What's the value of $\int_0^1\frac{1}{2y} \ln(y) \ln^2(1-y) \, dy$? I came across this integral while doing a different problem: 

$$ \int_0^1\frac{1}{2y} \ln (y)\ln^2(1-y) \, dy$$

I think we can evaluate this integral by differentiating the common integral representation of the beta function, but it seems to get a bit messy.
 A: There is a closed form antiderivative that can be found using repeated integration by parts:
$$\begin{align}\int\frac{\ln y\cdot\ln^2(1-y)}{2y}dy=\frac{\ln^4(1-y)}8+\frac{\ln y\cdot\ln ^3(1-y)}6+\left(\frac{\pi^2}{12}-\frac{\ln ^2y}2\right)\cdot\ln ^2(1-y)\\
+\left[\left(\frac{\pi^2}6+\operatorname{Li}_2\left(\tfrac y{y-1}\right)\right)\cdot\ln y-\operatorname{Li}_3(1-y)-\operatorname{Li}_3\left(\tfrac y{y-1}\right)\right]\cdot\ln(1-y)\\
+\left[\vphantom{\Large|}\zeta(3)-\operatorname{Li}_3(1-y)\right]\cdot\ln y+\operatorname{Li}_4(1-y)-\operatorname{Li}_4(y)-\operatorname{Li}_4\left(\tfrac y{y-1}\right)\color{gray}{+C}\end{align}$$
A: Another one:
Write $\log^2(1-y)=\sum_{n,m=1}^{\infty}\frac{y^{m+n}}{mn}$
we obtain (exchanging summation and integration)
$$
2I=\sum_{n,m=1}^{\infty}\frac{1}{mn}\int_0^1\log(y)y^{m+n-1}=\\
-\underbrace{\sum_{n,m=1}^{\infty}\frac{1}{mn}\frac{1}{(m+n)^2}}_{S}
$$
the double sum can be tackeled by writing (i shamelessly benefit from this awesome answer)
$$
S=\sum_{n,m=1}^{\infty}\frac{1}{mn}\frac{1}{(m+n)^2}=\frac{1}{2}\sum_{n,m=1}^{\infty}\frac{1}{m^2n^2}-\frac{1}{2}\sum_{n,m=1}^{\infty}\frac{1}{(m+n)^2n^2}-\frac{1}{2}\sum_{n,m=1}^{\infty}\frac{1}{(m+n)^2m^2}
$$
shifting arguments in the last two sums gives 
$$
2S=\sum_{n,m=1}^{\infty}\frac{1}{mn}\frac{1}{(m+n)^2}-\sum_{m=1,n<m}^{\infty}\frac{1}{n^2m^2}-\sum_{m=1,n>m}^{\infty}\frac{1}{n^2m^2}
$$
which yields
$$
2S=\sum_{n=m=1}^{\infty}\frac{1}{m^2n^2}=\zeta(4)=\frac{\pi^4}{90}
$$
and therefore
$$
I=-S/2=-\frac{\pi^4}{360}
$$
A: We start by introducing the integral
$$
I(a,b)=\frac{1}{2}\int_0^1y^{a-1}(1-y)^b\,dy=\frac{1}{2}B(a,1+b),
$$
where $B$ denotes the beta function. Note that this integral is singular at $a=0$ and $b=-1$. Since $\partial_a y^a=y^a\ln y$ we are
led to calculate
$$
\partial_{a,b,b}I(a,b)=\frac{1}{2}\int_0^1 y^{a-1}(1-y)^b\ln y\bigl(\ln(1-y)\bigr)^2\,dy
$$
as $a$ and $b$ tend to $0$. We will below inser the "non-dangerous" point $b=0$. In other words, we want to calculate
$$
\partial_{a,b,b}B(a,1+b)\mid_{a\to 0^+,b\to 0}.
$$
When differentiating the beta function, polygammas appear. Indeed,
$$
\begin{aligned}
\partial_bB(a,1+b)&=B(a,1+b)\bigl(\psi_0(1+b)-\psi_0(1+a+b)\bigr)\\
\partial_{b,b}B(a,1+b)&=B(a,1+b)\Bigl(\bigl(\psi_0(1+b)-\psi_0(1+a+b)\bigr)^2
+\psi_1(1+b)-\psi_1(1+a+b)\Bigr).
\end{aligned}
$$
Next, we can actually insert $b=0$ before we differentiate with respect to 
$a$ and take the limit $a\to 0$. We should differentiate the function (here we have used the facts that $\psi_0(1)=-\gamma$ (Euler's constant) and that $\psi_1(1)=\pi^2/6$)
$$
f(a)=B(a,1)\Bigl(\bigl(\gamma+\psi_0(1+a)\bigr)^2+\frac{\pi^2}{6}-\psi_1(1+a)\Bigr)
$$
and calculate $\lim_{a\to 0^+}f'(a)$. We get that
$$
\begin{aligned}
f'(a)&=B(a,1)\bigl(\psi_0(a)-\psi_0(1+a)\bigr)\Bigl(\bigl(\gamma+\psi_0(1+a)\bigr)^2+\frac{\pi^2}{6}-\psi_1(1+a)\Bigr)\\
&\quad+B(a,1)\Bigl(2\bigl(\gamma+\psi_0(1+a)\bigr)\psi_1(1+a)-\psi_2(1+a)\Bigr)
\end{aligned}
$$
Next, we use the (non-obvious) expansions around $a=0$
$$
\begin{aligned}
B(a,1)&=\frac{1}{a}+O(1)\\
\psi_0(a)&=-\frac{1}{a}-\gamma+O(a)\\
\psi_0(1+a)&=-\gamma+\frac{\pi^2}{6}a+O(a^2)\\
\psi_1(1+a)&=\frac{\pi^2}{6}+\psi_2(1)a+\frac{\pi^4}{30}a^2+O(a^3)\\
\psi_2(1+a)&=\psi_2(1)+\frac{\pi^4}{15}a+O(a^2).
\end{aligned}
$$
to find that, as $a\to0^+$,
$$
\begin{aligned}
f'(a)&\approx -\frac{1}{a^2}\Bigl(\bigl(\frac{\pi^2}{6}a\bigr)^2-\psi_2(1)a-\frac{\pi^4}{30}a^2\Bigr)+\frac{1}{a}\Bigl(2\frac{\pi^2}{6}a\frac{\pi^2}{6}-\psi_2(1)-\frac{\pi^4}{15}a\Bigr)+O(a)\\
&=-\frac{\pi^4}{180}+O(a)
\end{aligned}
$$
as $a\to 0^+$. We conclude that
$$
\partial_{a,b,b}B(a,1+b)\mid_{a\to 0^+,b\to 0}=-\frac{\pi^4}{180}.
$$
Finally, dividing by $2$ (remember, we had a one-half in front of the beta function in the beginning), we get that
$$
\int_0^1\frac{1}{2y}\ln y(\ln(1-y))^2\,dy=-\frac{1}{360}\pi^4.
$$
A: This question, for some reason, popped up in the Top Questions tab, and I thought I'd share another way to solve this integral using Harmonic Numbers. I hope you guys don't mind!
First, we use integration by parts on $u=\log^2(1-x)$ to get$$\begin{align*}\int\limits_0^1dx\,\frac {\log x\log^2(1-x)}{x} & =\frac 12\log^2x\log^2(1-x)\,\Biggr\rvert_0^1+\int\limits_0^1dx\,\frac {\log^2x\log(1-x)}{1-x}\\ & =\int\limits_0^1dx\,\frac {\log^2x\log(1-x)}{1-x}\end{align*}$$Now use the fact that

$$H(x)=-\frac {\log(1-x)}{1-x}=\sum\limits_{n\geq1}H_nx^n$$

And substitute to get$$\begin{align*}\int\limits_0^1dx\,\frac {\log x\log^2(1-x)}{x} & =-\sum\limits_{n\geq1}H_n\int\limits_0^1dx\,x^n\log^2x\\ & =-\lim\limits_{\mu\to0}\sum\limits_{n\geq1}H_n\frac {\partial^2}{\partial\mu^2}\frac 1{n+\mu+1}\\ & =-2\sum\limits_{n\geq1}\frac {H_n}{(n+1)^3}\end{align*}$$Split the sum up and use a well-known formula due to Euler

$$\sum\limits_{n\geq1}\frac {H_n}{n^m}=\frac 12(m+2)\zeta(m+1)-\frac 12\sum\limits_{n=1}^{m-2}\zeta(m-n)\zeta(n+1)$$

Therefore$$I=2\zeta(4)-5\zeta(4)+\zeta^2(2)=-\frac {\pi^4}{180}$$Our desired integral is half of that, so take half and the answer is$$\int\limits_0^1dx\,\frac {\log x\log^2(1-x)}{2x}\color{blue}{=-\frac {\pi^4}{360}}$$
A: I offer up yet another approach, this one relying on the Maclaurin series expansion for $\ln^2 (1 - x)$. It is similar to that used by @Frank W.
As was shown here
$$\ln^2 (1 - x) = 2 \sum_{n = 2}^\infty \frac{H_{n - 1} x^n}{n}, \qquad |x| < 1.$$
Here $H_n$ denotes the Harmonic number. So for the integral we may write
\begin{align}
\int_0^1 \frac{1}{2x} \ln x \ln^2 (1 - x) \, dx &= \sum_{n = 2}^\infty \frac{H_{n - 1}}{n} \int_0^1 x^{n - 1} \ln x \, dx = -\sum_{n = 2}^\infty \frac{H_{n - 1}}{n^3},
\end{align}
after integrating by parts. From properties for the Harmonic number, since
$$H_n = H_{n - 1} + \frac{1}{n},$$
the infinite sum can be rewritten as
\begin{align}
\int_0^1 \frac{1}{2x} \ln x \ln^2 (1 - x) \, dx &= \sum_{n = 2}^\infty \frac{1}{n^4} - \sum_{n = 2}^\infty \frac{H_n}{n^3} = \sum_{n = 1}^\infty \frac{1}{n^4} - \sum_{n = 1}^\infty \frac{H_n}{n^3}.
\end{align}
Values for each of these sums are well known. For the first
$$\sum_{n = 1}^\infty \frac{1}{n^4} = \zeta (4) = \frac{\pi^4}{90}.$$
For the second sum
$$\sum_{n = 1}^\infty \frac{H_n}{n^3} = \frac{\pi^4}{72}.$$
(for various proofs of this result, see here) Thus
$$\int_0^1 \frac{1}{2x} \ln x \ln^2 (1 - x) \, dx = \frac{\pi^4}{90} - \frac{\pi^4}{72} = -\frac{\pi^4}{360}.$$
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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$\ds{{1 \over 2}\int_{0}^{1}{\ln\pars{y}\ln^{2}\pars{1 - y} \over y}\,\dd y =
-\,{\pi^{4} \over 360}:\
{\large ?}}$.

\begin{align}
&\bbox[10px,#ffd]{\ds{{1 \over 2}\int_{0}^{1}{\ln\pars{y}\ln^{2}\pars{1 - y} \over y}\,\dd y}} =
\left.{1 \over 2}\,{\partial^{3} \over \partial\nu^{2}\partial\mu}\int_{0}^{1}{y^{\mu}\bracks{\pars{1 - y}^{\nu} - 1} \over y}\,\dd y\,\right\vert_{\ {\large\mu\ =\ 0^{+}} \atop {\large\nu\ =\ 0}}
\\[5mm] = &\
{1 \over 2}\,{\partial^{3} \over \partial\nu^{2}\partial\mu}
\bracks{{\Gamma\pars{\mu}\Gamma\pars{\nu + 1} \over \Gamma\pars{\mu + \nu + 1}} - {1 \over \mu}}_{\ {\large\mu\ =\ 0^{+}} \atop {\large\nu\ =\ 0}}\quad
\pars{~\Gamma:\ Gamma\ Function~}
\\[5mm] = &\
{1 \over 2}\,{\partial^{3} \over \partial\nu^{2}\partial\mu}
\bracks{{\pi \over \Gamma\pars{1 - \mu}\sin\pars{\pi\mu}}\,{\Gamma\pars{\nu + 1} \over \Gamma\pars{\mu + \nu + 1}} - {1 \over \mu}}_{\ {\large\mu\ =\ 0^{+}} \atop {\large\nu\ =\ 0}}
\\[5mm] = &\
{1 \over 2}\,{\partial^{3} \over \partial\nu^{2}\partial\mu}
\bracks{{1 \over \mu}\,{\Gamma\pars{\nu + 1} \over \Gamma\pars{1 - \mu}\Gamma\pars{\mu + \nu + 1}} + {\pi^{2} \over 6}\,\mu}
_{\ {\large\mu\ =\ 0^{+}} \atop {\large\nu\ =\ 0}}
\\[5mm] = &\
{1 \over 2}\,{\partial^{3} \over \partial\nu^{2}\partial\mu}
\bracks{\left.{1 \over 2}\,\partiald[2]{}{x}{\Gamma\pars{\nu + 1} \over \Gamma\pars{1 - x}\Gamma\pars{x + \nu + 1}}\,\right\vert_{\ x\ =\ 0^{+}}\mu + {\pi^{2} \over 6}\,\mu}_{\ {\large\mu\ =\ 0^{+}} \atop {\large\nu\ =\ 0}}
\\[5mm] = &\
{1 \over 2}\,{\partial^{2} \over \partial\nu^{2}}
\bracks{\left.{1 \over 2}\,\partiald[2]{}{x}{\Gamma\pars{\nu + 1} \over \Gamma\pars{1 - x}
\Gamma\pars{x + \nu + 1}}\,\right\vert_{\ x\ =\ 0^{+}} + {\pi^{2} \over 6}}
_{\ {\large\mu\ =\ 0^{+}} \atop {\large\nu\ =\ 0}} =
{1 \over 4}\,{\partial^{4} \over \partial\nu^{2}\partial\mu^{2}}
{\nu \choose \mu + \nu}_{\ {\large\mu\ =\ 0^{+}} \atop {\large\nu\ =\ 0}}
\\[5mm] = &\
{1 \over 4}\,\partiald[2]{}{\nu}
\bracks{-\,{\pi^{2} \over 6} + H^{2}_{\nu}  - \Psi\, '\pars{1 + \nu}}
_{\ \nu\ =\ 0}\quad
\pars{~H_{z}:\ Harmonic\ Number~}
\\[5mm] = &\
{1 \over 4}\bracks{2\Psi\, '^{2}\pars{1}  + 2H_{0}\,\Psi\,''\pars{1}- \Psi\, '''\pars{1}}
\qquad\qquad\qquad\qquad
\left\{\begin{array}{lcr}
\ds{\Psi\, '\pars{1}} & \ds{=} & \ds{\pi^{2} \over 6}
\\
\ds{\Psi\, '''\pars{1}} & \ds{=} & \ds{\pi^{4} \over 15}
\\
\ds{H_{0}} & \ds{=} & \ds{0}
\end{array}\right.
\\[5mm] = &\
{1 \over 4}\bracks{2\pars{\pi^{2} \over 6}^{2} + 0 - {\pi^{4} \over 15}} =
\bbx{-\,{\pi^{4} \over 360}}
\end{align}
A: Here is a solution without using Beta function:
Lets start with subbing $1-y=x$, we get $$\int_0^1\frac{\ln y\ln^2(1-y)}{y}dy=\int_0^1\frac{\ln^2x\ln(1-x)}{1-x}dx$$
We have the identity
$$\int_0^1\frac{x^{n}\ln^m(x)\ln(1-x)}{1-x}\ dx=\frac12\frac{\partial^m}{\partial n^m}\left(H_n^2+H_n^{(2)}\right)$$
Set $m=2$ then let $n$ approach $0$ we get
$$\int_0^1 \frac{\ln^2x\ln(1-x)}{1-x}dx=\frac12\frac{\partial^2}{\partial n^2}\left(H_n^2+H_n^{(2)}\right)_{n\to 0}\\=\frac12\left(4H_nH_n^{(3)}+2\left(H_n^{(2)}\right)^2+6H_n^{(4)}-4\zeta(2)H_n^{(2)}-4\zeta(3)H_n-\zeta(4)\right)_{n\to 0}\\=-\frac12\zeta(4)$$
