*Appendix B.1. Demonstration*

We use the following notation *α* = ∑*<sup>p</sup> <sup>i</sup>*=<sup>1</sup> *mi* to alleviate the writing of equations. Knowing that *<sup>∂</sup> <sup>∂</sup><sup>c</sup>* (*c*)*<sup>k</sup>* = (*c*)*<sup>k</sup> ψ*(*c* + *k*) − *ψ*(*c*) , *<sup>ψ</sup>*(*<sup>c</sup>* <sup>+</sup> *<sup>k</sup>*) <sup>−</sup> *<sup>ψ</sup>*(*c*) = <sup>∑</sup>*k*−<sup>1</sup> =0 1 *<sup>c</sup>*+ and (*c*)*<sup>k</sup>* = ∏*k*−<sup>1</sup> *<sup>i</sup>*=<sup>0</sup> (*c* + *i*) we can state that

$$\frac{\partial}{\partial a} \left\{ \frac{(a)\_a}{(a + \frac{1+p}{2})\_a} \right\} = \frac{(a)\_a [\psi(a+a) - \psi(a) - \psi(a + \frac{1+p}{2} + a) + \psi(a + \frac{1+p}{2})]}{(a + \frac{1+p}{2})\_a}$$

$$= \frac{\prod\_{k=0}^{a-1} (a+k) \left( \sum\_{k=0}^{a-1} \frac{1}{a+k} - \frac{1}{a + \frac{1+p}{2} + k} \right)}{(a + \frac{1+p}{2})\_a}.\tag{A10}$$

Using the fact that

$$\prod\_{k=0}^{a-1} (a+k) \sum\_{k=0}^{a-1} \frac{1}{a+k} = \prod\_{k=1}^{a-1} (a+k) + \prod\_{k=0, k \neq 1}^{a-1} (a+k) + \dots + \prod\_{k=0}^{a-2} (a+k) \tag{A11}$$

we can state that

$$\frac{\partial}{\partial a} \left\{ \frac{(a)\_a}{(a + \frac{1+p}{2})\_a} \right\} \Big|\_{a=0} = \frac{(a-1)!}{(\frac{1+p}{2})\_a} = \frac{(1)\_a}{(\frac{1+p}{2})\_a} \frac{1}{a}.\tag{A12}$$

*Appendix B.2. Demonstration*

$$\frac{\partial}{\partial a} \left\{ \frac{(a)\_a (a + \frac{1}{2})\_{m\_p}}{(a + \frac{1 + p}{2})\_a} \right\} = \frac{(a + \frac{1}{2})\_{m\_p} (a)\_a [\Psi(a+a) - \Psi(a) + \Psi(a + \frac{1}{2} + m\_p) - \Psi(a + \frac{1}{2})]}{(a + \frac{1 + p}{2})\_a} - 1$$

$$\frac{(a)\_a (a + \frac{1}{2})\_{m\_p} [\Psi(a + \frac{1 + p}{2} + a) - \Psi(a + \frac{1 + p}{2})]}{(a + \frac{1 + p}{2})\_a} \tag{A13}$$

$$= \frac{(a + \frac{1}{2})\_{m\_p} \prod\_{k=0}^{n-1} (a + k) \left[ \sum\_{k=0}^{n-1} \frac{1}{a + k} - \frac{1}{a + \frac{1 + p}{2} + k} + \sum\_{k=0}^{m\_p - 1} \frac{1}{a + \frac{1}{2} + k} \right]}{(a + \frac{1 + p}{2})\_a} . \tag{A14}$$

By developing the previous expression we can state that

$$\frac{\partial}{\partial a} \left\{ \frac{(a)\_a (a + \frac{1}{2})\_{m\_p}}{(a + \frac{1 + p}{2})\_a} \right\} \Big|\_{a = 0} = \frac{(\frac{1}{2})\_{m\_p} (a - 1)!}{(\frac{1 + p}{2})\_a} = \frac{(\frac{1}{2})\_{m\_p} (1)\_a}{(\frac{1 + p}{2})\_a} \frac{1}{a}. \tag{A15}$$

*Appendix B.3. Demonstration*

$$\frac{\partial}{\partial a} \left\{ \frac{(a + \frac{1}{2})\_{m\_p}}{(a + \frac{1 + p}{2})\_a} \right\} = \frac{(a + \frac{1}{2})\_{m\_p}}{(a + \frac{1 + p}{2})\_a} \left[ \sum\_{k=0}^{m\_p - 1} \frac{1}{a + \frac{1}{2} + k} - \sum\_{k=0}^{a - 1} \frac{1}{a + \frac{1 + p}{2} + k} \right]. \tag{A16}$$

As a consequence,

$$\frac{\partial}{\partial a} \left\{ \frac{(a + \frac{1}{2})\_{m\_p}}{(a + \frac{1 + p}{2})\_a} \right\} \Big|\_{a = 0} = \frac{(\frac{1}{2})\_{m\_p}}{(\frac{1 + p}{2})\_a} \left[ \sum\_{k = 0}^{m\_p - 1} \frac{1}{\frac{1}{2} + k} - \sum\_{k = 0}^{n - 1} \frac{1}{\frac{1 + p}{2} + k} \right]. \tag{A17}$$

*Appendix B.4. Demonstration*

$$\frac{\partial}{\partial a} \left\{ \frac{1}{(a + \frac{1+p}{2})\_a} \right\} = -\frac{\psi(a + \frac{1+p}{2} + a) - \psi(a + \frac{1+p}{2})}{(a + \frac{1+p}{2})\_a} \tag{A18}$$

$$=\frac{-1}{(a+\frac{1+p}{2})\_\alpha} \sum\_{k=0}^{\alpha-1} \frac{1}{a+\frac{1+p}{2}+k}.\tag{A19}$$

Finally,

$$\frac{\partial}{\partial a} \left\{ \frac{1}{(a + \frac{1+p}{2})\_a} \right\} \Big|\_{a=0} = \frac{-1}{(\frac{1+p}{2})\_a} \sum\_{k=0}^{a-1} \frac{1}{\frac{1+p}{2} + k}. \tag{A20}$$
