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 $$\int_{-1}^1\log^2(1-t)\log^2(1+t)dt=24−8\zeta(2)−8\zeta(3)−\zeta(4)+8\log(2)\zeta(2)-4\log^2(2)\zeta(2)+8\log(2)\zeta(3)−24\log(2)+12\log^2(2)-4\log^3(2)+\log^4(2)$$ I have basically two queries-
- How exactly is the partial differentiation done in $(1)$?
- How is the calculation of the step, the author tells us to perform by Mathematica, actually performed?
Crossposted on MathOverflow.
For the partial differentiation, the author is performing "differentiation under the integral sign", which involves application of the Leibniz integral rule. For example, taking a single derivative with respect to $x$ would look something like:
$$\begin{eqnarray} \frac{\partial}{\partial x} 2^{x+y-1} B(x, y) & = & \frac{\partial}{\partial x} \int_{-1}^1 (1-t)^{x-1} (1+t)^{y-1} dx \\ & = & \int_{-1}^1 \frac{\partial}{\partial x} \left((1-t)^{x-1} (1+t)^{y-1} \right) dx \\ & = & \int_{-1}^1 (1-t)^{x-1} \ln(1-t) (1+t)^{y-1} dx \end{eqnarray}$$
Using the standard derivative $\frac{d}{dx} a^x = a^x \ln a$.
As for how the final integral is calculated, the glib answer is "there's a reason they say to use Mathematica". And indeed the actual calculation is slightly beyond my skills, but I suspect there's a little trickery involving converting the $\ln^2$ terms into power series, integrating them and then turning the resulting sums back into the values shown.