TITLE: Solving higher order logarithms integrals without the beta function
QUESTION [8 upvotes]: Prove the following

$$\int^1_0 \frac{\log(x)^3}{\sqrt{x(1-x)}}dx= -\pi (12 \zeta(3) +
 \log^3(4) + \pi^2 \log(4))$$

The basic approach is differentiating the beta function three times which I am trying to avoid here.A contour method would be nice.
Question 1
Hence how to solve the integral without having to differentiate the beta function ? 
My question can be generalized to any problem involving higher order logarithms.
Question 2
Is there a general formula for
$$\frac{\partial^n \partial^m}{\partial x^n \partial y^m} B(x,y)$$

REPLY [2 votes]: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
&\int_{0}^{1}{\ln^{3}\pars{x} \over \root{x\pars{1 - x}}}\,\dd x
\,\,\,\stackrel{x\ =\ \sin^{2}\pars{\theta}}{=}\,\,\,
16\int_{0}^{\pi/2}\ln^{3}\pars{\sin\pars{\theta}}\,\dd\theta
\\[5mm] = &\
\left.16\,\Re\int_{0}^{\pi/2}\ln^{3}\pars{{1 - z^{2} \over 2z}\ic}
\,{\dd z \over \ic z}\,\right\vert_{\ z\ =\ \exp\pars{\ic\theta}} =
\left.16\,\Im\int_{0}^{\pi/2}\ln^{3}\pars{{1 - z^{2} \over 2z}\ic}
\,{\dd z \over z}\,\right\vert_{\ z\ =\ \exp\pars{\ic\theta}}
\\[5mm] \stackrel{\mrm{as}\ \epsilon\ \to\ 0^{+}}{\sim} &
-16\,\Im\int_{\pi/2}^{0}
\bracks{-\ln\pars{2\epsilon} + \pars{{\pi \over 2} - \theta}\ic}^{\,3}\
\ic\,\dd\theta -
16\,\Im\int_{\epsilon}^{1}
\bracks{\ln\pars{1 - x^{2} \over 2x} + {\pi \over 2}\,\ic}^{3}\,{\dd x \over x}
\\[5mm] = &\
16\int_{0}^{\pi/2}\bracks{-\ln^{3}\pars{2\epsilon} + 3\ln\pars{2\epsilon}\,\theta^{2}}\,\dd\theta -
16\int_{\epsilon}^{1}\bracks{3\ln^{2}\pars{1 - x^{2} \over 2x}\,{\pi \over 2} -{\pi^{3} \over 8}}\,{\dd x \over x}
\\[1cm] = &\
-8\pi\ln^{3}\pars{2\epsilon} + \color{#f00}{2\pi^{3}\ln\pars{2\epsilon}} -
24\pi\int_{\epsilon}^{1}{\ln^{2}\pars{1 - x^{2}} -
2\ln\pars{1 - x^{2} }\ln\pars{2x} + \ln^{2}\pars{2x} \over x}\,\dd x
\\[5mm] &\
\color{#f00}{\mbox{} - 2\pi^{3}\ln\pars{\epsilon}}
\\ \stackrel{\mrm{as}\ \epsilon\ \to\ 0^{+}}{\sim} &\
\color{#f00}{-8\pi\ln^{3}\pars{2\epsilon}} + 2\pi^{3}\ln\pars{2} -
12\pi\
\overbrace{\int_{0}^{1}{\ln^{2}\pars{1 - x} \over x}\,\dd x}
^{\ds{2\zeta\pars{3}}}\ +\
24\pi\ln\pars{2}\
\overbrace{\int_{0}^{1}{\ln\pars{1 - x} \over x}\,\dd x}
^{\ds{-\,\mrm{Li}_{2}\pars{1} = -\,{\pi^{2} \over 6}}}
\\[5mm] &\
\mbox{} +
12\pi\
\underbrace{\int_{0}^{1}{\ln\pars{1 - x}\ln\pars{x} \over x}\,\dd x}
_{\ds{\zeta\pars{3}}}\ -\
24\pi\
\underbrace{\int_{2\epsilon}^{2}{\ln^{2}\pars{x} \over x}\,\dd x}
_{\ds{{\ln^{3}\pars{2} \over 3} -
\color{#f00}{\ln^{3}\pars{2\epsilon} \over 3}}}
\label{1}\tag{1}
\end{align}

Note that

$$
\left\{\begin{array}{rcl}
\ds{\int_{0}^{1}{\ln\pars{1 - x}\ln\pars{x} \over x}\,\dd x} & \ds{=} &
\ds{-\int_{0}^{1}\mrm{Li}_{2}'\pars{x}\ln\pars{x}\,\dd x =
\int_{0}^{1}{\mrm{Li}_{2}\pars{x} \over x}\,\dd x =
\int_{0}^{1}\mrm{Li}_{3}'\pars{x}\,\dd x}
\\[1mm] & \ds{=} &
\ds{\mrm{Li}_{3}\pars{1} = \bbx{\ds{\zeta\pars{3}}}}
\\[5mm]
\ds{\int_{0}^{1}{\ln^{2}\pars{1 - x} \over x}\,\dd x} & \ds{=} &
\ds{\int_{0}^{1}{\ln^{2}\pars{x} \over 1 - x}\,\dd x =
2\int_{0}^{1}{\ln\pars{1 - x}\ln\pars{x} \over x}\,\dd x = \bbx{\ds{2\,\zeta\pars{3}}}}
\end{array}\right.
$$

Expression \eqref{1} becomes

\begin{align}
&\int_{0}^{1}{\ln^{3}\pars{x} \over \root{x\pars{1 - x}}}\,\dd x
\,\,\,\stackrel{\mrm{as}\ \epsilon\ \to\ 0^{+}}{\to}\,\,\,
2\pi^{3}\ln\pars{2} - 24\pi\,\zeta\pars{3} - 4\pi^{3}\ln\pars{2} + 12\pi\,\zeta\pars{3} - 8\pi\ln^{3}\pars{2}
\\[5mm] = &\,\,\,
\bbox[15px,#ffe,border:1px dotted navy]{\ds{-2\pi\bracks{\vphantom{\Large A}6\,\zeta\pars{3} + 4\ln^{3}\pars{2} + \pi^{2}\ln\pars{2}}}} \approx -96.6701
\end{align}