integration - What's the value of $int_0^1frac{1}{2y} ln(y) ln^2(1-y) , dy$?

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.

Answer

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.
$$