How to calculate the value of the special integral

Question

I get $${\left. \frac{\partial ^2}{\partial n^2} \left( \frac{\partial ^2}{\partial m^2} B(m,n) \right) \right|_{m = \frac{1}{2},n = 0}} = \int_0^1 \frac{\ln^2 x \ln^2 (1 - x)}{\sqrt x (1 - x)} \, dx =\text{ ?}$$ but how to calculate the value of beta function derivative. where $B(m,n)$ is beta function.

Answer 1

how to calculate the value of beta function derivative ?

By using its well-known relation to the $\Gamma$ function, and the well-known properties of the digamma, trigamma, and polygamma functions.

Answer 2

For the integral: $${\left. \frac{\partial ^2}{\partial n^2} \left( \frac{\partial ^2}{\partial m^2} B(m,n) \right) \right|_{m = \frac{1}{2},n = 0}} = \int_0^1 \frac{\ln^2 x \ln^2 (1 - x)}{\sqrt x (1 - x)} \, dx $$

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Since \begin{align} \partial_{m} B(m,n) = B(m,n) \left[ \psi(m) - \psi(m+n) \right] \end{align} then \begin{align} \partial_{m}^{2} B(m,n) = B(m,n) \left[ \psi'(m) - \psi'(m+n) + \left( \psi(m) - \psi(m+n) \right)^{2} \right] \end{align} and \begin{align} \partial_{m}^{2} \left. B(m,n) \right|_{m=\frac{1}{2}} = B(1/2,n) \left[ \psi'(1/2) - \psi'(n+1/2) + \left( \psi(1/2) - \psi(n+1/2) \right)^{2} \right]. \end{align} Now \begin{align} \partial_{n}\partial_{m}^{2} \left. B(m,n) \right|_{m=\frac{1}{2}} &= B(1/2,n) \left[ - \psi''(n+1/2) -2 \psi'(n+1/2) \left( \psi(1/2) - \psi(n+1/2) \right) \right. \\ & \left. + (\psi(n) - \psi(n+1/2))(\psi'(1/2) - \psi'(n+1/2) + \left( \psi(1/2) - \psi(n+1/2) \right)^{2} )\right]. \end{align} Also \begin{align} \partial_{n}^{2} \partial_{m}^{2} \left. B(m,n) \right|_{m=\frac{1}{2}} &= B(1/2,n) \left[ - \psi'''(n+1/2) -2 \psi''(n+1/2) \left( \psi(1/2) - \psi(n+1/2) \right) \right. \\ & \left. + 2 \left( \psi'(n+1/2)\right)^{2} + (\psi(n) - \psi(n+1/2))( - \psi''(n+1/2) -2 \psi'(n+1/2) \cdot \right. \\ & \left. \left( \psi(1/2) - \psi(n+1/2) \right) )\right] \\ & + B(1/2,n) (\psi(n) - \psi(n+1/2)) \left[ - \psi''(n+1/2) -2 \psi'(n+1/2) \left( \psi(1/2) - \psi(n+1/2) \right) \right. \\ & \left. + (\psi(n) - \psi(n+1/2))(\psi'(1/2) - \psi'(n+1/2) + \left( \psi(1/2) - \psi(n+1/2) \right)^{2} )\right] \end{align} When $n=0$ this becomes \begin{align} \partial_{n}^{2} \partial_{m}^{2} \left. B(m,n) \right|_{m=\frac{1}{2}}^{n=0} &= B(1/2,0) \left[ - \psi'''(1/2) + 2 \left( \psi'(1/2)\right)^{2} + 2(\psi(0) - \psi(1/2))( - \psi''(1/2)) \right]. \end{align}

This is value trying to be sought. Notice that $\Gamma(0) = \infty$ and thus the result is invalid.

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Correct Process

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Consider the integral \begin{align} I = \int_{0}^{1} \frac{\ln^2 x \ln^2 (1 - x)}{(1-x)\sqrt{x}} \, dx \end{align} for which integration by parts, in the form, \begin{align} I &= \int_{0}^{1} \frac{\ln^2 x}{\sqrt{x}} \cdot \frac{\ln^2 (1 - x)}{(1-x)} \, dx \\ &= \left[ - \frac{1}{3} \ln^{3}(1-x) \cdot \frac{\ln^2 x}{\sqrt{x}} \right]_{0}^{1} + \frac{1}{3} \int_{0}^{1} \ln^{3}(-x) \, D\left( x^{-1/2} \, \ln^{2}(x) \right) \, dx \\ &= \frac{1}{3} \int_{0}^{1} \ln^{3}(-x) \, D\left( x^{-1/2} \, \ln^{2}(x) \right) \, dx \\ &= \frac{2}{3} \int_{0}^{1} x^{-3/2} \ln(x) \, \ln^{3}(1-x) \, dx - \frac{1}{6} \int_{0}^{1} x^{-3/2} \ln^{2}(x) \, \ln^{3}(1-x) \, dx \\ &= \frac{2}{3} \partial_{x} \partial_{y}^{3} \left. B(x,y) \right|_{x=-1/2}^{y=1} - \frac{1}{6} \partial_{x}^{2} \partial_{y}^{3} \left. B(x,y) \right|_{x=-1/2}^{y=1}. \end{align}