Wednesday, 23 September 2015

integration - What's the value of int10frac12yln(y)ln2(1y),dy?



I came across this integral while doing a different problem:




1012yln(y)ln2(1y)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)=1210ya1(1y)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)=1210ya1(1y)blny(ln(1y))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)a0+,b0.

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 a0. 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+π26ψ1(1+a))


and calculate lima0+f(a). We get that
f(a)=B(a,1)(ψ0(a)ψ0(1+a))((γ+ψ0(1+a))2+π26ψ1(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)=1aγ+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 a0+,
f(a)1a2((π26a)2ψ2(1)aπ430a2)+1a(2π26aπ26ψ2(1)π415a)+O(a)=π4180+O(a)

as a0+. We conclude that
a,b,bB(a,1+b)a0+,b0=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(1y))2dy=1360π4.


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