Is there a close-form for $\int_{0}^{1} \arctan{x}\log{(1-x)}\log x\>dx$

Question

Is there a closed form for the given integral ? $$\int_{0}^{1} \arctan{x}\log{(1-x)}\log{(x)}dx$$

Answer 1

Notice that $$\arctan(x)=\sum_{n=0}^{\infty}\frac{(-1)^nx^{2n+1}}{2n+1}.$$ Therefore, $$\int_0^1\arctan(x)\ln(1-x)\ln(x)\,\mathrm{d}x=\int_0^1\ln(1-x)\ln(x)\sum_{n=0}^{\infty}\frac{(-1)^nx^{2n+1}}{2n+1}\,\mathrm{d}x$$ $$=\sum_{n=0}^{\infty}\frac{(-1)^n}{2n+1}\int_0^1x^{2n+1}\ln(1-x)\ln(x)\,\mathrm{d}x.$$ To take care of the integral: $$\int_0^1x^{2n+1}\ln(1-x)\ln(x)\,\mathrm{d}x=-\frac1{2n+1}\int_0^1-\frac{x^{2n+1}\ln(x)}{1-x}+\frac{x^{2n+1}\ln(1-x)}{x}\,\mathrm{d}x$$ $$=-\frac1{2n+1}\left[\int_0^1\frac{x^{2n+1}\ln(x)}{x-1}\,\mathrm{d}x+\int_0^1x^{2n}\ln(x)\,\mathrm{d}x\right].$$ Notice that $$\frac{x^{2n+1}\ln(x)}{x-1}=\frac{x^{2n+1}-1}{x-1}\ln(x)+\frac{\ln(x)}{x-1}=\sum_{m=0}^{2n}x^m\ln(x)-\frac{\ln(x)}{1-x}.$$ Therefore, $$\int_0^1\frac{x^{2n+1}\ln(x)}{x-1}\,\mathrm{d}x=\sum_{m=0}^{2n}\int_0^1x^m\ln(x)\,\mathrm{d}x-\int_0^1\frac{\ln(x)}{1-x}\,\mathrm{d}x.$$ Remember that $$-\int_0^1\frac{\ln(x)}{1-x}\,\mathrm{d}x=-\int_0^1\frac{\ln(1-x)}{x}\,\mathrm{d}x=\operatorname{Li}_2(1)=\frac{\pi^2}6$$ and $$\int_0^1x^m\ln(x)\,\mathrm{d}x=\frac{1^{m+1}}{m+1}\ln(1)-\lim_{x\to0}\frac{x^{m+1}}{m+1}\ln(x)-\frac1{m+1}\int_0^1x^{m+1}\frac1{x}\,\mathrm{d}x=-\frac1{(m+1)^2},$$ hence $$\sum_{m=0}^{2n}\int_0^1x^m\ln(x)\,\mathrm{d}x-\int_0^1\frac{\ln(x)}{1-x}\,\mathrm{d}x=-\sum_{m=0}^{2n}\frac1{(m+1)^2}+\frac{\pi^2}6=\frac{\pi^2}6-\sum_{m=1}^{2n+1}\frac1{m^2}.$$ Meanwhile, $$\int_0^1x^{2n}\ln(x)\,\mathrm{d}x=-\frac1{(2n+1)^2}.$$ Therefore, $$\int_0^1x^{2n+1}\ln(1-x)\ln(x)\,\mathrm{d}x=-\frac1{2n+1}\left[\frac{\pi^2}6-\sum_{m=1}^{2n+1}\frac1{m^2}-\frac1{(2n+1)^2}\right].$$ Keep in mind that $$\frac{\pi^2}6-\sum_{m=1}^{2n+1}\frac1{m^2}-\frac1{(2n+1)^2}=\sum_{m=1}^{\infty}\frac1{m^2}-\sum_{m=1}^{2n+1}\frac1{m^2}-\frac1{(2n+1)^2}=\sum_{2n+2}^{\infty}\frac1{m^2}-\frac1{(2n+1)^2}=\sum_{2n+1}^{\infty}\frac1{m^2}-\frac2{(2n+1)^2},$$ thus $$\int_0^1x^{2n+1}\ln(1-x)\ln(x)\,\mathrm{d}x=\frac2{(2n+1)^3}-\frac1{2n+1}\sum_{m=2n+1}^{\infty}\frac1{m^2}.$$ Therefore, $$\int_0^1\arctan(x)\ln(1-x)\ln(x)\,\mathrm{d}x=\sum_{n=0}^{\infty}\left[\frac{2(-1)^n}{(2n+1)^4}-\frac{(-1)^n}{(2n+1)^2}\sum_{m=2n+1}^{\infty}\frac1{m^2}\right]$$ $$=2\sum_{m=0}^{\infty}\frac{(-1)^n}{(2n+1)^4}-\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n+1)^2}\sum_{m=2n+1}^{\infty}\frac1{m^2}.$$ Now, for $\mathfrak{Re}(s)\gt0,$ the Dirichlet Beta function is defined $$\beta(s)=\sum_{m=0}^{\infty}\frac{(-1)^n}{(2n+1)^s}.$$ Thus $$\int_0^1\arctan(x)\ln(1-x)\ln(x)\,\mathrm{d}x=2\beta(4)-\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n+1)^2}\sum_{m=2n+1}^{\infty}\frac1{m^2}.$$ Yet again, $$\sum_{m=2n+1}^{\infty}\frac1{m^2}=\frac{\pi^2}6-\sum_{m=1}^{2n}\frac1{m^2},$$ so $$\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n+1)^2}\sum_{m=2n+1}^{\infty}\frac1{m^2}=\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n+1)^2}\frac{\pi^2}6-\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n+1)^2}\sum_{m=1}^{2n}\frac1{m^2}$$ $$=\frac{\pi^2}6\beta(2)-\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n+1)^2}\sum_{m=1}^{2n}\frac1{m^2},$$ and as such, $$\int_0^1\arctan(x)\ln(1-x)\ln(x)\,\mathrm{d}x=2\beta(4)-\frac{\pi^2}6\beta(2)+\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n+1)^2}\sum_{m=1}^{2n}\frac1{m^2}.$$ This is the most I know how to simplify the integral.

Edit: This can be simplified further. According to Compute $\sum_{n=1}^\infty (-1)^{n-1}\frac{H_{2n+1}}{(2n+1)^3}$ and $\sum_{n=1}^\infty (-1)^{n-1}\frac{H_{2n+1}^{(2)}}{(2n+1)^2}$, it is the case that $$\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n+1)^2}\sum_{m=1}^{2n+1}\frac1{m^2}=-\frac18G\zeta(2)+\frac{35}{64}\pi\zeta(3)+\frac{15}{16}\zeta(4)-\frac1{768}\psi^{(3)}\left(\frac14\right).$$ Notice that $G=\beta(2),$ $\zeta(2)=\frac{\pi^2}6,$ and $\zeta(4)=\frac{\pi^4}{90},$ so $$\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n+1)^2}\sum_{m=1}^{2n+1}\frac1{m^2}=-\frac{\pi^2}{108}\beta(2)+\frac{35}{64}\pi\zeta(3)+\frac{\pi^4}{96}-\frac1{768}\psi^{(3)}\left(\frac14\right).$$ On the other hand, $$\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n+1)^2}\sum_{m=1}^{2n+1}\frac1{m^2}=\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n+1)^2}\left[\sum_{m=1}^{2n}\frac1{m^2}+\frac1{(2n+1)^2}\right]$$ $$=\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n+1)^2}\sum_{m=1}^{2n}\frac1{m^2}+\beta(4).$$ Therefore, $$\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n+1)^2}\sum_{m=1}^{2n}\frac1{m^2}=-\beta(4)-\frac{\pi^2}{108}\beta(2)+\frac{35}{64}\pi\zeta(3)+\frac{\pi^4}{96}-\frac1{768}\psi^{(3)}\left(\frac14\right),$$ and $$\int_0^1\arctan(x)\ln(1-x)\ln(x)\,\mathrm{d}x=2\beta(4)-\frac{\pi^2}6\beta(2)+\left[-\beta(4)-\frac{\pi^2}{108}\beta(2)+\frac{35}{64}\pi\zeta(3)+\frac{\pi^4}{96}-\frac1{768}\psi^{(3)}\left(\frac14\right)\right]$$ $$=\beta(4)-\frac{19\pi^2}{108}\beta(2)+\frac{35}{64}\pi\zeta(3)+\frac{\pi^4}{96}-\frac1{768}\psi^{(3)}\left(\frac14\right).$$ Finally, notice that $$\beta(4)=\frac1{768}\psi^{(3)}\left(\frac14\right)-\frac{\pi^4}{96},$$ which is equivalent to $$\beta(4)+\frac{\pi^4}{96}-\frac1{768}\psi^{(3)}\left(\frac14\right)=0.$$ Therefore, $$\int_0^1\arctan(x)\ln(1-x)\ln(x)\,\mathrm{d}x=\frac{19\pi^2}{108}\beta(2)+\frac{35}{64}\pi\zeta(3).$$ This is the final answer, and the simplest it can be.