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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-

  1. How exactly is the partial differentiation done in $(1)$?
  2. How is the calculation of the step, the author tells us to perform by Mathematica, actually performed?

Crossposted on MathOverflow.

Souparna
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1 Answers1

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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.

ConMan
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