Closed form for an infinite series of Bessel functions

Question

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.

Answer 1

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.)

UPDATE 3:

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 $a = 2 \pi$).

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UPDATE 2:

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.

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UPDATE 1:

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

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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 $\infty$.

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

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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 $a$.