Roots of Digamma

We use the identity $F(z) \; = \; \frac{\psi(z)}{\Gamma(z)} \; = \; -e^{-2\gamma z} \prod_{\alpha \in \Psi} \left(1 - \frac{z}{\alpha}\right)e^{\frac{z}{\alpha}} \;,$ which is valid for all $z \in \mathbb{C}$ , to deduce that $F(z)F(\omega z)F(\omega^2z) \; = \; -\prod_{\alpha \in \Psi}\left(1 - \frac{z^3}{\alpha^3}\right) \; = \; -1 + \left(\sum_{\alpha \in \Psi}\frac{1}{\alpha^3}\right)z^3 + \cdots$ where $\omega = e^{\frac{2\pi i}{3}}$ is a primitive cube root of unity. Standard results tell us that $F(z) \; = \; -1 - 2\gamma z + \tfrac14(\pi^2 - 6\gamma^2)z^2 - \tfrac13\big(2\gamma^3 - \gamma \pi^2 + 4\zeta(3)\big)z^3 + \cdots$ and hence it is easy to show that $F(z)F(\omega z)F(\omega^2 z) \; = \; -1 - \big(\gamma^3 + \tfrac12\gamma \pi^2 + 4\zeta(3)\big)z^3 + \cdots$ so that $\sum_{\alpha \in \Psi} \frac{1}{\alpha^3} \; = \; -4\zeta(3) - \gamma^3 - \tfrac12\gamma \pi^2$ making the answer $4 + 3 + 3 + 2 + 2 = \boxed{14}$ .

before diving into this solution give a read to this . from here we have two easy to derive identities $\sum a_k^{-2}= \gamma^2+\dfrac{\pi^2}{2} , \sum \dfrac{1}{a_k^2-a_k z}= \dfrac{-\dfrac{\psi '(z)}{\psi(z)}+\psi(z)+2\gamma}{z}$ subtracting the first from the second and dividing by z $\sum \dfrac{1}{a_k^3-a_k^2 z}=\dfrac{-\dfrac{\psi '(z)}{\psi(z)}+\psi(z)+2\gamma-z\left( \gamma^2+\dfrac{\pi^2}{2}\right)}{z^2}\\ \sum \dfrac{1}{a_k^3}=\lim_{z\to 0}\dfrac{-\dfrac{\psi '(z)}{\psi(z)}+\psi(z)+2\gamma-z\left( \gamma^2+\dfrac{\pi^2}{2}\right)}{z^2}$ we can use laurent series(easy to derive by taking deravatives and setting it in the taylor series) $\psi(x)=-\dfrac{1}{x}-\gamma+\zeta(2)x-\zeta(3)x^2+O(x^3)$ to solve the limit. now this is very tedious to do by hand so i used my favourite math engine to just do it all at once. we have or in our case $\sum \dfrac{1}{a_k^3}= -4\zeta(3)-\gamma^3 -\dfrac{\gamma \pi^2}{2}$ and the answer follows.