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Wolfram Mathworld gives the following values for the beta function derivative.

$$\beta'(-1) = \frac{2K}{\pi},\quad \beta'(0) = \ln \left[\frac{\Gamma^{2}(\frac{1}{4})}{2\pi\sqrt{2}} \right],\quad \beta'(1) = \frac{\pi}{4}\left(\gamma +2\ln 2+3\ln\pi-4\ln \left(\Gamma \left(\frac{1}{4}\right) \right)\right)$$

I can see how $\beta'(1)$ could be derived from $\beta'(0)$ using the functional equation (or vice versa) but how was one of them found originally. The same for $\beta'(-1)$ and $\beta'(2)$. Is there any other known values of the derivative? Also can I have links to any papers on the subject, I cannot seem to find any?

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$\beta'(-1)$ can actually be found directly from the functional equation. We have $$ \beta'(-1) = - \frac{\mathrm{d}}{\mathrm{d}s} \left[\left(\frac{2}{\pi}\right)^s \sin \left(\frac{\pi}{2} s\right) \Gamma(s) \beta(s)\right] ~ \Bigg\rvert_{s=2} \, .$$ Now note that the only non-vanishing term on the right-hand side is the one containing the derivative of the sine. Therefore $$\beta'(-1) = \left(\frac{2}{\pi}\right)^2 \frac{\pi}{2} [-\cos(\pi)] \Gamma(2) \beta(2) = \frac{2 \mathrm{K}}{\pi} \, .$$ However, this also means that $\beta'(2)$ cannot be calculated from $\beta'(-1)$.


$\beta'(1)$ is not as easy to find. We can start from the integral \begin{align} \frac{1}{2} \int \limits_0^\infty \frac{-\ln(x)}{\cosh(x)} \, \mathrm{d} x &= - \frac{1}{2} \frac{\mathrm{d}}{\mathrm{d}s} \int \limits_0^\infty \frac{x^{s-1}}{\cosh(x)} \, \mathrm{d} x ~ \Bigg\rvert_{s=1} = - \frac{1}{2} \frac{\mathrm{d}}{\mathrm{d}s} [2 \Gamma(s) \beta(s)] ~ \Bigg\rvert_{s=1} \\ &= \frac{\pi \gamma}{4} - \beta'(1) \, . \end{align} Since the left-hand side is a representation of Vardi's integral $$ \int \limits_0^1 \frac{-\ln(-\ln(x))}{1+x^2} \, \mathrm{d} x = \frac{\pi}{4} \ln\left(\frac{\operatorname{\Gamma}^4\left(\frac{1}{4}\right)}{4\pi^3}\right) \, ,$$ we obtain the desired result, from which the value of $\beta'(0)$ follows by the functional equation.

Various ways to evaluate the integral are discussed in this paper by Blagouchine. By the way, the author convincingly argues that the integral should actually be called Malmsten's integral, since Malmsten first computed it in 1842, which is in line with the claim from the Wikipedia page.


As an alternative route, here is a brief sketch of how I have calculated $\beta'(1)$ :

Plugging $z=\frac{1}{4}$ into Binet's first log-gamma formula, we obtain (after some algebra and integration by parts/substitutions) \begin{align} \ln\left(\frac{\operatorname{\Gamma}^4\left(\frac{1}{4}\right)}{4\pi^2}\right) &= 2 \ln(2) - 1 + \int \limits_0^\infty \frac{\tanh(x)}{x} \mathrm{e}^{-x} \, \mathrm{d} x \\ &\phantom{=} + \int \limits_0^\infty \left[\frac{\coth(x)}{x} -\frac{1}{\sinh^2 (x)}\right] \mathrm{e}^{-x} \, \mathrm{d} x - \int \limits_0^\infty \left[\coth(x)-\frac{1}{x}\right] \mathrm{e}^{-x} \, \mathrm{d} x \\ &\equiv 2\ln(2) - 1 + I_1 + I_2 - I_3 \, . \end{align}

We can then compute \begin{align} I_1 &= \gamma + \ln \left(\frac{\pi}{2}\right) - \frac{4}{\pi} \beta'(1) \, ,\\ I_2 &= \gamma \, ,\\ I_3 &= \gamma + \ln(2) - 1 \, \end{align} and solve for $\beta'(1)$ .

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  • $\begingroup$ Are any other values known? $\endgroup$ – Joshua Farrell Aug 23 '18 at 9:02
  • $\begingroup$ @JoshuaFarrell Other values seem to have expressions in terms of derivatives of other special functions only. Mathematica gives $$ \beta'(2) = \frac{\partial_1 \zeta\left(2,\frac{1}{4}\right)-\partial_1 \zeta\left(2,\frac{3}{4}\right)}{16} - 2 \mathrm{K} \ln(2)$$ for example, which can be derived from the series definition of $\beta$ . Apparently there is no nicer form. $\endgroup$ – ComplexYetTrivial Aug 23 '18 at 10:03

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