오늘의 계산 59 - A Crazy Integral

밤새서 푼 적분 하나. 하라는 공부는 안 하고…!

 

Proposition. The following holds:[1]

$$\begin{array}{r}{\int }_0^1\frac{\log x\log (1-x){\log }^2(1+x)}{x}dx=\frac{7}{8}\zeta (2)\zeta (3)-\frac{25}{16}\zeta (5).\end{array}$$

 

Proof. Below is my proof of the above identity.

 

Step 1. Let $I$ be the integral in question:
$$\begin{array}{rl}I& ={\int }_0^1\frac{\log x\log (1-x){\log }^2(1+x)}{x}dx.\end{array}$$ By the simple algebraic formula $(a+b{)}^3+(a-b{)}^3-2a^3=6a{b}^2$, it follows that
$$\begin{array}{rl}I& =\frac{1}{6}{\int }_0^1\frac{\log x}{x}\{{\log }^3(1-x^2)+{\log }^3\left(\frac{1-x}{1+x}\right)-2{\log }^3(1-x)\}dx\\ \text{(1)}& & =\frac{1}{6}{\int }_0^1\frac{\log x{\log }^3(1-x^2)}{x}dx+\frac{1}{6}{\int }_0^1\frac{\log x{\log }^3\left(\frac{1-x}{1+x}\right)}{x}dx-\frac{1}{3}{\int }_0^1\frac{\log x{\log }^3(1-x)}{x}dx\end{array}$$ Applying the substitution $x^2\mapsto x$, the first integral reduces to
$$\begin{array}{rl}\frac{1}{6}{\int }_0^1\frac{\log x{\log }^3(1-x^2)}{x}dx& =\frac{1}{24}{\int }_0^1\frac{\log x{\log }^3(1-x)}{x}dx=\frac{1}{24}\frac{{\partial }^4\beta }{\partial z\partial w^3}(0^{+},1)\\ & =\frac{1}{2}\zeta (5)-\frac{1}{4}\zeta (2)\zeta (3).\end{array}$$ So $\text{(1)}$ can be written as
$$\begin{array}{}\text{(2)}& I& =\frac{7}{4}\zeta (2)\zeta (3)-\frac{7}{2}\zeta (5)+I_2,\end{array}$$ where $I_2$ is given by
$$\begin{array}{r}{I}_2=\frac{1}{6}{\int }_0^1\frac{\log x{\log }^3\left(\frac{1-x}{1+x}\right)}{x}dx\end{array}$$

Step 2. Now we calculate $I_2$. Applying integrating by parts, followed by the substitution $x\mapsto {\displaystyle \frac{1-y}{1+y}}$, we have
$$\begin{array}{rl}{I}_2& ={[\frac{1}{12}{\log }^{2}x{\log }^3\left(\frac{1-x}{1+x}\right)]}_0^1+\frac{1}{2}{\int }_0^1\frac{{\log }^{2}x{\log }^2\left(\frac{1-x}{1+x}\right)}{1-x^2}dx\\ & =\frac{1}{4}{\int }_0^1\frac{{\log }^{2}y{\log }^2\left(\frac{1-y}{1+y}\right)}{y}dy\\ & ={\int }_0^1\frac{{\log }^{2}y}{y}{\{\frac{1}{2}{\log }^2\left(\frac{1+y}{1-y}\right)\}}^{2}dy\\ & =\sum _{m=0}^{\mathrm{\infty }}\sum _{n=0}^{\mathrm{\infty }}\frac{1}{(2m+1)(2n+1)}{\int }_0^1{y}^{2m+2n+1}{\log }^{2}ydy\\ & =\sum _{m=0}^{\mathrm{\infty }}\sum _{n=0}^{\mathrm{\infty }}\frac{2}{(2m+1)(2n+1)(2m+2n+2{)}^3}.\end{array}$$ By the following partial fraction decomposition
$$\begin{array}{r}\frac{1}{ab(a+b{)}^3}=\frac{a+b}{ab(a+b{)}^4}=\frac{1}{a(a+b{)}^4}+\frac{1}{b(a+b{)}^4},\end{array}$$
together with the symmetry in the role of $m$ and $n$, it follows that
$$\begin{array}{rl}{I}_2& =4\sum _{m=0}^{\mathrm{\infty }}\sum _{n=0}^{\mathrm{\infty }}\frac{1}{(2n+1)(2m+2n+2{)}^4}=\frac{1}{4}\sum _{m=0}^{\mathrm{\infty }}\sum _{n=0}^{\mathrm{\infty }}\frac{1}{(2n+1)(m+n+1{)}^4}\\ & =\frac{1}{4}\sum _{n=0}^{\mathrm{\infty }}\sum _{m=n+1}^{\mathrm{\infty }}\frac{1}{2n+1}\cdot \frac{1}{m^4}=\frac{1}{4}\sum _{n=1}^{\mathrm{\infty }}\sum _{m=n}^{\mathrm{\infty }}\frac{1}{2n-1}\cdot \frac{1}{m^4}\\ & =\frac{1}{4}\sum _{m=1}^{\mathrm{\infty }}\left(\sum _{n=1}^m\frac{1}{2n-1}\right)\frac{1}{m^4}.\end{array}$$ Now by exploiting the following identity
$$\begin{array}{rl}& {\int }_0^1\frac{1-x^k}{1-x}dx=1+\frac{1}{2}+\cdots +\frac{1}{k}\\ ⟹& {\int }_0^1\frac{1+x^k-2x^{2k}}{2(1-x)}dx=1+\frac{1}{3}+\cdots +\frac{1}{2k-1},\end{array}$$ we can rephrase $I_2$ as follows:
$$\begin{array}{rl}{I}_2& =\frac{1}{8}\sum _{m=1}^{\mathrm{\infty }}\frac{1}{m^4}{\int }_0^1\frac{1+x^m-2x^{2m}}{1-x}dx\\ & =\frac{1}{8}{\int }_0^1\frac{1}{1-x}(\zeta (4)+{\mathrm{Li}}_4(x)-2{\mathrm{Li}}_4(x^2))dx.\end{array}$$ Writing ${\displaystyle \frac{1}{1-x}=\frac{{\mathrm{Li}}_0(x)}{x}}$ and applying the simple differentiation rule ${\displaystyle \frac{d}{dx}{\mathrm{Li}}_s(x^p)=\frac{p{\mathrm{Li}}_{s-1}(x^p)}{x}}$, by applying integrating parts 3 times,
$$\begin{array}{rl}{I}_2& =\frac{1}{8}{[{\mathrm{Li}}_1(x)(\zeta (4)+{\mathrm{Li}}_4(x)-2{\mathrm{Li}}_4(x^2))]}_0^1-\frac{1}{8}{\int }_0^1\frac{{\mathrm{Li}}_1(x)}{x}({\mathrm{Li}}_3(x)-4{\mathrm{Li}}_3(x^2))dx\\ & =-\frac{1}{8}{[{\mathrm{Li}}_2(x)({\mathrm{Li}}_3(x)-4{\mathrm{Li}}_3(x^2))]}_0^1+\frac{1}{8}{\int }_0^1\frac{{\mathrm{Li}}_2(x)}{x}({\mathrm{Li}}_2(x)-8{\mathrm{Li}}_2(x^2))dx\\ & =\frac{3}{8}\zeta (2)\zeta (3)+\frac{1}{8}{[{\mathrm{Li}}_3(x)({\mathrm{Li}}_2(x)-8{\mathrm{Li}}_2(x^2))]}_0^1-\frac{1}{8}{\int }_0^1\frac{{\mathrm{Li}}_3(x)}{x}({\mathrm{Li}}_1(x)-16{\mathrm{Li}}_1(x^2))dx\\ & =-\frac{1}{2}\zeta (2)\zeta (3)+\frac{1}{8}{\int }_0^1\frac{{\mathrm{Li}}_3(x)}{x}(\log (1-x)-16\log (1-x^2))dx\\ & =-\frac{1}{2}\zeta (2)\zeta (3)+\frac{1}{8}{\int }_0^1\frac{{\mathrm{Li}}_3(x)\log (1-x)}{x}dx-2{\int }_0^1\frac{{\mathrm{Li}}_3(x)\log (1-x^2)}{x}dx.\end{array}$$ Now applying the substitution $x\mapsto x^2$ to the first integral, from the identity ${\mathrm{Li}}_3(x^2)=4{\mathrm{Li}}_3(x)+4{\mathrm{Li}}_3(-x)$ it follows that
$$\begin{array}{rl}\frac{1}{8}{\int }_0^1\frac{{\mathrm{Li}}_3(x)\log (1-x)}{x}dx& =\frac{1}{4}{\int }_0^1\frac{{\mathrm{Li}}_3(x^2)\log (1-x^2)}{x}dx\\ & ={\int }_0^1\frac{\{{\mathrm{Li}}_3(x)+{\mathrm{Li}}_3(-x)\}\log (1-x^2)}{x}dx.\end{array}$$ This finally gives
$$\begin{array}{r}{I}_2=-\frac{1}{2}\zeta (2)\zeta (3)-{\int }_{-1}^1\frac{{\mathrm{Li}}_3(x)\log (1-x^2)}{x}dx.\end{array}$$ Plugging this back to $\text{(2)}$, we obtain
$$\begin{array}{}\text{(3)}& I_1=\frac{5}{4}\zeta (2)\zeta (3)-\frac{7}{2}\zeta (5)+I_3,\end{array}$$ where $I_3$ is given by
$$\begin{array}{r}{I}_3=-{\int }_{-1}^1\frac{{\mathrm{Li}}_3(x)\log (1-x^2)}{x}dx.\end{array}$$

Step 3. Finally, we evaluate $I_3$ into a closed form and thus complete the calculation. Since the integrand is holomorphic on the unit disc, we can shift up the contour of integration to the clockwise semicircular arc:
$$\begin{array}{rl}{I}_3& =-i{\int }_{\pi }^0{\mathrm{Li}}_3(e^{i\theta })\log (1-e^{2i\theta })d\theta =-\frac{1}{i}{\int }_0^{\pi }{\mathrm{Li}}_3(e^{i\theta })\log (1-e^{2i\theta })d\theta \end{array}$$ Now expanding with Taylor series and taking advantage of the fact that $I_3$ is real, we have
$$\begin{array}{rl}{I}_3& =\sum _{n=1}^{\mathrm{\infty }}\sum _{m=1}^{\mathrm{\infty }}\frac{1}{n^3}\frac{1}{m}{\int }_0^{\pi }\{\cos (n\theta )\sin (2m\theta )+\sin (n\theta )\cos (2m\theta )\}d\theta \\ & =\sum _{n=1}^{\mathrm{\infty }}\sum _{m=1}^{\mathrm{\infty }}\frac{1}{n^3}\frac{1}{m}{\int }_0^{\pi }\cos (n\theta )\sin (2m\theta )d\theta \\ & +\sum _{n=1}^{\mathrm{\infty }}\sum _{m=1}^{\mathrm{\infty }}\frac{1}{n^3}\frac{1}{m}{\int }_0^{\pi }\sin (n\theta )\cos (2m\theta )d\theta \\ & =:S_1+S_2.\end{array}$$ To evaluate $S_1$, we note the following simple identity
$$\begin{array}{r}{\int }_0^{\pi }\cos (n\theta )\sin (2m\theta )d\theta =\frac{1+(-1{)}^{n-1}}{2}{\int }_0^{\pi }\cos (\frac{1}{2}n\theta )\sin (m\theta )d\theta .\end{array}$$ Then we can write
$$\begin{array}{rl}{S}_1& =\sum _{n=0}^{\mathrm{\infty }}\frac{1}{(2n+1{)}^3}{\int }_0^{\pi }\left\{\sum _{m=1}^{\mathrm{\infty }}\frac{\sin (m\theta )}{m}\right\}\cos ((n+\frac{1}{2})\theta )d\theta ,\\ S_2& =\sum _{m=1}^{\mathrm{\infty }}\frac{1}{m}{\int }_0^{\pi }\left\{\sum _{n=1}^{\mathrm{\infty }}\frac{\sin (n\theta )}{n^3}\right\}\cos (2m\theta )d\theta .\end{array}$$ But it is not hard to check that the following Fourier expansion holds:
$$\begin{array}{r}\sum _{n=1}^{\mathrm{\infty }}\frac{\sin n\theta }{n}=\mathrm{\Im }\sum _{n=1}^{\mathrm{\infty }}\frac{e^{i\theta }}{n}=-\mathrm{\Im }\log (1-e^{i\theta })=\frac{\pi -\theta }{2},0<\theta <\pi ,\end{array}$$ $$\begin{array}{r}\sum _{n=1}^{\mathrm{\infty }}\frac{\sin n\theta }{n^3}=\frac{{\theta }^3}{12}-\frac{\pi {\theta }^2}{4}+\frac{{\pi }^2\theta }{6},0<\theta <\pi .\end{array}$$ Plugging these to $S_1$, $S_2$ and performing some tedious calculation, we obtain
$$\begin{array}{r}{S}_1=\sum _{n=0}^{\mathrm{\infty }}\frac{2}{(2n+1{)}^5}=\frac{31}{16}\zeta (5)\text{and}{S}_2=-\frac{{\pi }^2}{16}\sum _{m=1}^{\mathrm{\infty }}\frac{1}{m^3}=-\frac{3}{8}\zeta (2)\zeta (3).\end{array}$$ Therefore we have
$$\begin{array}{r}I=\frac{7}{8}\zeta (2)\zeta (3)-\frac{25}{16}\zeta (5).\end{array}$$

풀이는 아래 참고문헌 [1]에 있습니다. 참고로 이 적분의 값은 실험적으로 발견되었다고 하네요. 그럼 제가 처음 푼 건가… 라고 말하기에는, 꼴이 뭔가 그럴듯하니 이미 누군가 풀어놨겠죠 -.-;;

References

1. Shobhit, A crazy integral - Integrals and Series