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In this post M.N.C.E gave the equality below $$\frac{\partial ^{5}}{\partial a^{3}\partial b^{2}}\mathrm{B}\left ( 1,0^{+} \right )=\left [ \frac{1}{b}+O\left ( 1 \right ) \right ]\left [ \left ( 12\zeta ^{2}\left ( 3 \right )-\frac{23\pi ^{6}}{1260} \right )b+O\left ( b^{2} \right ) \right ]_{b=0}$$ where $\mathrm B(a,b)$ is the Beta function.

But how to get this equality? Can the same method be used to evaluate $$\frac{\partial ^{6}}{\partial a^{3}\partial b^{3}}\mathrm{B}\left ( 1,0^{+} \right )~~,~~\frac{\partial ^{6}}{\partial a^{4}\partial b^{2}}\mathrm{B}\left ( 1,0^{+} \right )?$$

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Renascence_5
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Hint: The ingredients are $$B\left(a,b\right)=\frac{\Gamma\left(a\right)\Gamma\left(b\right)}{\Gamma\left(a+b\right)} $$the link between Gamma and Polygamma functions $$\Gamma'\left(z\right)=\Gamma\left(z\right)\psi^{\left(0\right)}\left(z\right),\,\Gamma''\left(z\right)=\Gamma\left(z\right)\left(\psi^{\left(0\right)}\left(z\right)^{2}+\psi^{\left(1\right)}\left(z\right)\right),\dots $$ the relation between Polygamma functions and Riemann (and Hurwitz) Zeta function $$\psi^{\left(n\right)}\left(z\right)=\left(-1\right)^{n+1}n!\zeta\left(n+1,z\right) $$ and the asymptotic $$\Gamma\left(x\right)=\frac{1}{x}+O\left(1\right) $$ as $x\rightarrow0$.

Marco Cantarini
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  • sorry, I'm not quite understand, could you please show me more details, for example, how to get the first one in my question? – Renascence_5 May 07 '16 at 01:21