I've found the below equivalence.

For $a,b,x\in\mathbb{C}$, provided that there are no singularities on the right-hand side:

\begin{multline}\sum _{k=2}^{\infty}\sum _{j=1}^{\infty}\frac{x^k}{(a j+b)^k}=-\frac{x^2}{2b(b-x)}+\frac{\pi x}{2a}\csc{\frac{b\pi }{a}}\csc{\frac{\pi (b-x)}{a}}\sin{\frac{\pi x}{a}}\\-\frac{x \pi}{a}\int _0^1\left(\csc{\frac{2 \pi (b-x)}{a}}\sin{\frac{2 \pi (b-x)u}{a}}-\csc{\frac{2 \pi b}{a}}\sin{\frac{2 \pi b u}{a}}\right)\cot{\pi u}\,du\end{multline}

Even when the left-hand side may not converge, the right-hand side may still converge.

I've notice that $b=1/2$ causes the integral (the imaginary part for real $a,b$) to vanish for $x=1$ and $a=-i$ (though the real part doesn't vanish).

The integral will always vanish for $x=2b$, so it's a matter of finding a triple $(a,b,2b)$ that zero out the other part.

What are the zeros of this equation?


For $|x| < |i\Im(b)+1+ \max(\Re(b),0)|$ and $b \not \in \mathbb{Z}_{\le 1}$ then

$$\sum_{k=2}^{\infty}\sum _{j=1}^{\infty}\frac{x^k}{( j+b)^k}=\sum _{j=1}^{\infty}\sum_{k=2}^{\infty}\frac{x^k}{( j+b)^k} = \sum_{j=1}^\infty \frac{1}{1-\frac{x}{j+b}}-1-\frac{x}{j+b}$$ $$= \sum_{j=1}^\infty \frac{x}{j+b-x}-\frac{x}{j+b}=x (\frac{\Gamma'(b+1)}{\Gamma(b+1)}-\frac{\Gamma'(b-x+1)}{\Gamma(b-x+1)})$$

  • $\begingroup$ I'm not a fan of that function. Anyway, what about the zeros? $\endgroup$ – user630964 Feb 20 at 0:49
  • $\begingroup$ No worries, I'm calculating the limits for the cases that have singularities. I don't think it will be as restrictive as your formula. $\endgroup$ – user630964 Feb 20 at 5:16
  • $\begingroup$ @JRS. I don't understand what you are asking. $\psi = \Gamma'/\Gamma$ is a well-known non-elementary function. Are you asking about the zeros of $z \mapsto \psi(z)-c$ or about the zeros of $z \mapsto \psi(z+d)-\psi(z)$. The approximate location of both of them can be found from the argument principle and the asymptotic of $\psi(z)$ (the one giving the Stirling approximation for $\Gamma(z)$). And why do you care about the zeros. $\endgroup$ – reuns Feb 20 at 16:29
  • $\begingroup$ I'm asking what are the zeros of the above equation, the equtaion posted on the OP? Oh, yes, I know the psi function very well, I don't like it, too artificial for my taste. $\endgroup$ – user630964 Feb 20 at 22:18
  • $\begingroup$ They are where they are :) What I said is that for any fixed $c,d$ you can find the approximate location of the zeros of $z \mapsto \psi(z)-c$ and $z \mapsto \psi(z+d)-\psi(z)$ using the argument principle and the asymptotic of $\psi(z)$. Then with more work you can obtain an approximate location of the zeros of $\psi(z)-\psi(z+w)$. You can also search on the internet as some people probably already made a huge part of the work, given how common is $\psi(z)$. $\endgroup$ – reuns Feb 20 at 22:25

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