How can you approach $int_0^{pi/2} xfrac{ln(cos x)}{sin x}dx$

Mathematics Asked on December 10, 2020

Here is a new challenging problem:

Show that

$$I=int_0^{pi/2} xfrac{ln(cos x)}{sin x}dx=2ln(2)G-frac{pi}{8}ln^2(2)-frac{5pi^3}{32}+4Imleft{text{Li}_3left(frac{1+i}{2}right)right}$$

My attempt:

With Weierstrass substitution we have

$$I=2int_0^1frac{arctan x}{x}lnleft(frac{1-x^2}{1+x^2}right)dxoverset{xto frac{1-x}{1+x}}{=}4int_0^1frac{frac{pi}{4}-arctan x}{1-x^2}lnleft(frac{2x}{1+x^2}right)dx$$

$$=piunderbrace{int_0^1frac{1}{1-x^2}lnleft(frac{2x}{1+x^2}right)dx}_{I_1}-4underbrace{int_0^1frac{arctan x}{1-x^2}lnleft(frac{2x}{1+x^2}right)dx}_{I_2}$$

By setting $xto frac{1-x}{1+x}$ in the first integral we have



For the second integral, write $frac{1}{1-x^2}=frac{1}{2(1-x)}+frac{1}{2(1+x)}$

$$I_2=frac12int_0^1frac{arctan x}{1-x}lnleft(frac{2x}{1+x^2}right)dx+frac12int_0^1frac{arctan x}{1+x}lnleft(frac{2x}{1+x^2}right)dx$$

The first integral is very similar to this one

lnleft(frac{2x^2}{1+x^2}right),mathrm{d}x =
-frac{pi}{16}ln^{2}left(2right) –
frac{11}{192},pi^{3} +
text{Li}_{3}left(frac{1 + mathrm{i}}{2}right)right}$$

So we are left with only $int_0^1frac{arctan xln(1+x^2)}{1+x}dx$ as $int_0^1frac{arctan xln x}{1+x}dx$ is already nicely calculated by FDP here. Any idea?

I noticed that if we use $xtofrac{1-x}{1+x}$ in $int_0^1frac{arctan xln(1+x^2)}{1+x}dx$ 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 $int_0^{pi/2} frac{xcos(2nx)}{sin x}dx$. I would like to see different approaches if possible.

Thank you.

3 Answers

Many ways to go are possible!

A simple way would be to exploit the known result,

$$int_0^1 frac{arctan(x)}{x}logleft(frac{1+x^2}{(1-x)^2}right)=frac{pi^3}{16},tag 1$$

since with the Weierstrass subs the main integral reduces to

$$mathcal{I}=2int_0^1frac{arctan(x)}{x}logleft(frac{1-x^2}{1+x^2}right)textrm{d}x$$ $$=-2 int_0^1 frac{ arctan(x)}{x}log left(frac{1+x^2}{(1-x)^2}right) textrm{d}x-2 int_0^1 frac{arctan(x)log (1-x)}{x} textrm{d}x$$ $$+2 int_0^1 frac{arctan(x)log (1+x) }{x} textrm{d}x$$ $$=2log(2)G-frac{pi}{8}log^2(2)-frac{5}{32}pi^3+4Imleft{text{Li}_3left(frac{1+i}{2}right)right},$$

where the last two integrals are calculated by Ali Shather in this answer

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 $17$,

$$ int_0^x frac{arctan(t)log(1+t^2)}{t} textrm{d}t-2 int_0^1 frac{arctan(xt)log (1-t)}{t}textrm{d}t$$ $$=2sum_{n=1}^{infty} (-1)^{n-1} frac{x^{2n-1}}{(2n-1)^3}, |x|le1.$$

Correct answer by user97357329 on December 10, 2020

$$ int_0^1 frac{arctan x ln(1+x^2)}{1+x} dx=frac{pi}{16}ln^{2}left(2right) - frac{11}{192},pi^{3} + 2Imleft{% text{Li}_{3}left(frac{1 + mathrm{i}}{2}right)right}+{Gln2}$$ $$int_0^1frac{arctan xln(frac{2x}{1+x^2})}{1-x}dx=frac{pi^3}{192}-dfrac{Gln 2}{2}$$ $$int_0^1frac{arctan xln(frac{2x}{1+x^2})}{1+x}dx=frac{pi}{16}ln^{2}left(2right) + frac{pi^3}{24} - 2Imleft{%} text{Li}_{3}left(frac{1 + mathrm{i}}{2}right)right}-dfrac{Gln 2}{2}$$

Answered by user178256 on December 10, 2020

$newcommand{bbx}[1]{,bbox[15px,border:1px groove navy]{displaystyle{#1}},} newcommand{braces}[1]{leftlbrace,{#1},rightrbrace} newcommand{bracks}[1]{leftlbrack,{#1},rightrbrack} newcommand{dd}{mathrm{d}} newcommand{ds}[1]{displaystyle{#1}} newcommand{expo}[1]{,mathrm{e}^{#1},} newcommand{ic}{mathrm{i}} newcommand{mc}[1]{mathcal{#1}} newcommand{mrm}[1]{mathrm{#1}} newcommand{pars}[1]{left(,{#1},right)} newcommand{partiald}[3][]{frac{partial^{#1} #2}{partial #3^{#1}}} newcommand{root}[2][]{,sqrt[#1]{,{#2},},} newcommand{totald}[3][]{frac{mathrm{d}^{#1} #2}{mathrm{d} #3^{#1}}} newcommand{verts}[1]{leftvert,{#1},rightvert}$ begin{align} I & equiv int_{0}^{pi/2}x{lnpars{cospars{x}} over sinpars{x}},dd x \[5mm] & = bbox[5px,#ffd]{2lnpars{2},mrm{G} - {pi over 8}ln^{2}pars{2} - {5pi^{3} over 32} + 4,Impars{mrm{Li}_3pars{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{lnpars{cospars{x}} over sinpars{x}},dd x} \[5mm] & = left. Reint_{x = 0}^{x = pi/2}bracks{-iclnpars{z}}{lnpars{bracks{z + 1/z}/2} over pars{z - 1/z}/pars{2ic}},{dd z over ic z},rightvert_{ z = exppars{ic x}} \[5mm] & = left. -2,Imint_{x = 0}^{x = pi/2}lnpars{z}, lnpars{1 + z^{2} over 2z} ,{dd z over 1 - z^{2}},rightvert_{ z = exppars{ic x}} \[5mm] & = 2,Imint_{1}^{0}bracks{lnpars{y} + {pi over 2},ic}, bracks{lnpars{1 - y^{2} over 2y} - {pi over 2},ic} ,{ic,dd y over 1 + y^{2}} \[5mm] & = -2int_{0}^{1}bracks{lnpars{y}lnpars{1 - y^{2} over 2y} + {pi^{2} over 4}}, ,{dd y over 1 + y^{2}} \[5mm] & = -2 overbrace{int_{0}^{1}{lnpars{y}lnpars{1 - y} over 1 + y^{2}},dd y}^{ds{I_{1}}} - 2 overbrace{int_{0}^{1}{lnpars{y}lnpars{1 + y} over 1 + y^{2}},dd y}^{ds{I_{2}}} \[2mm] & + 2lnpars{2} underbrace{int_{0}^{1}{lnpars{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} + 2lnpars{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} + Impars{mrm{Li}_{3}pars{1 + ic over 2}} \[2mm] ds{I_{2}} & ds{=} & ds{phantom{-}2mrm{G}lnpars{2} + {3pi over 32},ln^{2}pars{2}} + {11pi^{3} over 128} - 3,Impars{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}).

Answered by Felix Marin on December 10, 2020

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