I am doing a long physics calculation and have arrived at the following sum:


Does this have an analytic closed-form solution? An added complication is that in my problem, $a$ can take complex values, and the square root should be treated as having a branch cut along $(-\infty,0]$ (and I am only interested in the real part).

Note that a simpler version of my problem is the case where $b=0$, for which the above becomes


which can be written in closed form as $\gamma+\psi(a+1)$, where $\gamma$ is the Euler-Mascheroni constant and $\psi$ is the digamma function (this comes from a standard series representation of the digamma function). So my problem might be viewed as a generalized form the digamma series representation.

Mathematica can't solve this, and I haven't seen integrands with square roots like this in any standard formulas I've looked at. Maybe there's no closed-form solution in terms of standard functions? This wouldn't surprise me since the square root seems to make things pretty messy. But on the other hand there are a lot of special functions and tricks I don't know about...

  • $\begingroup$ can you do the case $a=0$? $\endgroup$ Jun 23, 2019 at 5:57
  • $\begingroup$ No I can't, and neither can Mathematica $\endgroup$
    – WillG
    Jun 23, 2019 at 5:58
  • $\begingroup$ Do you want an approximation for small or large value of parameters? $\endgroup$
    – Nosrati
    Jun 23, 2019 at 7:58
  • $\begingroup$ I think that that does not have finite closed-form expression in terms of very large class of special functions. Why do you want closed-form expression? $\endgroup$
    – Grešnik
    Jun 23, 2019 at 9:18
  • 2
    $\begingroup$ $$\sum _{n=1}^{\infty } \left(\frac{1}{n}-\frac{1}{\sqrt{(n+a)^2+b^2}}\right)=\int_0^{\infty } \left(-\frac{1}{1-e^x}+\frac{e^{-a x} J_0(b x)}{1-e^x}\right) \, dx$$ no hope... $\endgroup$ Jun 23, 2019 at 11:31

1 Answer 1


I don't know this might be useful or not, anyway by generating function for Legendre polynomials $$\dfrac{1}{\sqrt{(n+a)^2+b^2}}=\dfrac{1}{n}\dfrac{1}{\sqrt{1+2\frac{a}{n}+\left(\frac{\sqrt{(a^2+b^2}}{n}\right)^2}}=\dfrac{1}{n}\sum_{k=0}^\infty \left(\frac{\sqrt{a^2+b^2}}{n}\right)^kP_k\left(\frac{-a}{\sqrt{a^2+b^2}}\right)$$ then $$\sum_{n=1}^{\infty}\left(\frac{1}{n}-\frac{1}{\sqrt{(n+a)^2+b^2}}\right)= \color{blue}{-\sum_{k=1}^\infty \left(a^2+b^2\right)^{\frac{k}{2}}\zeta(k)P_k\left(\frac{-a}{\sqrt{a^2+b^2}}\right)}$$


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