}_3\left(-\frac{1}{2}\right). \end{align} After that, we introduce two identities:

\begin{align} &(1)\quad\mathbf{Li}_2(x)+\mathbf{Li}_2\left(\frac{x}{x-1}\right)=-\frac{1}{2}\ln^2(1-x)\;(x\lt1);\\ &(2)\quad\mathbf{Li}_2(x)+\mathbf{Li}_2(1-x)=\frac{\pi^2}{6}-\ln x\ln(1-x). \end{align} Substitute $x=\frac{1}{3}$ into indentites (1) and (2), we have

\begin{align} \mathbf{Li}_2\left(\frac{1}{3}\right)+\mathbf{Li}_2\left(-\frac{1}{2}\right)&=-\frac{1}{2}\ln^2\left(\frac{2}{3}\right);\\ \mathbf{Li}_2\left(\frac{1}{3}\right)+\mathbf{Li}_2\left(\frac{2}{3}\right)&=\frac{\pi^2}{6}+\ln 3\ln\left(\frac{2}{3}\right),\end{align} thus

\begin{align} \mathbf{Li}_2\left(\frac{2}{3}\right)-\mathbf{Li}_2\left(-\frac{1}{2}\right)&=\frac{\pi^2}{6}+\ln 3\ln\left(\frac{2}{3}\right)+\frac{1}{2}\ln^2\left(\frac{2}{3}\right).\end{align} Similarly, we introduce more identities:

\begin{align} &(3)\quad\mathbf{Li}_3(x)+\mathbf{Li}_3\left(-x\right)=\frac{1}{4}\mathbf{Li}_3\left(x^2\right);\\ &(4)\quad\mathbf{Li}_3\left(x\right)+\mathbf{Li}_3\left(1-x\right)+\mathbf{Li}_3\left(\frac{x}{x-1}\right)\\ &\quad=\zeta(3)+\frac{\pi^2\ln(1-x)}{6}-\frac{\ln x\ln^2(1-x)}{2}+\frac{\ln^3(1-x)}{6};\\ &(5)\quad\mathbf{Li}_3\left(\frac{1-x}{1+x}\right)-\mathbf{Li}_3\left(\frac{x-1}{x+1}\right)-2\mathbf{Li}_3\left(1-x\right)\\ &\qquad-2\mathbf{Li}_3\left(\frac{1}{1+x}\right)+\frac{1}{2}\mathbf{Li}_3\left(1-x^2\right)\\ &\quad=-\frac{7\zeta(3)}{4}+\frac{\pi^2\ln(1+x)}{6}-\frac{\ln^3(1+x)}{3}. \end{align} Substitute $x=-\frac{1}{2}$ and $x=\frac{1}{3}$ into identities $(3)\sim(5)$, we can get the relationship between $\mathbf{Li}_3\left(-\frac{1}{2}\right)4$ and $\mathbf{Li}_3\left(\frac{2}{3}\right).$

Finally, substitute all the results of integrals $I_1\sim I_4$ followed by some cumbersome calculation, we can obtain

\begin{align} I&=\frac{1}{2}I_1-\frac{1}{2}I_2-\frac{1}{2}I_3-I_4\\ &=\frac{13\zeta(3)}{24}. \end{align} By the way, all the identites can be proved by taking the derivation on $x.$