Evaluating $\int_0^1 x \tan(\pi x) \log(\sin(\pi x))dx$

Solution 1:

\begin{align} \mathcal{I}=&\int^1_0x\tan(\pi x)\ln(\sin(\pi x))\ {\rm d}x\\ =&\left(\int^{1/2}_0+\int_{1/2}^1\right)x\tan(\pi x)\ln(\sin(\pi x))\ {\rm d}x\tag1\\ =&\int^{1/2}_0(2x-1)\tan(\pi x)\ln(\sin(\pi x))\ {\rm d}x\tag2\\ =&-\frac{2}{\pi^2}\int^{\pi/2}_0x\cot{x}\ln(\cos{x})\ {\rm d}x\tag3\\ =&-\frac{1}{\pi}\int^{\pi/2}\tan{x}\ln(\sin{x})\ {\rm d}x+\frac{2}{\pi^2}\int^{\pi/2}_0x\tan{x}\ln(\sin{x})\ {\rm d}x\tag4\\ =&\frac{2}{\pi^2}\int^{\pi/2}_0\ln(\sin{x})\ln(\cos{x})\ {\rm d}x-\frac{2}{\pi^2}\int^{\pi/2}_0x\tan{x}\ln(\sin{x})\ {\rm d}x\tag5\\ =&\frac{1}{\pi^2}\int^{\pi/2}_0\ln(\sin{x})\ln(\cos{x})\ {\rm d}x-\frac{1}{2\pi}\int^{\pi/2}_0\tan{x}\ln(\sin{x})\ {\rm d}x\tag6\\ =&\frac{1}{8\pi^2}\frac{\partial^2{\rm B}}{\partial a\partial b}\left(\frac{1}{2},\frac{1}{2}\right)-\frac{1}{8\pi}\frac{\partial{\rm B}}{\partial b}\left(0^+,1\right)\tag7\\ =&\frac{1}{8\pi^2}\left(4\pi\ln^2{2}-\frac{\pi^3}{6}\right)-\frac{1}{8\pi}\left(-\frac{\pi^2}{6}\right)=\boxed{\Large{\color{red}{\dfrac{\ln^2{2}}{2\pi}}}}\\ \end{align}

Explanation:

$(1)$: Split the integral at $\displaystyle\frac12$.

$(2)$: Substituted $\displaystyle x\mapsto1-x$ in the second integral.

$(3)$: Substituted $\displaystyle x\mapsto\frac{1}{2}-\frac{x}{\pi}$.

$(4)$: Substituted $\displaystyle x\mapsto\frac{\pi}{2}-x$.

$(5)$: Integrated by parts.

$(6)$: Took the average of $(4)$ and $(5)$.

$(7)$: $\displaystyle {\rm B}(a,b)=2\int^{\pi/2}_0\sin^{2a-1}{x}\cos^{2b-1}{x}\ {\rm d}x$

Solution 2:

Integrating by parts, we get $$ \begin{align} &\int_0^1x\tan(\pi x)\log(\sin(\pi x))\,\mathrm{d}x\\ &=\int_0^{1/2}(2x-1)\tan(\pi x)\log(\sin(\pi x))\,\mathrm{d}x\tag{1a} \\ &=-\frac1\pi\int_0^{1/2}(2x-1)\log(\sin(\pi x))\,\mathrm{d}\log(\cos(\pi x))\tag{1b}\\ &=\small\frac2\pi\int_0^{1/2}\log(\sin(\pi x))\log(\cos(\pi x))\,\mathrm{d}x +\int_0^{1/2}(2x-1)\cot(\pi x)\log(\cos(\pi x))\,\mathrm{d}x\tag{1c}\\ &=\small\frac2\pi\int_0^{1/2}\log(\sin(\pi x))\log(\cos(\pi x))\,\mathrm{d}x +\int_0^{1/2}(-2x)\tan(\pi x)\log(\sin(\pi x))\,\mathrm{d}x\tag{1d}\\ &=\small\frac1\pi\int_0^{1/2}\log(\sin(\pi x))\log(\cos(\pi x))\,\mathrm{d}x -\frac12\int_0^{1/2}\tan(\pi x)\log(\sin(\pi x))\,\mathrm{d}x\tag{1e}\\ \end{align} $$ Explanation:
$\text{(1a)}$: Subtract $\frac12\int_0^1\tan(\pi x)\log(\sin(\pi x))\,\mathrm{d}x=0$ and use the symmetry of $x\mapsto1-x$
$\text{(1b)}$: Apply $\mathrm{d}\log(\cos(\pi x))=\pi\tan(\pi x)\,\mathrm{d}x$
$\text{(1c)}$: Integrate by Parts
$\text{(1d)}$: Substitute $x\mapsto\frac12-x$
$\text{(1e)}$: Average $\text{(1a)}$ and $\text{(1d)}$

Using $$ \log(\sin(x))=-\log(2)-\sum_{k=1}^\infty\frac{\cos(2kx)}{k}\tag{2} $$ $$ \log(\cos(x))=-\log(2)-\sum_{k=1}^\infty(-1)^k\frac{\cos(2kx)}{k}\tag{3} $$ we get $$ \begin{align} &\int_0^{1/2}\log(\sin(\pi x))\log(\cos(\pi x))\,\mathrm{d}x\\ &=\color{#C00000}{\int_0^{1/2}(-\log(2))^2\,\mathrm{d}x}\\ &-\color{#00A000}{\log(2)\int_0^{1/2}\sum_{k=1}^\infty\left[-1-(-1)^k\right]\frac{\cos(2\pi kx)}{k}\,\mathrm{d}x}\\ &+\color{#0000FF}{\int_0^{1/2}\sum_{j=1}^\infty\sum_{k=1}^\infty(-1)^k\frac{\cos(2\pi jx)}{j}\frac{\cos(2\pi kx)}{k}\,\mathrm{d}x}\\ &=\color{#C00000}{\frac{\log(2)^2}2}-\color{#00AA00}{0}+\color{#0000FF}{\frac14\sum_{k=1}^\infty\frac{(-1)^k}{k^2}}\\ &=\frac{\log(2)^2}2-\frac{\pi^2}{48}\tag{4} \end{align} $$ since $\int_0^{1/2}\cos(2\pi jx)\cos(2\pi kx)\,\mathrm{d}x=0$ when $j\ne k$.

Substituting $t=\sin(\pi x)$ and $e^{-u/2}=t$, we get $$ \begin{align} \int_0^{1/2}\tan(\pi x)\log(\sin(\pi x))\,\mathrm{d}x &=\frac1\pi\int_0^1\frac{t}{1-t^2}\log(t)\,\mathrm{d}t\\ &=-\frac1{4\pi}\int_0^\infty\frac{e^{-u}}{1-e^{-u}}u\,\mathrm{d}u\\ &=-\frac1{4\pi}\int_0^\infty\sum_{k=1}^\infty ue^{-ku}\,\mathrm{d}u\\ &=-\frac1{4\pi}\sum_{k=1}^\infty\frac1{k^2}\\ &=-\frac{\pi}{24}\tag{6} \end{align} $$ Combining $(1)$, $(4)$, and $(6)$ yields $$ \int_0^1x\tan(\pi x)\log(\sin(\pi x))\,\mathrm{d}x =\frac{\log(2)^2}{2\pi}\tag{7} $$

Solution 3:

$\def\Li{{\rm{Li}_2\,}}$Denote the considered integral as $I$ and set $y=\pi x$, we have \begin{equation} I=\frac{1}{\pi^2}\int_0^\pi \frac{y\sin y}{\cos y}\,\ln(\sin y)\,dy \end{equation} Perform integration by parts by taking $u=y$, we have \begin{align} I&=-\left.\frac{y}{2\pi^2}\int\frac{\ln\left(1-\cos^2y\right)}{\cos y}\,d(\cos y)\right|_0^\pi+\frac{1}{2\pi^2}\int_0^\pi\int\frac{\ln\left(1-\cos^2y\right)}{\cos y}\,d(\cos y)\,dy\\ &=\left.\frac{y}{4\pi^2}\Li\left(\cos^2y\right)\,\right|_0^\pi-\frac{1}{4\pi^2}\int_0^\pi\Li\left(\cos^2y\right)\,dy\\ &=\frac{\Li\left(1\right)}{4\pi}-\frac{1}{4\pi^2}\int_0^{\pi}\sum_{k=1}^\infty\frac{\cos^{2k}y}{k^2}\,dy\quad\Rightarrow\quad\color{red}{\mbox{use series representation of dilogarithm}}\\ &=\frac{\pi}{24}-\frac{1}{4\pi^2}\sum_{k=1}^\infty\int_0^{\pi}\frac{\cos^{2k}y}{k^2}\,dy\quad\Rightarrow\quad\color{red}{\mbox{justified by Fubini-Tonelli theorem}}\\ &=\frac{\pi}{24}-\frac{1}{2\pi^2}\sum_{k=1}^\infty\frac{1}{k^2}\int_0^{\pi/2}\cos^{2k}y\,\,dy\quad\Rightarrow\quad\color{red}{\mbox{by symmetry argument}}\\ &=\frac{\pi}{24}-\frac{1}{4\pi^2}\sum_{k=1}^\infty\frac{\Gamma\left(k+\frac{1}{2}\right)\Gamma\left(\frac{1}{2}\right)}{k^2\,\Gamma\left(k+1\right)}\quad\Rightarrow\quad\color{red}{\mbox{Wallis' integrals}}\\ &=\frac{\pi}{24}-\frac{1}{4\pi}\sum_{k=1}^\infty\frac{(2k)!}{4^k\,k^2\,(k!)^2}\tag1 \end{align} $\def\arctanh{{\rm{\,arctanh}\,}}$Here is the tedious part (and also the hardest part). I use Mathematica to help me out to find generating function of the following series. Let us start with \begin{equation} \sum_{k=0}^\infty\frac{\Gamma\left(k+\frac{1}{2}\right)\Gamma\left(\frac{1}{2}\right)x^k}{\Gamma\left(k+1\right)}=\pi\sum_{k=0}^\infty\frac{(2k)!\,\,x^k}{4^k\,(k!)^2}=\frac{\pi}{\sqrt{1-x}} \end{equation} Divide by $x$ and then integrate it, we have \begin{equation} \sum_{k=1}^\infty\frac{(2k)!\,\,x^k}{4^k\,k\,(k!)^2}=\int\left[\frac{1}{x\,\sqrt{1-x}}-\frac{1}{x}\right]\,dx=-2\arctanh\left(\sqrt{1-x}\,\right)-\ln x+C_1 \end{equation} Taking the limit as $x\to0$, we obtain \begin{equation} C_1=\lim_{x\to0}\left(2\arctanh\left(\sqrt{1-x}\,\right)+\ln x\right)=\ln4 \end{equation} Hence \begin{equation} \sum_{k=1}^\infty\frac{(2k)!}{4^k\,k\,(k!)^2}x^k=-2\arctanh\left(\sqrt{1-x}\,\right)-\ln x+\ln4 \end{equation} Repeat the process once more, we obtain \begin{align} \sum_{k=1}^\infty\frac{(2k)!\,\,x^k}{4^k\,k^2\,(k!)^2}=&\,-2\int\frac{\arctanh\left(\sqrt{1-x}\,\right)}{x}\,dx-\int\frac{\ln x}{x}\,dx+\ln4\int\frac{dx}{x}\\ =&\,\,2\arctanh\left(\sqrt{1-x}\,\right)\left[\arctanh\left(\sqrt{1-x}\,\right)-2\ln\left(\frac{\sqrt{1-x}+1}{2}\right)\right]\\&\,-2\,\Li\left(\frac{\sqrt{1-x}-1}{\sqrt{1-x}+1}\right)-\frac{\ln^2x}{2}+\ln4\ln x+C_2\\ \end{align} Taking the limit as $x\to0$, we obtain \begin{equation} C_2=-2\ln^22 \end{equation} Hence \begin{align} \sum_{k=1}^\infty\frac{(2k)!\,\,x^k}{4^k\,k^2\,(k!)^2} =&\,\,2\arctanh\left(\sqrt{1-x}\,\right)\left[\arctanh\left(\sqrt{1-x}\,\right)-2\ln\left(\frac{\sqrt{1-x}+1}{2}\right)\right]\\&\,-2\,\Li\left(\frac{\sqrt{1-x}-1}{\sqrt{1-x}+1}\right)-\frac{\ln^2x}{2}+\ln4\ln x-2\ln^22\tag2 \end{align} Thus, by putting $x=1$ to $(2)$ then $(1)$ becomes \begin{equation} I=\frac{\pi}{24}-\frac{1}{4\pi}\left(\frac{\pi^2}{6}-2\ln^2 2\right)=\frac{\ln^22}{2\pi} \end{equation} and we are done.