A rare integral involving $\operatorname{Li}_2$

Question

A rare but interesting integral problem: $$\int_{0}^{1} \frac{\operatorname{Li}_2(-x)- \operatorname{Li}_2(1-x)+\ln(x)\ln(1+x)+\pi x\ln(1+x) -\pi x\ln(x)}{1+x^2}\frac{\text{d}x}{\sqrt{1-x^2} } =\frac{\pi^3}{48\sqrt{2} }.$$ Where $\operatorname{Li}_2$ is dilogarithm.
The integral without the factor $\frac{1}{\sqrt{1-x^2} }$ is much easier(it can be expressed using $\operatorname{Li}_3$,$\ln(2)$,Catalan's constant and $\pi$). However, the same idea dosen't work for this one. Any suggestion will be much appreciated.

Answer 1

Following the answer I wrote on your slightly more advanced integral, I now present two methods. First is the original method resulting in heavy use of differentiation under the integral sign, and the second is the new approach of simplifying the given integral using polylogarithm identities.

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Method 1:

We begin by separating the integral into three parts:

$$I = \pi \color{red}{\overbrace{\int_{0}^{1} \frac{x \ln (1+x)}{(1+x^2) \sqrt{1-x^2}} \, dx}^J} - \pi \color{blue}{\overbrace{\int_{0}^{1} \frac{ x \ln (x)}{(1+x^2) \sqrt{1-x^2}} \, dx}^K} + \color{purple}{\underbrace{\int_{0}^{1}\frac{\operatorname{Li}_2(-x)-\operatorname{Li}_2(1-x)+\ln(x)\ln(1+x)}{(x^2+1)\sqrt{1-x^2}} \, dx}_L}$$

We begin by evaluating $\color{red}{J}$: $$J := \int_{0}^{1}\frac{x\ln(x+1)}{(x^2+1)\sqrt{1-x^2}} \, dx$$ Consider

\begin{align*}\int_{0}^{1}\frac{x(\ln(x+1)+\ln(1-x))}{(x^2+1)\sqrt{1-x^2}} \, dx&=-\frac{2}{\sqrt{2}}\int_{0}^{1}\frac{x}{1-x^2}\operatorname{artanh}\left(\frac{\sqrt{1-x^2}}{\sqrt2}\right) \, dx\\&=-\frac{2}{\sqrt{2}}\int_{0}^{1}\frac{1}{u}\operatorname{artanh}\left(\frac{u}{\sqrt2}\right)\,du \\&= -\frac{2}{\sqrt2}\left(-\frac{\pi^2}{48}+\frac{\ln^2(2)}{8}+\operatorname{Li}_2\left(\frac{1}{\sqrt2}\right)\right)\end{align*}

Consider now

\begin{align*}\int_{0}^{1}\frac{x(\ln(x+1)-\ln(1-x))}{(x^2+1)\sqrt{1-x^2}} \,dx&=\frac{2}{\sqrt{2}}\int_{0}^{1}\frac{1}{1-x^2}\operatorname{artanh}\left(\frac{\sqrt{1-x^2}}{\sqrt2}\right)\,dx\\&=\frac{2}{\sqrt{2}}\int_{0}^{1}\frac{1}{u\sqrt{1-u^2}}\operatorname{artanh}\left(\frac{u}{\sqrt2}\right) \,du\\&=\frac{\pi^2}{4\sqrt2}\end{align*}

Thus upon summing the two we have $$\color{red}{\implies J =\frac{\pi^2}{8\sqrt2}-\frac{1}{\sqrt2}\left(-\frac{\pi^2}{48}+\frac{\ln^2(2)}{8}+\operatorname{Li}_2\left(\frac{1}{\sqrt2}\right)\right)}$$

Now we evaluate $\color{blue}{K}$: $$K := \int_{0}^{1} \frac{x \ln (x)}{(1+x^2)\sqrt{1-x^2}} \, dx$$

\begin{align*}\int_{0}^{1}\frac{x \ln(x)}{\sqrt{1-x^2}(1+x^2)}\,dx &=\frac{1}{\sqrt{2}}\int_{0}^{1}\frac{1}{x}\left(\operatorname{artanh}\left(\frac{\sqrt{1-x^2}}{\sqrt2}\right)-\operatorname{artanh}\left(\frac{1}{\sqrt2}\right)\right)\,dx\\&=\frac{1}{\sqrt{2}}\int_{0}^{1}\frac{u}{1-u^2}\left(\operatorname{artanh}\left(\frac{u}{\sqrt2}\right)-\operatorname{artanh}\left(\frac{1}{\sqrt2}\right)\right)\,du\end{align*}

We shall now approach this with differentiation under the integral sign.

Define $$f(b):=\frac{1}{\sqrt{2}}\int_{0}^{1}\frac{u}{1-u^2}\left(\operatorname{artanh}\left(\frac{u}{b}\right)-\text{arctanh}\left(\frac{1}{b}\right)\right)\,du$$

$$f'(b)=\frac{1}{2 \sqrt{2}}\left(\frac{b(\ln(b+1)-\ln(b-1))-2\ln(2)}{b^2-1}\right)$$

Notice that $\lim_{b \to \infty} f(b) = 0 \implies f(\sqrt{2})-f(\infty)=K$

$$\color{blue}{\implies K=-\frac{1}{4\sqrt{2}}\left(\frac{\pi^2}{3}+2\operatorname{Li}_2\left(\frac{1}{2}-\frac{1}{\sqrt{2}}\right)+\ln^2(2)-\operatorname{arsinh}(1)^2-2\ln(2)\operatorname{arsinh}(1)\right)}$$

Now we evaluate $\color{purple}{L}$: $$L := \int_{0}^{1}\frac{\operatorname{Li}_2(-x)-\operatorname{Li}_2(1-x)+\ln(x)\ln(1+x)}{(x^2+1)\sqrt{1-x^2}} \,dx \\= -\frac{\pi^3}{24\sqrt2}+\frac{2}{\sqrt{2}}\int_{0}^{1}\frac{x\ln(x)}{1-x^2}\arctan \left(\frac{\sqrt{2}x}{\sqrt{1-x^2}}\right) \,dx$$ $$=-\frac{\pi^3}{24 \sqrt{2}}+\frac{2}{\sqrt{2}}\int_{0}^{\infty}\frac{u}{1+u^2}\ln\left(\frac{u}{\sqrt{u^2+1}}\right)\arctan\left(\sqrt{2}u\right)\,du$$ $$=-\frac{\pi^3}{12\sqrt{2}} + \frac{2}{\sqrt{2}}\int_{0}^{\infty}\frac{u}{1+u^2}\ln\left(\frac{u}{\sqrt{u^2+1}}\right)\left(\arctan (\sqrt{2}u)-\frac{\pi}{2}\right)\,du$$ Now separate the logarithm and consider the two integrals $$\color{green}{A:=\int_{0}^{\infty}\frac{u}{1+u^2}\ln\left(u^2+1\right)\left(\arctan (\sqrt{2}u)-\frac{\pi}{2}\right) \, du}$$ $$\color{brown}{B:=\int_{0}^{\infty}\frac{u}{1+u^2}\ln\left(u\right)\left(\arctan (\sqrt{2}u)-\frac{\pi}{2}\right) \, du}$$ such that $\color{purple}{L} = -\frac{\pi^3}{12 \sqrt{2}} + \frac{2}{\sqrt{2}}\color{brown}{B}-\frac{1}{\sqrt{2}} \color{green}{A}$.

We begin by evaluating $\color{green}{A}$ via differentiation under the integral sign:

$$A=\int_{0}^{\infty}\frac{u}{1+u^2}\ln\left(u^2+1\right)\left(\arctan (\sqrt{2}u)-\frac{\pi}{2}\right) \, du$$

Recall

$$\ln(u^2+1)=2\int_{0}^{\infty}\frac{1-\cos(ux)}{x}e^{-x}\,dx$$

Define

$$g(b) := \int_{0}^{\infty}\frac{u}{1+u^2}\ln\left(u^2+1\right)\left(\arctan\left(\frac{u}{b}\right)-\frac{\pi}{2}\right)\,du$$ $$g'(b) = -2\int_{0}^{\infty}\frac{u^2}{(1+u^2)(b^2+u^2)}\int_{0}^{\infty}\frac{1-\cos(ux)}{x}e^{-x}\,dx\,du$$ $$=-\frac{\pi}{(1-b^2)}\int_{0}^{\infty}\frac{e^{-x}}{x}\left(1-e^{-x}-b+be^{-xb}\right)\,dx =\frac{\pi b\ln(1+b)-\pi\ln(2)}{1-b^2}$$

Notice $\lim_{b \to 0} g(b) = 0 \implies g \left(\frac{1}{\sqrt{2}}\right) - g(0) = A$

$$\color{green}{\implies A = -\frac{\pi^3}{24}-\pi \, \operatorname{arcoth}(\sqrt{2}) \,\ln(2)-\frac{\pi}{2}\operatorname{artanh}\left(\frac{1}{7}+\frac{2\sqrt2}{7}\right)\ln\left(\frac{2+\sqrt2}{16}\right)+\frac{\pi}{2}\operatorname{Li}_2\left(\frac{2-\sqrt2}{4}\right)}$$

Similarly, we evaluate $\color{brown}{B}$ via differentiation under the integral sign:

$$B=\int_{0}^{\infty}\frac{u}{1+u^2}\ln\left(u\right)\left(\arctan (\sqrt{2}u)-\frac{\pi}{2}\right) \, du$$

Define

$$h(b) := \int_{0}^{\infty}\frac{u}{1+u^2}\ln\left(u\right)\left(\arctan\left(\frac{u}{b}\right)-\frac{\pi}{2}\right) \, du$$ $$h'(b) = -\int_{0}^{\infty}\frac{u^2\ln(u)}{(u^2+1)(b^2+u^2)} \, du = \frac{\pi \, b \ln(b)}{2(1-b^2)}$$

Notice $\lim_{b \to 0} h(b) = 0 \implies h \left(\frac{1}{\sqrt{2}}\right) - h(0) = B$

$$\color{brown}{\implies B = -\frac{\pi^3}{96}-\frac{\pi \ln^2(2)}{16}}$$

Thus we have $$\color{purple}{L} = -\frac{\pi^3}{12 \sqrt{2}} + \frac{2}{\sqrt{2}} \color{brown}{\left(-\frac{\pi^3}{96}-\frac{\pi \ln^2(2)}{16}\right)} - \frac{1}{\sqrt{2}} \color{green}{\left( -\frac{\pi^3}{24}-\pi \, \operatorname{arcoth}(\sqrt{2}) \,\ln(2)-\frac{\pi}{2}\operatorname{artanh}\left(\frac{1}{7}+\frac{2\sqrt2}{7}\right)\ln\left(\frac{2+\sqrt2}{16}\right)+\frac{\pi}{2}\operatorname{Li}_2\left(\frac{2-\sqrt2}{4}\right) \right)}$$

$$\color{purple}{\implies L = -\frac{\pi\operatorname{Li}_2\left(\frac{1}{2}-\frac{1}{2\sqrt{2}}\right)}{2\sqrt{2}}-\frac{\pi^3}{16\sqrt{2}}-\frac{\pi\ln^2(2)}{8\sqrt{2}}-\frac{\pi\ln\left(16-8\sqrt{2}\right)\, \operatorname{artanh}\left(\frac{1}{7}+\frac{2\sqrt{2}}{7}\right)}{2\sqrt{2}}+\frac{\pi\ln(2)\,\operatorname{arcoth}\left(\sqrt{2}\right)}{\sqrt{2}}}$$ $$\color{purple}{\implies L = -\frac{\pi\operatorname{Li}_2\left(\frac{1}{2}-\frac{1}{2\sqrt{2}}\right)}{2\sqrt{2}}-\frac{\pi^3}{16\sqrt{2}}-\frac{7\pi\ln^2(2)}{8\sqrt{2}}+\frac{\pi\ln^2\left(2-\sqrt{2}\right)}{4\sqrt{2}}+\frac{\pi\ln(2)\ln\left(2+\sqrt{2}\right)}{2\sqrt{2}}}$$

Thus finally, we have

$$I = \pi \color{red}{\left(\frac{\pi^2}{8\sqrt2}-\frac{1}{\sqrt2}\left(-\frac{\pi^2}{48}+\frac{\ln^2(2)}{8}+\operatorname{Li}_2\left(\frac{1}{\sqrt2}\right)\right)\right)} - \pi \color{blue}{\left( -\frac{1}{4\sqrt{2}}\left(\frac{\pi^2}{3}+2\operatorname{Li}_2\left(\frac{1}{2}-\frac{1}{\sqrt{2}}\right)+\ln^2(2)-\operatorname{arsinh}(1)^2-2\ln(2)\operatorname{arsinh}(1)\right) \right)} + \color{purple}{\left( -\frac{\pi\operatorname{Li}_2\left(\frac{1}{2}-\frac{1}{2\sqrt{2}}\right)}{2\sqrt{2}}-\frac{\pi^3}{16\sqrt{2}}-\frac{7\pi\ln^2(2)}{8\sqrt{2}}+\frac{\pi\ln^2\left(2-\sqrt{2}\right)}{4\sqrt{2}}+\frac{\pi\ln(2)\ln\left(2+\sqrt{2}\right)}{2\sqrt{2}} \right) }$$

After some simplification

$$\implies I = -\frac{\pi\operatorname{Li}_2\left(\frac{1}{\sqrt{2}}\right)}{\sqrt{2}}+\frac{\pi\operatorname{Li}_2\left(\frac{1}{2}-\frac{1}{\sqrt{2}}\right)}{2\sqrt{2}}-\frac{\pi\operatorname{Li}_2\left(\frac{1}{2}-\frac{1}{2\sqrt{2}}\right)}{2\sqrt{2}}+\frac{\pi^3}{6\sqrt{2}}-\frac{\pi\ln^2(2)}{2\sqrt{2}}+\frac{\pi\ln^2\left(2-\sqrt{2}\right)}{4\sqrt{2}}-\frac{\pi\ln^2\left(1+\sqrt{2}\right)}{4\sqrt{2}}$$

We now finally prove the following equality: $$2\operatorname{Li}_2 \left(\frac{\sqrt{2}}{2}\right)-\operatorname{Li}_2 \left(\frac{1-\sqrt{2}}{2}\right)+\operatorname{Li}_2 \left(-\frac{\sqrt{2}}{2} \left(\frac{1-\sqrt{2}}{2}\right)\right)\\=\frac{7\pi^2}{24} - \ln^2 (2) + \frac{1}{2} \ln^2 (2-\sqrt{2})-\frac{1}{2} \ln^2 (1+\sqrt{2})$$

By setting $z=-\frac{\sqrt{2}}{2}$ and $w = \frac{1-\sqrt{2}}{2}$ into this identity referenced to CJ.Hill, 1830 we determine that $$\operatorname{Li}_2 \left(-\frac{\sqrt{2}}{2} \left(\frac{1-\sqrt{2}}{2}\right)\right)=\operatorname{Li}_2 \left(-\frac{\sqrt{2}}{2}\right)+\operatorname{Li}_2 \left(\frac{1-\sqrt{2}}{2}\right)+\operatorname{Li}_2 \left(\frac{2-\sqrt{2}}{2}\right)+\operatorname{Li}_2 \left(\frac{1}{2}\right)+\frac{1}{8} \ln^2 (2)$$

Substituting this into the LHS of the equality we wish to prove, and using the well-known value $\operatorname{Li}_2 \left(\frac{1}{2}\right)=\frac{\pi^2}{12}-\frac{1}{2}\ln^2 (2)$, the LHS becomes $$2\operatorname{Li}_2 \left(\frac{\sqrt{2}}{2}\right)+\operatorname{Li}_2 \left(-\frac{\sqrt{2}}{2}\right)+\operatorname{Li}_2 \left(\frac{2-\sqrt{2}}{2}\right)+\frac{\pi^2}{12}-\frac{3}{8}\ln^2 (2)$$ which further simplifies, by setting $z=\frac{\sqrt{2}}{2}$ into equation (3), to become $$\operatorname{Li}_2 \left(\frac{\sqrt{2}}{2}\right)+\operatorname{Li}_2 \left(\frac{2-\sqrt{2}}{2}\right)+\frac{\pi^2}{8}-\frac{5}{8}\ln^2 (2)$$ which simplifies, by setting $z=\frac{\sqrt{2}}{2}$ into equation (5), to become $$\frac{7\pi^2}{24}-\frac{5}{8}\ln^2 (2) + \frac{1}{2} \ln (2) \ln\left(\frac{2-\sqrt{2}}{2}\right)=\frac{7\pi^2}{24}-\frac{9}{8}\ln^2 (2) + \frac{1}{2} \ln (2) \ln\left(2-\sqrt{2}\right)$$

Now if we consider the RHS of the equality we are trying to prove, we can factor \begin{align*}\frac{1}{2}\ln^2 (2-\sqrt{2})-\frac{1}{2}\ln^2 (1+\sqrt{2}) &= \frac{1}{2}\left(\ln(2-\sqrt{2})+\ln(1+\sqrt{2})\right)\left(\ln(2-\sqrt{2})-\ln(1+\sqrt{2})\right)\\&=\frac{1}{4}\ln (2)\ln(3\sqrt{2}-4)\end{align*} Since $3\sqrt{2}-4 = \frac{1}{\sqrt{2}} \left(2-\sqrt{2}\right)^2$, the given equality is true. $\square$

Thus upon multiplying the dilogarithm equality, we have proven by $\frac{\pi}{2\sqrt{2}}$ and rearranging, we have $$I = \frac{\pi^3}{48 \sqrt{2}}.$$ $\square$

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Method 2:

We shall make use of the following representation: $$\tag{1}\int_{0}^{1}\frac{\ln\left|1-2t+t^2(1-x^2)\right|}{t}\, dt=-\Re\left(\operatorname{Li}_2(1+x)+\operatorname{Li}_2(1-x)\right).$$

First notice that through dilogarithm identities, we can express the given integral in the following way:

$$I=\frac{\pi^2}{6}\int_{0}^{1}\frac{1}{(1+x^2)\sqrt{1-x^2}}\, dx +\int_{0}^{1}\frac{-\Re\left(\operatorname{Li}_2(1+x)+\operatorname{Li}_2(1-x)\right)}{(1+x^2)\sqrt{1-x^2}}\,dx+\int_{0}^{1}\frac{\pi x\ln\left(1+\frac{1}{x}\right)}{(1+x^2)\sqrt{1-x^2}}\, dx$$

Now use $(1)$ and enforce the substitution $u=\frac{x}{\sqrt{1-x^2}}$ on the first two integrals to determine \begin{align*} I&=\frac{\pi^2}{6}\int_{0}^{\infty}\frac{1}{2u^2+1}\, du+\int_{0}^{1}\int_{0}^{\infty}\frac{\ln\left|1-2t+\frac{t^2}{1+u^2}\right|}{t(2u^2+1)}\, du\, dt+\int_{0}^{1}\frac{\pi x\ln\left(1+\frac{1}{x}\right)}{(1+x^2)\sqrt{1-x^2}}\, dx \\ &=\frac{\pi^3}{12\sqrt{2}}+\frac{\pi}{\sqrt{2}}\int_{0}^{1}\frac{\ln\left|\frac{1+\frac{\sqrt{2}}{2}\sqrt{1-2t}-t}{1+\frac{\sqrt{2}}{2}}\right|}{t}\, dt+\int_{0}^{1}\frac{\pi x\ln\left(1+\frac{1}{x}\right)}{(1+x^2)\sqrt{1-x^2}}\, dx \end{align*} where we use $\int_{0}^{\infty}\frac{1}{2u^2+1}\, du=\frac{\pi}{2\sqrt{2}}$ and $$\int_{0}^{\infty}\frac{\ln\left|1-2t+\frac{t^2}{1+u^2}\right|}{2u^2+1}\, du=\frac{\pi}{\sqrt{2}}\ln\left|\frac{1+\frac{\sqrt{2}}{2}\sqrt{1-2t}-t}{1+\frac{\sqrt{2}}{2}}\right|$$ as proven in my answer to your other question.

Continuing similarly then, \begin{align*} I &= \frac{\pi^3}{12\sqrt{2}}+\frac{\pi}{2\sqrt{2}}\int_{0}^{1}\frac{2\ln\left|\frac{1+\frac{\sqrt{2}}{2}\sqrt{1-2t}-t}{1+\frac{\sqrt{2}}{2}}\right|+\ln\left|\frac{1-\frac{\sqrt{2}}{2}\sqrt{1-2t}-t}{1-\frac{\sqrt{2}}{2}}\right|-\ln\left|\frac{1-\frac{\sqrt{2}}{2}\sqrt{1-2t}-t}{1-\frac{\sqrt{2}}{2}}\right|}{t}\,dt+\int_{0}^{1}\frac{\pi x\ln\left(1+\frac{1}{x}\right)}{(1+x^2)\sqrt{1-x^2}}\, dx \\ &\stackrel{\text{IBP}}{=}\frac{\pi^3}{12\sqrt{2}}+\frac{\pi}{2\sqrt{2}}\int_{0}^{1}\frac{\ln(2t^2-2t+1)}{t}\,dt+\Re\left(\frac{\pi}{2\sqrt{2}}\int_{0}^{1}\frac{2t\ln(t)}{\sqrt{\frac{1}{2}-t}\, (2t^2-2t+1)}\, dt\right)+\int_{0}^{1}\frac{\pi x\ln\left(1+\frac{1}{x}\right)}{(1+x^2)\sqrt{1-x^2}}\, dx \\ &= \frac{\pi^3}{12\sqrt{2}}+\frac{\pi}{2\sqrt{2}}\int_{0}^{1}\frac{\ln(2t^2-2t+1)}{t}\,dt+\frac{\pi}{\sqrt{2}}\int_{0}^{1/2}\frac{t\ln(t)}{\sqrt{\frac{1}{2}-t}\, (2t^2-2t+1)}\, dt+\int_{0}^{1}\frac{\pi x\ln\left(1+\frac{1}{x}\right)}{(1+x^2)\sqrt{1-x^2}}\, dx \end{align*}

Now enforce the substitution $\frac{1}{t}=1+\frac{1}{x}$ on the second integral to determine $$\frac{\pi}{\sqrt{2}}\int_{0}^{1/2}\frac{t\ln(t)}{\sqrt{\frac{1}{2}-t}\, (2t^2-2t+1)}\, dt=-\int_{0}^{1}\frac{\pi x\ln\left(1+\frac{1}{x}\right)}{(1+x^2)\sqrt{1-x^2}}\, dx,$$ meaning that we have reduced the given integral to $$I=\frac{\pi^3}{12\sqrt{2}}+\frac{\pi}{2\sqrt{2}}\int_{0}^{1}\frac{\ln(2t^2-2t+1)}{t}\,dt ,$$ and we leave showing $$\int_{0}^{1}\frac{\ln(2t^2-2t+1)}{t}\,dt=-\frac{\pi^2}{8}$$ as an exercise.