I came across this integral while doing a different problem:
∫1012yln(y)ln2(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)=12∫10ya−1(1−y)bdy=12B(a,1+b),
where B denotes the beta function. Note that this integral is singular at a=0 and b=−1. Since ∂aya=yalny we are
led to calculate
∂a,b,bI(a,b)=12∫10ya−1(1−y)blny(ln(1−y))2dy
as a and b tend to 0. We will below inser the "non-dangerous" point b=0. In other words, we want to calculate
∂a,b,bB(a,1+b)∣a→0+,b→0.
When differentiating the beta function, polygammas appear. Indeed,
∂bB(a,1+b)=B(a,1+b)(ψ0(1+b)−ψ0(1+a+b))∂b,bB(a,1+b)=B(a,1+b)((ψ0(1+b)−ψ0(1+a+b))2+ψ1(1+b)−ψ1(1+a+b)).
Next, we can actually insert b=0 before we differentiate with respect to
a and take the limit a→0. We should differentiate the function (here we have used the facts that ψ0(1)=−γ (Euler's constant) and that ψ1(1)=π2/6)
f(a)=B(a,1)((γ+ψ0(1+a))2+π26−ψ1(1+a))
and calculate lim. 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.
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