Vtgyjfy

In solving $displaystyleint_0^frac{pi}{4}dfrac{ln(sin x)ln(cos x)}{sin xcos x} dx,$ I have found that this is equal to $dfrac{1}{16}displaystyleint_0^1dfrac{ln^2(1-x)}{x} dx.$ WolframAlpha says that the desired value is $dfrac{zeta(3)}{8},$ so I suspect a conversion to a series is necessary.

How do I prove $displaystyleint_0^1dfrac{ln^2(1-x)}{x} dx=displaystylesum_{n=1}^inftydfrac{2}{n^3}$?

Note that the above integral can also be given as $displaystyleint_0^1dfrac{ln^2x}{1-x} dx$, which I know is equal to $displaystylesum_{n=0}^infty x^nln^2x.$

Also for reference, here is a picture of my original work to get to this point.

We have $$int_{0}^{1}frac{log^{2}left(1-xright)}{x}dxstackrel{xrightarrow1-x}{=}int_{0}^{1}frac{log^{2}left(xright)}{1-x}dx$$ $$stackrel{DCT}{=}
sum_{kgeq0}int_{0}^{1}log^{2}left(xright)x^{k}dxstackrel{IBP}{=}
2sum_{kgeq0}frac{1}{left(k+1right)^{3}}=color{red}{2zetaleft(3right)}.$$

I thought it might be instructive to present a way forward that exploits the Polylogarithm Functions. To that end, we proceed.

Note that integrating by parts with $u=log^2(1-x)$ and $v=log(x)$, we have

$$begin{align}
int_0^1 frac{log^2(1-x)}{x},dx=2int_0^1 frac{log(1-x)log(x)}{1-x},dx tag 1
end{align}$$

Integrating by parts the right-hand side of $(1)$ with $u=log(1-x)$ and $v=text{Li}_2(1-x)$ yields

$$begin{align}
2int_0^1 frac{log(1-x)log(x)}{1-x},dx&=2int_0^1 frac{text{Li}_2(1-x)}{1-x},dx\\
&=2int_0^1 frac{text{Li}_2(x)}{x},dx\\
&=2text{Li}_3(1)\\
&=2zeta(3)
end{align}$$

as expected!

$newcommand{angles}[1]{leftlangle,{#1},rightrangle}
newcommand{braces}[1]{leftlbrace,{#1},rightrbrace}
newcommand{bracks}[1]{leftlbrack,{#1},rightrbrack}
newcommand{dd}{mathrm{d}}
newcommand{ds}[1]{displaystyle{#1}}
newcommand{expo}[1]{,mathrm{e}^{#1},}
newcommand{half}{{1 over 2}}
newcommand{ic}{mathrm{i}}
newcommand{iff}{Longleftrightarrow}
newcommand{imp}{Longrightarrow}
newcommand{Li}[1]{,mathrm{Li}_{#1}}
newcommand{mc}[1]{,mathcal{#1}}
newcommand{mrm}[1]{,mathrm{#1}}
newcommand{ol}[1]{overline{#1}}
newcommand{pars}[1]{left(,{#1},right)}
newcommand{partiald}[3]{frac{partial^{#1} #2}{partial #3^{#1}}}
newcommand{root}[2]{,sqrt[#1]{,{#2},},}
newcommand{totald}[3]{frac{mathrm{d}^{#1} #2}{mathrm{d} #3^{#1}}}
newcommand{ul}[1]{underline{#1}}
newcommand{verts}[1]{leftvert,{#1},rightvert}$

1. This one is $ul{slightly different}$ of the straightforward @Dr. MV answer:
   begin{align}
   color{#f00}{{1 over 16}int_{0}^{1}{ln^{2}pars{1 - x} over x},dd x} &,,,
   stackrel{x mapsto pars{1 - x}}{=},,,
   {1 over 16}int_{0}^{1}{ln^{2}pars{x} over 1 - x},dd x
   end{align}
   

   ---

   
   Integrating by Parts a few times ( the main purpose is to 'sit' a
   $ds{lnpars{1 - x}}$-factor in the integrand numerator ):
   begin{align}
   color{#f00}{{1 over 16}int_{0}^{1}{ln^{2}pars{1 - x} over x},dd x} & =
   {1 over 16}int_{0}^{1}lnpars{1 - x}
   bracks{2lnpars{x},{1 over x}},dd x =
   -,{1 over 8}int_{0}^{1}Li{2}'pars{x}lnpars{x},dd x
   \[5mm] & =
   {1 over 8}int_{0}^{1}Li{2}pars{x},{1 over x},dd x =
   {1 over 8}int_{0}^{1}Li{3}'pars{x},dd x = {1 over 8},Li{3}pars{1}
   \[5mm] & = color{#f00}{{1 over 8},zetapars{3}}
   end{align}


2. Another approach uses the Beta Function
   $ds{mrm{B}pars{mu,nu} =
   int_{0}^{1}x^{mu - 1},pars{1 - x}^{nu - 1},,dd x =
   {Gammapars{mu}Gammapars{nu} over Gammapars{mu + nu}}}$ with
   $ds{Repars{mu} > 0,, Repars{nu} > 0}$. $ds{Gamma,}$: Gamma
   Function.
   begin{align}
   &color{#f00}{{1 over 16}int_{0}^{1}{ln^{2}pars{1 - x} over x},dd x} =
   {1 over 16},lim_{mu to 0},,partiald[2]{}{mu}
   int_{0}^{1}{pars{1 - x}^{mu} - 1 over x},dd x
   \[5mm] & =
   {1 over 16},lim_{mu to 0},,partiald[2]{}{mu}bracks{mu
   int_{0}^{1}lnpars{x}pars{1 - x}^{mu - 1},dd x} =
   {1 over 16},lim_{mu to 0 atop nu to 0},,{partial^{3} over partialmu^{2},partialnu}
   bracks{muint_{0}^{1}x^{nu}pars{1 - x}^{mu - 1},dd x}
   \[5mm] & =
   {1 over 16},lim_{mu to 0 atop nu to 0},,{partial^{3} over partialmu^{2},partialnu}bracks{mu,{Gammapars{nu + 1}Gammapars{mu} over Gammapars{mu + nu + 1}}} =
   {1 over 16},lim_{mu to 0 atop nu to 0},,{partial^{3} over partialmu^{2},partialnu}bracks{Gammapars{nu + 1}Gammapars{mu + 1} over Gammapars{mu + nu + 1}}
   \[5mm] & = color{#f00}{{1 over 8},zetapars{3}}
   end{align}

Here is an approach that makes use of an Euler sum.

We will first find a Maclaurin series expansion for $ln^2 (1 - x)$. As
$$ln (1 - x) = - sum_{n = 1}^infty frac{x^n}{n},$$
we have
begin{align*}
ln^2 (1 - x) &= left (- sum_{n = 1}^infty frac{x^n}{n} right ) cdot left (- sum_{n = 1}^infty frac{x^n}{n} right ).
end{align*}
Shifting the summation index $n mapsto n + 1$ gives
begin{align*}
ln^2 (1 - x) &= x^2 left (- sum_{n = 0}^infty frac{x^n}{n + 1} right ) cdot left (- sum_{n = 0}^infty frac{x^n}{n + 1} right )\
&= sum_{n = 0}^infty sum_{k = 0}^n frac{x^{n + 2}}{(k + 1)(n - k + 1)},
end{align*}
where the last line has been obtained by applying the Cauchy product.

Shifting the summation indices as follows: $n mapsto n - 2, k mapsto k - 1$ gives
begin{align*}
ln^2 (1 - x) &= sum_{n = 2}^infty sum_{k = 1}^{n - 1} frac{x^n}{k(n - k)}\
&= sum_{n = 2}^infty sum_{k = 1}^{n - 1} left (frac{1}{nk} + frac{1}{n(n - k)} right ) x^n\
&= 2 sum_{n = 2}^infty frac{x^n}{n} sum_{k = 1}^{n - 1} frac{1}{k}\
&= 2 sum_{n = 2}^infty frac{H_{n - 1} x^n}{n},
end{align*}
where $H_n$ is the $n$th harmonic number.

Now evaluating the integral. From the above Maclaurin series expansion for $ln^2 (1 - x)$ the integral can be written as
begin{align*}
int_0^1 frac{ln^2 (1 - x)}{x} , dx &= 2 sum_{n = 2}^infty frac{H_{n - 1}}{n} int_0^1 x^{n - 1} , dx = 2 sum_{n = 2}^infty frac{H_{n - 1}}{n^2}.
end{align*}

From properties of the harmonic numbers we have
$$H_n = H_{n - 1} + frac{1}{n},$$
thus
begin{align*}
int_0^1 frac{ln^2 (1 - x)}{x} , dx &= 2 sum_{n = 2}^infty frac{H_n}{n^2} - 2 sum_{n = 2}^infty frac{1}{n^3} = 2 sum_{n = 1}^infty frac{H_n}{n^2} - 2 sum_{n = 1}^infty frac{1}{n^3}.
end{align*}

Each sum can be readily found. They are:
$$sum_{n = 1}^infty frac{1}{n^3} = zeta (3) quad text{and} quad sum_{n = 1}^infty frac{H_n}{n^2} = 2 zeta (3).$$
A proof of the result for the second sum containing the harmonic number can, for example, be found here. Thus
$$int_0^1 frac{ln^2 (1 - x)}{x} , dx = 4 zeta (3) - 2 zeta (3) = 2 zeta (3),$$
as required.

Everything you did is good. Indeed, it suffices to show that $I_n:=int_0^1 x^n ln^2 x = frac{2}{(n+1)^3}$ to conclude. The interversion $int / Sigma$ is possible as everything here is positive.

To compute $I_n$, I tried an integration by parts (using that a primitive of $ln^2 x$ is $x(ln^2 x-2ln x)$) to obtain the relation $$I_n = frac{-2}{n+1}int_0^1 x^nln x.$$
Let us call this latter integral $J_n$. Once again with an integration by parts, you can show that $J_n = frac{-1}{(n+1)^2}$, and thus conclude.