I have come across an infinite series involving Bessel functions where the summation is over the argument inside the Bessel function (rather than over the index of the Bessel function, which seems to be the case usually studied).

Specifically I am wondering whether a closed form is known for $$\sum_{n=1}^{\infty} \frac{J_k(nz)}{n^k}.$$ Here, $k$ is an integer and $J_k$ is the Bessel function of the first kind.

  • $\begingroup$ have you read any of the questions on the left? this here, for example, might be inspiring: math.stackexchange.com/questions/937058/… $\endgroup$ – tired Feb 18 '17 at 11:10
  • $\begingroup$ @tired It relies on a representation for $J_0$ which is considerably simpler than the similar ones $J_k(nz) = \frac{1}{\pi} \int_0^{\pi} \cos(kt - nz \sin(t)) \, \mathrm{d}t$, $k > 0.$ I don't see where to go with that. $\endgroup$ – user404188 Feb 18 '17 at 17:14
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    $\begingroup$ at least alternating sums can be handeld by the contour approach @RV suggested: $$ S_3=\sum_{n\geq1}\frac{(-1)^nJ_3(n)}{n^3}=-\frac{1}{96}\\ S_4=\sum_{n\geq1}\frac{(-1)^nJ_4(n)}{n^4}=-\frac{1}{768}\\ S_5=\sum_{n\geq1}\frac{(-1)^nJ_5(n)}{n^5}=-\frac{1}{7680}\\ S_6=\sum_{n\geq1}\frac{(-1)^nJ_6(n)}{n^6}=-\frac{1}{92160}\\ $$ $\endgroup$ – tired Feb 19 '17 at 18:58
  • $\begingroup$ in general $S_k=\sum_{n\geq 1} \frac{(-1)^nJ_k(n)}{n^{k}}=-\frac{1}{2^k (k-1)!}$ for $k\geq1$. it is also, for a parameter range so that all contour integrals vanish, easily possible to generalize this to arbitrary arguments of Bessel's function $\endgroup$ – tired Feb 19 '17 at 19:24
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    $\begingroup$ there is also some numerical evidence that the non-alternating sums also have closed form expressions in terms of rational numbers for example $$ \mathcal{S}_1=\sum_{n\geq1}\frac{J_1(n)}{n}\approx\frac{3}{4}\\ \mathcal{S}_2=\sum_{n\geq1}\frac{J_2(n)}{n^2}\approx\frac{13}{48}\\ \mathcal{S}_3=\sum_{n\geq1}\frac{J_3(n)}{n^3}\approx\frac{9}{160} $$ $\endgroup$ – tired Feb 20 '17 at 0:06

In summary, for $k \ge 1$ and $0 < a \le 2 \pi$, we have $$\begin{align}\sum_{n=1}^{\infty} \frac{J_{k}(an)}{n^{k}} &= \int_{0}^{\infty} \frac{J_{k}(ax)}{x^{k}} \, dx - \frac{1}{2} \, \operatorname{Res} \left[\frac{\pi J_{k}(az) \cot(\pi z)}{z^{k}},0 \right] \\ &= a^{k-1} \, \frac{2^{-k} \, \Gamma(\frac{1}{2})}{\Gamma(k+ \frac{1}{2})}- \frac{1}{2} \frac{a^{k}}{2^{k}k!} \tag{1}\\ &=\frac{a^{k-1}}{(2k-1)!!} - \frac{a^{k}}{2^{k+1} k!} . \tag{2}\end{align}$$

$(1)$ The pole at the origin is a simple pole regardless of the value of $k$.

$(2)$ http://mathworld.wolfram.com/DoubleFactorial.html (2)

(The formula should seemingly also hold for $k=0$ if $0 < a \ {\color{red}{<}} \ 2 \pi$, but the argument is a bit more subtle.)


I think I finally understand what's going on.

For a positive parameter $a$, $J_{k}(az)$ behaves like a constant times $\displaystyle \frac{e^{-iaz}}{\sqrt{az}}$ as $\text{Im}(z) \to + \infty$, and a constant times $ \displaystyle \frac{e^{iaz}}{\sqrt{az}}$ as $\text{Im}(z) \to - \infty$.

As $\text{Im}(z) \to +\infty$, $$e^{-iaz} \left(\cot (\pi z) +i \right) = e^{-iaz} e^{i \pi z} \csc(\pi z) \to 0 $$ if $a < 2 \pi$ (and remains bounded if $a = 2 \pi$).

Similarly, as $\text{Im}(z) \to - \infty$, $$e^{iaz} \left(\cot(\pi z) - i \right)= e^{i az} e^{- i \pi z} \csc(\pi z) \to 0$$ if $a< 2 \pi$ (and remains bounded if $a = 2 \pi$).


The issue is most likely replacing $\cot(\pi z)$ with $\mp i$ when letting $y \to \pm \infty$. (Thanks to Daniel Fischer for pointing this out.)

Since $J_{k}(az)$ doesn't remain bounded as $\text{Im}(z) \to \pm \infty$, this is not immediately justified by the dominated convergence theorem.

If would appear that we can only do this when $a \le 2 \pi$, but the reason is unclear.


Until I figure out why the positive parameter $a$ must be less than or equal to $2 \pi$, this answer is incomplete.

I will first use contour integration to confirm that $$\sum_{n=1}^{\infty} \frac{J_{1}(n)}{n} = \frac{3}{4}.$$

I'm going to use the fact that $$\int_{-\infty}^{\infty} \frac{J_{1}(x)}{x} \, dx = 2 \int_{0}^{\infty} \frac{J_{1}(x)}{x} \, dx = 2.$$

(One can use Ramanujan's master theorem to find the Mellin transform of $J_{\nu}(x)$. See Example 4.4 HERE.)

Let's integrate the function $$f(z) = \frac{\pi J_{1}(z) \cot(\pi z)}{z}$$ counterclockwise around a rectangular contour with vertices at $ \pm \left(N+ \frac{1}{2}\right) \pm iy$, where $N$ is a positive integer and $y >0$. (Recall that $J_{1}(z)$ is an entire function.)

Doing so, we get $$\small \int_{-N - 1/2}^{N+1/2} \frac{\pi J_{1}(t-iy)\cot\left(\pi(t-iy)\right)}{t-iy} \, dt + \int_{-y}^{y} \frac{\pi J_{1} \left(N+ \frac{1}{2}+it \right)\cot \left(\pi (N+ \frac{1}{2}+it) \right)}{N+ \frac{1}{2}+it} \, i \, dt$$

$$ \small + \int_{N + 1/2}^{-N-1/2} \frac{\pi J_{1}(t+iy)\cot\left(\pi(t+iy)\right)}{t+iy} \, dt + \int_{y}^{-y} \frac{\pi J_{1} \left(-N- \frac{1}{2}+it \right)\cot \left(\pi (-N-\frac{1}{2}+it) \right)}{-N- \frac{1}{2}+it} \, i \, dt$$

$$ \small = 2 \pi i \sum_{n=-N}^{N} \operatorname{Res}[f(z), n].$$

If we let $N$ go to infinity through the positive integers, the second and fourth integrals will vanish since, among other things, the magnitude of $J_{1}\left(N+ \frac{1}{2} + it\right)$ and $J_{1}\left(- N- \frac{1}{2}+it\right) $ will remain bounded if $t$ remains bounded. (This follows from the asymptotic behavior of the $J_{1}(z)$ for large $z$.)

So we're left with $$\int_{-\infty}^{\infty} \frac{\pi J_{1}(t-iy)\cot\left(\pi(t-iy)\right)}{t-iy} \, dt -\int_{-\infty}^{\infty} \frac{\pi J_{1}(t+iy)\cot\left(\pi(t+iy)\right)}{t+iy} \, dt $$

$$ = 2 \pi i \sum_{n=-\infty}^{\infty}\operatorname{Res}[f(z), n] = 2 \pi i \left(2 \sum_{n=1}^{\infty} \frac{J_{1}(n)}{n} + \operatorname{Res}[f(z), 0]\right)$$

$$ = 2 \pi i \left(2 \sum_{n=1}^{\infty} \frac{J_{1}(n)}{n} + \frac{1}{2} \right).$$

Now we can let $y \to \infty$ and use the fact that $\cot(z) \to \mp i$ uniformly as $\operatorname{Im}(z) \to \pm \infty$ to conclude that $$\sum_{n=1}^{\infty} \frac{J_{1}(n)}{n} = \frac{1}{4} \lim_{y \to \infty} \int_{-\infty}^{\infty} \frac{J_{1}(t-iy)}{t+iy} \, dt +\frac{1}{4} \lim_{y \to \infty} \int_{-\infty}^{\infty} \frac{J_{1}(t+iy)}{t+iy} \, dt - \frac{1}{4}. $$

The last step is to argue that $$\int_{-\infty}^{\infty} \frac{J_{1}(t+iy)}{t+iy} \, dt = \int_{-\infty}^{\infty} \frac{J_{1}(t-iy)}{t-iy} \, dt = \int_{-\infty}^{\infty} \frac{J_{1}(x)}{x} \, dx =2 $$ for all $y>0$.

One can show this by integrating $\frac{J_{1}(z)}{z}$ around rectangular contours in the upper and lower half-planes, and then letting the widths of the rectangles go to $\infty$.

A similar approach shows that $$\begin{align} \sum_{n=1}^{\infty} \frac{J_{2}(n)}{n^{2}} &= \frac{1}{2} \int_{-\infty}^{\infty} \frac{J_{2}(x)}{x^{2}} \, dx - \frac{1}{2} \, \operatorname{Res}\left[\frac{\pi J_{2}(z) \cot(\pi z)}{z^{2}},0 \right] \\ &= \frac{1}{2} \left(\frac{2}{3} \right) - \frac{1}{2} \left(\frac{1}{8} \right) \\ &= \frac{13}{48}. \end{align}$$

In general, for $a>0$, it would seem that $$\begin{align} \sum_{n=1}^{\infty} \frac{J_{1}(an)}{n} &= \frac{1}{2} \int_{-\infty}^{\infty} \frac{J_{1}(ax)}{x} \, dx - \frac{1}{2} \, \operatorname{Res} \left[\frac{\pi J_{1}(az) \cot(\pi z)}{z},0 \right] \\ &= \frac{1}{2} \int_{-\infty}^{\infty} \frac{J_{1}(u)}{u} \, du - \frac{1}{2} \left(\frac{a}{2} \right) \\ &=1 - \frac{a}{4}, \end{align}$$

$$ \begin{align} \sum_{n=1}^{\infty} \frac{J_{2}(an)}{n^{2}} &= \frac{1}{2} \int_{-\infty}^{\infty} \frac{J_{2}(ax)}{x^{2}}- \frac{1}{2} \, \operatorname{Res} \left[\frac{\pi J_{2}(az)\cot (\pi z)}{z^{2}} ,0\right] \\ &= \frac{a}{2} \int_{-\infty}^{\infty} \frac{J_{2}(u)}{u^{2}} \, du - \frac{1}{2} \left(\frac{a^{2}}{8} \right) \\ &= \frac{a}{3} - \frac{a^{2}}{16}, \end{align} $$

$$\begin{align} \sum_{n=1}^{\infty} \frac{J_{3}(an)}{n^{3}} &= \frac{1}{2} \int_{-\infty}^{\infty} \frac{J_{3}(ax)}{x^{3}}- \frac{1}{2} \, \operatorname{Res} \left[\frac{\pi J_{3}(az)\cot (\pi z)}{z^{3}} ,0\right] \\ &= \frac{a^{2}}{2} \int_{-\infty}^{\infty} \frac{J_{3}(u)}{u^{3}} \, du - \frac{1}{2} \left(\frac{a^{3}}{48} \right) \\ &= \frac{a^{2}}{15} - \frac{a^{3}}{96}, \end{align} $$ etc.

But numerical approximations suggest that we need the additional restriction $a \le 2 \pi$.

(The first series converges very slowly when $a = 2 \pi$, but it does appear to be converging to $1- \frac{\pi}{2}$.)

I don't immediately see why this particular restriction on $a$ is needed, but the formulas themselves suggest that a restriction of some sort is needed since they don't make sense for large values of $a$.

  • $\begingroup$ i was very close but couldn't finish...congrats.. very, very nice (++1)! $\endgroup$ – tired Mar 14 '17 at 18:01
  • $\begingroup$ Can we find a formula for general $k$ like i did for the alternating sums :)? $\endgroup$ – tired Mar 14 '17 at 18:11
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    $\begingroup$ I don't know. But the pattern should hold for $k=0$. That is, $$\sum_{n=1}^{\infty} J_{0}(an) = \frac{1}{2} \int_{-\infty}^{\infty}J_{0}(ax) \, dx - \frac{1}{2} \operatorname{Res} \left[\pi J_{0}(az) \cot(\pi z), 0 \right] = \frac{1}{a} - \frac{1}{2}. $$ But the argument is a bit more subtle. And the formula doesn't hold for $a= 2 \pi$. In fact, according to the asymptotic expansion, the above series doesn't even converge for $a = 2 \pi$. $\endgroup$ – Random Variable Mar 14 '17 at 18:26
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    $\begingroup$ @tired I think it should be $$\frac{a^{k-1}}{(2k-1)!!} - \frac{a^{k}}{2^{k+1}k!}$$ Because the pole is always simple, calculating the residue for general $k$ is actually pretty easy. I don't know why I was thinking it would be difficult. $\endgroup$ – Random Variable Mar 14 '17 at 19:53
  • $\begingroup$ @RV i think you are right i was a bit too fast...this completes the question of OP in a very nice way i think you should just add it to your answer $\endgroup$ – tired Mar 14 '17 at 20:16

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