How to find partial derivatives of the Beta Function?

Question

I was reading the book (Almost) Impossible Integrals, Sums and Series. The author used a method involving taking partial derivatives of the Beta Function to solve some integrals. $$B(x,y)=\int_0^1u^{x-1}(1-u)^{y-1}du$$ Substituting $t=1-2u$, we get $$2^{x+y-1}B(x,y)=\int_{-1}^1(1-t)^{x-1}(1+t)^{y-1}dt$$ Differentiating, we get $$\dfrac{\partial^4}{\partial x^2\partial y^2}\left(2^{x+y-1}B(x,y)\right)=\int_{-1}^1(1-t)^{x-1}\log^2(1-t)(1+t)^{y-1}\log^2(1+t)dt\tag{1}$$ Taking the limits $x\to1$ and $y\to1$ on both sides, $$\lim_{x\to1, y\to1}\dfrac{\partial^4}{\partial x^2\partial y^2}\left(2^{x+y-1}B(x,y)\right)=\int_{-1}^1\log^2(1-t)\log^2(1+t)dt$$ $$\int_{-1}^1\log^2(1-t)\log^2(1+t)dt=\dfrac{1}{4}\lim_{x\to1, y\to1}\dfrac{\partial^4}{\partial x^2\partial y^2}\left(2^{x+y}B(x,y)\right)$$ Next, the author says to calculate the RHS using Mathematica, to obtain $$\int_{-1}^1\log^2(1-t)\log^2(1+t)dt=24−8\zeta(2)−8\zeta(3)−\zeta(4)+8\log(2)\zeta(2)-4\log^2(2)\zeta(2)+8\log(2)\zeta(3)−24\log(2)+12\log^2(2)-4\log^3(2)+\log^4(2)$$ How is the calculation of the step, the author tells us to perform by Mathematica, actually performed?

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I already posted this on Mathematics Stack Exchange, but didn't get any satisfactory response for 2 weeks, so I've crossposted it here.

Answer 1

The input you need to calculate second derivatives of Beta functions is $$\frac{\partial}{\partial x}B(x,y)=\bigl[\psi ^{(0)}(x)-\psi ^{(0)}(x+y)\bigr] B(x,y),$$ $$\frac{\partial^n}{\partial x^n}\psi ^{(0)}(x)=\psi ^{(n)}(x),$$ and then you need the limiting values at $(x,y)=(1,1)$, $$B(1,1)=1,\;\;\psi ^{(0)}(1)=-\gamma_{\rm{Euler}},\;\;\psi ^{(0)}(2)=1-\gamma_{\rm{Euler}},\;\;\psi ^{(1)}(1)=\pi^2/6,$$ $$\qquad\psi ^{(1)}(2)=\pi^2/6-1,\;\;\psi ^{(2)}(2)=2-2 \zeta (3),\;\;\psi ^{(3)}(2)=\pi ^4/15-6.$$ The rest is algebra.