3
$\begingroup$

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

$$\sum_{n=1}^{\infty}\left(\frac{1}{n}-\frac{1}{\sqrt{(n+a)^2+b^2}}\right).$$

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

$$\sum_{n=1}^{\infty}\left(\frac{1}{n}-\frac{1}{n+a}\right),$$

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...

$\endgroup$
6
  • $\begingroup$ can you do the case $a=0$? $\endgroup$ – mathworker21 Jun 23 '19 at 5:57
  • $\begingroup$ No I can't, and neither can Mathematica $\endgroup$ – WillG Jun 23 '19 at 5:58
  • $\begingroup$ Do you want an approximation for small or large value of parameters? $\endgroup$ – Nosrati Jun 23 '19 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 '19 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... $\endgroup$ – Mariusz Iwaniuk Jun 23 '19 at 11:31
1
$\begingroup$

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 $$\sum_{n=1}^{\infty}\left(\frac{1}{n}-\frac{1}{\sqrt{(n+a)^2+b^2}}\right)= \color{blue}{-\sum_{k=1}^\infty \left(a^2+b^2\right)^{\frac{k}{2}}\zeta(k)P_k\left(\frac{-a}{\sqrt{a^2+b^2}}\right)}$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.