How can you approach $\int_0^{\pi/2} x\frac{\ln(\cos x)}{\sin x}dx$
Here is a new challenging problem:
Show that
$$I=\int_0^{\pi/2} x\frac{\ln(\cos x)}{\sin x}dx=2\ln(2)G-\frac{\pi}{8}\ln^2(2)-\frac{5\pi^3}{32}+4\Im\left\{\text{Li}_3\left(\frac{1+i}{2}\right)\right\}$$
My attempt:
With Weierstrass substitution we have
$$I=2\int_0^1\frac{\arctan x}{x}\ln\left(\frac{1-x^2}{1+x^2}\right)dx\overset{x\to \frac{1-x}{1+x}}{=}4\int_0^1\frac{\frac{\pi}{4}-\arctan x}{1-x^2}\ln\left(\frac{2x}{1+x^2}\right)dx$$
$$=\pi\underbrace{\int_0^1\frac{1}{1-x^2}\ln\left(\frac{2x}{1+x^2}\right)dx}_{I_1}-4\underbrace{\int_0^1\frac{\arctan x}{1-x^2}\ln\left(\frac{2x}{1+x^2}\right)dx}_{I_2}$$
By setting $x\to \frac{1-x}{1+x}$ in the first integral we have
$$I_1=\frac12\int_0^1\frac{1}{x}\ln\left(\frac{1-x^2}{1+x^2}\right)dx$$
$$=\frac14\int_0^1\frac{1}{x}\ln\left(\frac{1-x}{1+x}\right)dx=\frac14\left[-\text{Li}_2(x)+\text{Li}_2(-x)\right]_0^1=-\frac38\zeta(2)$$
For the second integral, write $\frac{1}{1-x^2}=\frac{1}{2(1-x)}+\frac{1}{2(1+x)}$
$$I_2=\frac12\int_0^1\frac{\arctan x}{1-x}\ln\left(\frac{2x}{1+x^2}\right)dx+\frac12\int_0^1\frac{\arctan x}{1+x}\ln\left(\frac{2x}{1+x^2}\right)dx$$
The first integral is very similar to this one
$$\int_0^1\frac{\arctan\left(x\right)}{1-x}\, \ln\left(\frac{2x^2}{1+x^2}\right)\,\mathrm{d}x = -\frac{\pi}{16}\ln^{2}\left(2\right) - \frac{11}{192}\,\pi^{3} + 2\Im\left\{% \text{Li}_{3}\left(\frac{1 + \mathrm{i}}{2}\right)\right\}$$
So we are left with only $\int_0^1\frac{\arctan x\ln(1+x^2)}{1+x}dx$ as $\int_0^1\frac{\arctan x\ln x}{1+x}dx$ is already nicely calculated by FDP here. Any idea?
I noticed that if we use $x\to\frac{1-x}{1+x}$ in $\int_0^1\frac{\arctan x\ln(1+x^2)}{1+x}dx$ we will have a nice symmerty but still some annoying integrals appear.
In $I$, I also tried the Fourier series of $\ln(\cos x)$ but I stopped at $\int_0^{\pi/2} \frac{x\cos(2nx)}{\sin x}dx$. I would like to see different approaches if possible.
Thank you.
Solution 1:
Many ways to go are possible!
A simple way would be to exploit the known result,
$$\int_0^1 \frac{\arctan(x)}{x}\log\left(\frac{1+x^2}{(1-x)^2}\right)=\frac{\pi^3}{16},\tag 1$$
since with the Weierstrass subs the main integral reduces to
$$\mathcal{I}=2\int_0^1\frac{\arctan(x)}{x}\log\left(\frac{1-x^2}{1+x^2}\right)\textrm{d}x$$ $$=-2 \int_0^1 \frac{ \arctan(x)}{x}\log \left(\frac{1+x^2}{(1-x)^2}\right) \textrm{d}x-2 \int_0^1 \frac{\arctan(x)\log (1-x)}{x} \textrm{d}x$$ $$+2 \int_0^1 \frac{\arctan(x)\log (1+x) }{x} \textrm{d}x$$ $$=2\log(2)G-\frac{\pi}{8}\log^2(2)-\frac{5}{32}\pi^3+4\Im\left\{\text{Li}_3\left(\frac{1+i}{2}\right)\right\},$$
where the last two integrals are calculated by Ali Shather in this answer https://math.stackexchange.com/q/3261446.
End of story
Credit for this approach goes to Cornel.
A first note: Interestingly, different ways make the problem very difficult. It would be nice to have in place more ways to go.
A second note: The generalization of the key integral in $(1)$ may be found in the book, (Almost) Impossible Integrals, Sums, and Series, page $17$,
$$ \int_0^x \frac{\arctan(t)\log(1+t^2)}{t} \textrm{d}t-2 \int_0^1 \frac{\arctan(xt)\log (1-t)}{t}\textrm{d}t$$ $$=2\sum_{n=1}^{\infty} (-1)^{n-1} \frac{x^{2n-1}}{(2n-1)^3}, \ |x|\le1.$$
Solution 2:
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
\newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
\newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
\newcommand{\dd}{\mathrm{d}}
\newcommand{\ds}[1]{\displaystyle{#1}}
\newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
\newcommand{\ic}{\mathrm{i}}
\newcommand{\mc}[1]{\mathcal{#1}}
\newcommand{\mrm}[1]{\mathrm{#1}}
\newcommand{\pars}[1]{\left(\,{#1}\,\right)}
\newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
\newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
\newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
\newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
I & \equiv \int_{0}^{\pi/2}x{\ln\pars{\cos\pars{x}}
\over \sin\pars{x}}\,\dd x
\\[5mm] & =
\bbox[5px,#ffd]{2\ln\pars{2}\,\mrm{G} - {\pi \over 8}\ln^{2}\pars{2} - {5\pi^{3} \over 32} + 4\,\Im\pars{\mrm{Li}_3\pars{1 + \ic \over 2}}}:\ {\Large ?}\label{1}\tag{1}
\end{align}
$\ds{\mrm{G}}$ is the Catalan Constant and
$\ds{\mrm{Li}_{s}}$ is the polylogarithm.
\begin{align} I & \equiv \bbox[5px,#ffd]{\int_{0}^{\pi/2}x{\ln\pars{\cos\pars{x}} \over \sin\pars{x}}\,\dd x} \\[5mm] & = \left. \Re\int_{x\ =\ 0}^{x\ =\ \pi/2}\bracks{-\ic\ln\pars{z}}{\ln\pars{\bracks{z + 1/z}/2} \over \pars{z - 1/z}/\pars{2\ic}}\,{\dd z \over \ic z}\,\right\vert_{\ z\ =\ \exp\pars{\ic x}} \\[5mm] & = \left. -2\,\Im\int_{x\ =\ 0}^{x\ =\ \pi/2}\ln\pars{z}\, \ln\pars{1 + z^{2} \over 2z} \,{\dd z \over 1 - z^{2}}\,\right\vert_{\ z\ =\ \exp\pars{\ic x}} \\[5mm] & = 2\,\Im\int_{1}^{0}\bracks{\ln\pars{y} + {\pi \over 2}\,\ic}\, \bracks{\ln\pars{1 - y^{2} \over 2y} - {\pi \over 2}\,\ic} \,{\ic\,\dd y \over 1 + y^{2}} \\[5mm] & = -2\int_{0}^{1}\bracks{\ln\pars{y}\ln\pars{1 - y^{2} \over 2y} + {\pi^{2} \over 4}}\, \,{\dd y \over 1 + y^{2}} \\[5mm] & = -2\ \overbrace{\int_{0}^{1}{\ln\pars{y}\ln\pars{1 - y} \over 1 + y^{2}}\,\dd y}^{\ds{I_{1}}}\ -\ 2\ \overbrace{\int_{0}^{1}{\ln\pars{y}\ln\pars{1 + y} \over 1 + y^{2}}\,\dd y}^{\ds{I_{2}}} \\[2mm] & + 2\ln\pars{2}\ \underbrace{\int_{0}^{1}{\ln\pars{y} \over 1 + y^{2}}\,\dd y} _{\ds{I_{3}}}\ +\ 2\ \underbrace{\int_{0}^{1}{\ln^{2}\pars{y} \over 1 + y^{2}}\,\dd y} _{\ds{I_{4}}}\ -\ \underbrace{{\pi^{2} \over 2}\int_{0}^{1}{\dd y \over 1 + y^{2}}} _{\ds{\pi^{3} \over 8}} \\ & = -2I_{1} -2I_{2} + 2\ln\pars{2}\, I_{3} +2I_{4} - {\pi^{3} \over 8} \label{2}\tag{2} \end{align} Those integrals are well known or/and very -laboriously- doable: \begin{equation} \left\{\begin{array}{rcl} \ds{I_{1}} & \ds{=} & \ds{-\,{\pi \over 32}\,\ln^{2}\pars{2}} - {\pi^{3} \over 128} + \Im\pars{\mrm{Li}_{3}\pars{1 + \ic \over 2}} \\[2mm] \ds{I_{2}} & \ds{=} & \ds{\phantom{-}2\mrm{G}\ln\pars{2} + {3\pi \over 32}\,\ln^{2}\pars{2}} + {11\pi^{3} \over 128} - 3\,\Im\pars{\mrm{Li}_{3}\pars{1 + \ic \over 2}} \\[2mm] \ds{I_{3}} & \ds{=} & \ds{-\,\mrm{G}} \\[2mm] \ds{I_{4}} & \ds{=} & \ds{\phantom{-}{\pi^{3} \over 16}} \end{array}\right.\label{3}\tag{3} \end{equation} (\ref{2}) and (\ref{3}) lead to the coveted result (\ref{1}).
Solution 3:
$$ \int_0^1 \frac{\arctan x \ln(1+x^2)}{1+x} dx=\frac{\pi}{16}\ln^{2}\left(2\right) - \frac{11}{192}\,\pi^{3} + 2\Im\left\{% \text{Li}_{3}\left(\frac{1 + \mathrm{i}}{2}\right)\right\}+{G\ln2}$$ $$\int_0^1\frac{\arctan x\ln(\frac{2x}{1+x^2})}{1-x}dx=\frac{\pi^3}{192}-\dfrac{G\ln 2}{2}$$ $$\int_0^1\frac{\arctan x\ln(\frac{2x}{1+x^2})}{1+x}dx=\frac{\pi}{16}\ln^{2}\left(2\right) + \frac{\pi^3}{24} - 2\Im\left\{%} \text{Li}_{3}\left(\frac{1 + \mathrm{i}}{2}\right)\right\}-\dfrac{G\ln 2}{2}$$