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 lima→0+f′(a). We get that
f′(a)=B(a,1)(ψ0(a)−ψ0(1+a))((γ+ψ0(1+a))2+π26−ψ1(1+a))+B(a,1)(2(γ+ψ0(1+a))ψ1(1+a)−ψ2(1+a))
Next, we use the (non-obvious) expansions around a=0
B(a,1)=1a+O(1)ψ0(a)=−1a−γ+O(a)ψ0(1+a)=−γ+π26a+O(a2)ψ1(1+a)=π26+ψ2(1)a+π430a2+O(a3)ψ2(1+a)=ψ2(1)+π415a+O(a2).
to find that, as a→0+,
f′(a)≈−1a2((π26a)2−ψ2(1)a−π430a2)+1a(2π26aπ26−ψ2(1)−π415a)+O(a)=−π4180+O(a)
as a→0+. We conclude that
∂a,b,bB(a,1+b)∣a→0+,b→0=−π4180.
Finally, dividing by 2 (remember, we had a one-half in front of the beta function in the beginning), we get that
∫1012ylny(ln(1−y))2dy=−1360π4.
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