Why does this nominally divergent limit of an infinite sum of bessel functions converge

Question

The setup isn't important, but in case you're curious, this is a physically relevant thing I'm trying to calculate that should have a finite value. This equation gives the electric field of a small patch of voltage $V$ of length $l$ on the surface of a grounded conducting cylinder of radius $R$ in the center of the cylinder: $$ |\mathbf{E}|=\frac{Vl^2}{2\pi R^3}\lim_{\epsilon\rightarrow 0^+}\sum_{m=1}^\infty \frac{\alpha_{1m}\exp\left(-\alpha_{1m}\epsilon\right)}{J_2(\alpha_{1m})}\approx 1.77578 \frac{Vl^2}{2\pi R^3} $$ $J_1$ is the first Bessel function of the first kind, and $\alpha_{1m}$ is it's $m$'th zero (not including the one at $x=0$). The $\approx$ comes from me evaluating this numerically. I understand that I cannot move the limit into the sum because if I do, the sum no longer converges. What I don't understand is why it even has a well defined limit with the limit outside of the sum.

For large $m$, the zeros go toward $\alpha_{1m}\rightarrow (m+1/4)\pi$. For large inputs $J_2(x)$ asymptotes to: $$ J_2(x)\rightarrow -\sqrt{\frac{2}{\pi}}\sin\left(x+\frac{\pi}{4}\right)\sqrt\frac{1}{x} $$ So it seems to me that the sum for large $m$ turns into: $$ \sum_{m=N}^\infty m^{3/2} (-1)^m\exp(-m\pi\epsilon) $$ And even if I add even and odd terms together, I should get $$ \sum_{m'=N/2}^\infty m'^{1/2} \exp(-2m'\pi\epsilon) $$ So I think as $\epsilon$ goes to zero, my sum will go to infinity. Certainly the above sum diverges, so I must have done something too handwavy with limits here. Strangely though, as $\epsilon$ goes to zero, the sum goes to a finite value.

Answer 1

New Answer. In this answer, we derive an integral representation of the limit. Using the special case of DLMF 10.6.2,

$$ J_1'(z) = -J_2(z) + \frac{J_1(z)}{z}, $$

we may write the sum as

\begin{align*} \sum_{m=1}^{\infty} \frac{\alpha_{1,m}e^{-\epsilon \alpha_{1,m}}}{J_2(\alpha_{1,m})} &= -\sum_{m=1}^{\infty} \frac{\alpha_{1,m}e^{-\epsilon \alpha_{1,m}}}{J_1'(\alpha_{1,m})} = - \frac{1}{2\pi i} \int_{\mathcal{C}} \frac{z e^{-\epsilon z}}{J_1(z)} \, \mathrm{d}z. \end{align*}

Here, $\mathcal{C}$ denotes the Hankel contor traversed in the positive sense. Now by noting that $J_1(z)$ grows exponentially as $\operatorname{Im}(z) \to \pm\infty$ in the region $\left|\arg(z)\right| \leq \frac{\pi}{2}$, we can deform the contour $\mathcal{C}$ to the imaginary axis traversed from top to bottom. Hence,

\begin{align*} \sum_{m=1}^{\infty} \frac{\alpha_{1,m}e^{-\epsilon \alpha_{1,m}}}{J_2(\alpha_{1,m})} &= - \frac{1}{2\pi i} \int_{i\infty}^{-i\infty} \frac{z e^{-\epsilon z}}{J_1(z)} \, \mathrm{d}z \\ &= \frac{1}{2\pi} \int_{-\infty}^{\infty} \frac{xe^{-i\epsilon x}}{I_1(x)} \, \mathrm{d}x \\ &\to \bbox[color:blue; padding:8px; border:1px dotted navy;]{ \frac{1}{2\pi} \int_{-\infty}^{\infty} \frac{x}{I_1(x)} \, \mathrm{d}x } \quad \text{as $\epsilon \to 0^+$}, \end{align*}

where we substituted $z = ix$ in the second step and $I_1(z)$ is the modified Bessel function of the first kind and order $1$. Thanks to the exponential growth of $I_1(x)$ as $x \to \pm\infty$, this integral can be efficiently approximated, yielding the numerical value of

$$ \approx 1.7757859970218740923 $$

as observed by OP.

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Old Answer. In fact, turns out that

$$ \lim_{\varepsilon \to 0^+} \sum_{n=1}^{\infty} \frac{(-1)^n}{n^s} e^{-\varepsilon n} = -(1 - 2^{1-s}) \zeta(s) \tag{1} $$

for any $s \in \mathbb{C}$. In particular, when $s = -\frac{3}{2}$, we get

$$ \lim_{\varepsilon \to 0^+} \sum_{n=1}^{\infty} (-1)^n n^{3/2} e^{-\varepsilon n} = -(1 - 2^{5/2}) \zeta(-3/2) \approx -0.118681. $$

Although I am no expert of Bessel functions, I guess an analogous behavior is occurring in OP's sum limit as well.

Finally, let me demonstrate why we expect the limit in $\text{(1)}$ to have a finite value. To this end, let us consider, for example, the sum $ \sum_{n=1}^{\infty} (-1)^n n^2 $. Although this does not converge,

$$ \sum_{n=1}^{\infty} (-1)^n n^2 e^{-\varepsilon n} = - \frac{(1 - e^{-\varepsilon})e^{-\varepsilon}}{(1+e^{-\varepsilon})^3}, \qquad \operatorname{Re}(\varepsilon) > 0 $$

and hence letting $\varepsilon \to 0^+$ makes this converge to a finite value, which is $0$ in this case. This shows that this "exponential regularization" (or more commonly, Abel summation) is capable of facilitating cancellations between $(-1)^n n^2$'s. A similar thing occurs when $n^2$ is replaced by any power of $n$.

Answer 2

The following is a supplement to the second part of Sangchul Lee's answer that is labeled "Old Answer."

Let $\eta(s)$ be the Dirichlet eta function, which can be defined in terms of the Riemann zeta function as $ \eta(s)= \left(1-2^{1-s} \right) \zeta(s).$

To prove that $$\lim_{\varepsilon \to 0^+} \sum_{n=1}^{\infty} (-1)^n n^{3/2} e^{-\varepsilon n} = \operatorname{Li}_{-3/2}(-1) = - \eta(-3/2), $$ we'll use the dominated convergence theorem to show that the polylogarithm function $\operatorname{Li}_{-3/2}(x)$, as a function of the real variable $x$, is a continuous function on $[-1, 0]$. (We can show more, but that's all that's needed here.)

An integral representation for $\operatorname{Li}_{s}(x)$ that is valid for $\Re(s) >-2$ is $$\operatorname{Li}_{s}(x) = \frac{x}{\Gamma(s+2)} \int_{0}^{\infty} \frac{t^{s+1} e^{t}(e^{t}+x)}{(e^{t}-x)^{3}} \, \mathrm dt, \quad \Re(s)>-2, x \notin[1, \infty). \tag{$\spadesuit$}$$

This integral representation can be obtained by performing two integrations by parts on the Bose–Einstein integral $$\operatorname{Li}_{s}(x) = \frac{x}{\Gamma(s)} \int_{0}^{\infty} \frac{t^{s-1}}{e^{t}-x} \, \mathrm dt, \quad \Re(s)>0, x \notin[1, \infty).$$

At $x=-1$, $(\spadesuit)$ is an integral representation for $-\eta(s)$ for $\Re(s) >-2$.

Plugging $s= - \frac{3}{2}$ into $(\spadesuit)$ gives $$\operatorname{Li}_{-3/2}(x) = \frac{x}{\sqrt{\pi}} \int_{0}^{\infty} \frac{t^{-1/2}e^{t}(e^{t}+x)}{(e^{t}-x)^{3}} \, \mathrm dt. $$

But for $x \in [-1,0] $, $\frac{x}{\sqrt{\pi}}\frac{t^{-1/2}e^{t}(e^{t}+x)}{(e^{t}-x)^{3}} $ is dominated by the integrable function $$\frac{t^{-1/2}e^{-t}}{\sqrt{\pi}},$$ showing that $\operatorname{Li}_{-3/2}(x)$ is a continuous function on $[-1,0]$.

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More generally, $\spadesuit$ can be used to show that $\operatorname{Li}_{s}(x)$ is a continuous function on $[-1,0]$ for any fixed value of $s$ such that $\Re(s) > -2$, showing that $$\lim_{\varepsilon \to 0^+} \sum_{n=1}^{\infty} (-1)^n \frac{e^{-\varepsilon n}}{n^{s}} = \operatorname{Li}_{s}(-1) = - \eta(-s) $$ for at least $\Re(s) >-2$.