A Summation of Series Involving Bessel Functions and Order Derivatives of Bessel Functions

In this note, we find that the sum of a series involving Bessel functions and the order derivatives of Bessel functions:

$\displaystyle P(x)\equiv\sum_{n=0}^{\infty}nJ_{n}(x)\frac{\partial J_{n}(x)}{\partial n}$

(1)

can be expressed in a closed-form.

The motivation of finding this sum rule arises from the study of entanglement entropy in two dimensional bosonic free field. In Calabrese_2004 , the authors provide a way to compute such entanglement entropy. The key point is the Green’s function in the $n$-sheeted geometry $G_{n}(\mathbf{r},\mathbf{r})$. It obeys the Helmholtz equation in polar coordinates and can be expressed as

$\displaystyle G_{n}(r,\theta;m^{2})=\frac{1}{2n\pi}\sum_{k=0}^{\infty}\varepsilon_{k}\int_{0}^{\infty}\lambda d\lambda\frac{J^{2}_{k/n}(\lambda r)}{\lambda^{2}+m^{2}},$

(2)

where $\varepsilon_{0}=1$ or $\varepsilon_{n}=2$ otherwise. With the Green’s function in the $n$-sheeted geometry, the Rényi entropy can be expressed as

$\displaystyle\Tr\rho^{n}=\exp{-\frac{1}{2}\int_{a^{2}}^{m^{2}}dm^{{}^{\prime}2}\int d^{2}r\left(G_{n}(r,\theta;m^{{}^{\prime}2})-nG_{1}(r,\theta;m^{{}^{\prime}2})\right)}.$

(3)

Then the von Neumann entropy

$\displaystyle S=-\Tr\rho\log\rho=\left.-\frac{\partial}{\partial n}\Tr\rho^{n}\right|_{n=1}=\frac{1}{2}\int_{a^{2}}^{m^{2}}dm^{{}^{\prime}2}\int d^{2}r\left(\left.\frac{\partial}{\partial n}G_{n}(r,\theta;m^{{}^{\prime}2})\right|_{n=1}-G_{1}(r,\theta;m^{{}^{\prime}2})\right).$

(4)

It is not hard to see the summation over the product of Bessel functions in (2) playing an important role. In the last step calculating the von Neumann entropy, the derivative with respect to $n$ requires the order derivatives and the summation now becomes

$\displaystyle\left.\frac{\partial}{\partial n}\sum_{k=0}^{\infty}\varepsilon_{k}J^{2}_{k/n}(\lambda r)\right|_{n=1}=\left.2\sum_{k=0}^{\infty}\varepsilon_{k}\frac{-k}{n^{2}}J_{k/n}(\lambda r)\frac{\partial J_{k/n}(\lambda r)}{\partial(k/n)}\right|_{n=1}=-4P(\lambda r).$

(5)

Let us start with this formula found in Order

$\displaystyle\int^{\infty}_{x}\frac{J^{2}_{\nu}(t)}{t}dt=\frac{1}{2\nu}-\frac{1}{2\nu}\sum_{n=0}^{\infty}\varepsilon_{n}J^{2}_{\nu+n}(x),$

(6)

where $\varepsilon_{0}=1$ and $\varepsilon_{n}=1$ otherwise. By multiplying $\nu$ on both sides and taking the derivative with respect to $\nu$, we have

$\displaystyle\int^{\infty}_{x}\frac{J^{2}_{\nu}(t)+2\nu J_{\nu}(t)\hat{J}_{\nu}(t)}{t}dt=-\sum_{n=0}^{\infty}\varepsilon_{n}J_{\nu+n}(x)\hat{J}_{\nu+n}(x)=-2\sum_{n=1}^{\infty}J_{\nu+n}(x)\hat{J}_{\nu+n}(x)-J_{\nu}(x)\hat{J}_{\nu}(x),$

(7)

where $\hat{J}_{\nu}=\partial J_{\nu}/\partial\nu$. Then we set $\nu\in\mathbb{N}$ and sum over $\nu$ form $0$ to $\infty$. (7) becomes

$\displaystyle\int^{\infty}_{x}\frac{\sum_{\nu=0}^{\infty}J^{2}_{\nu}(t)+2P(t)}{t}dt+\sum^{\infty}_{\nu=0}J_{\nu}(x)\hat{J}_{\nu}(x)=-2\sum^{\infty}_{\nu=0}\sum^{\infty}_{n=1}J_{\nu+n}(x)\hat{J}_{\nu+n}(x).$

(8)

Notice that the RHS gives $-2P(x)$. From (3.2) in Order , the second term in the LHS of (8) gives

$\displaystyle\sum^{\infty}_{\nu=0}J_{\nu}(x)\hat{J}_{\nu}(x)=-\frac{1}{2}\int^{\infty}_{x}\frac{J^{2}_{0}(t)}{t}dt+\frac{1}{2}J_{0}(x)\hat{J}_{0}(x),$

(9)

and also

$\displaystyle\sum_{\nu=0}^{\infty}J^{2}_{\nu}(t)=\frac{1}{2}\left(\sum_{\nu=-\infty}^{\infty}J^{2}_{\nu}(t)+J^{2}_{0}(t)\right)=\frac{J^{2}_{0}(t)+1}{2}.$

(10)

Thus, (8) reduces to

$\displaystyle\int^{\infty}_{x}\left(\frac{J^{2}_{0}(t)+1}{2}+2P(t)\right)\frac{dt}{t}-\frac{1}{2}\int^{\infty}_{x}\frac{J^{2}_{0}(t)}{t}dt+\frac{1}{2}J_{0}(x)\hat{J}_{0}(x)=-2P(x).$

(11)

After taking the derivative with respect to $x$ and substituting $\hat{J}_{0}(x)=\frac{\pi}{2}Y_{0}(x)$ (8.486(1) in GR ), we have the differential equation

$\displaystyle P^{\prime}(x)-\frac{P(x)}{x}=\frac{1}{4x}-\frac{\pi}{8}(J_{0}(x)Y_{0}(x))^{\prime}.$

(12)

This equation can be easily solved by the method of variation of parameters. The solution

	$\displaystyle P(x)$	$\displaystyle=-x\left(\int^{\infty}_{x}\left(\frac{1}{4t^{2}}-\frac{\pi}{8}(J_{0}(t)Y_{0}(t))^{\prime}\right)dt+C\right)$
		$\displaystyle=-x\left(\frac{1}{4x}-\left.\frac{\pi}{8t}J_{0}(t)Y_{0}(t)\right|_{x}^{\infty}-\frac{\pi}{8}\int^{\infty}_{x}\frac{J_{0}(t)Y_{0}(t)}{t^{2}}dt+C\right)$
		$\displaystyle=-\frac{1}{8}\left(2+\pi J_{0}(x)Y_{0}(x)-\pi x\int^{\infty}_{x}\frac{J_{0}(t)Y_{0}(t)}{t^{2}}dt+Cx\right).$		(13)

To determine the integral constant $C$, we can analyse the asymptotic behavior of $P(x)$ at large $x$. When $x\to\infty$, we have $P(x)\sim-\frac{1}{8}(2+Cx)$, which means when $x$ is sufficiently large, $P(x)$ is approximately a linear function and the slope is $C$. However, the asymptotic form of $\hat{J}_{n}$ can be written as Order

$\displaystyle\hat{J}_{n}(x)\sim\sqrt{\frac{\pi}{2x}}\sin\left(x-\frac{n\pi}{2}-\frac{\pi}{4}\right)\ \ (x\to\infty).$

(14)

Combining with the asymptotic approximation of $J_{n}$ in (III), it is easy to find that $P(x)\sim\frac{a}{x}+b\sim b$ when $x\to\infty$, where $a$ and $b$ are constants. More detailed discussion about the asymptotic behavior can be found in the next section. So the integral constant $C=0$ and finally we obtain a closed-form

$\displaystyle P(x)=-\frac{1}{8}\left(2+\pi J_{0}(x)Y_{0}(x)-\pi x\int^{\infty}_{x}\frac{J_{0}(t)Y_{0}(t)}{t^{2}}dt\right).$

(15)

Also, we can evaluate the integral (see Appendix), which can be expressed by a Meijer G function

$\displaystyle\int^{\infty}_{x}\frac{J_{0}(t)Y_{0}(t)}{t^{2}}dt=-\frac{1}{2\sqrt{\pi}}G^{20}_{13}\left(x^{2}\middle|\begin{array}[]{c}1\\ -\frac{1}{2},-\frac{1}{2},-\frac{1}{2}\end{array}\right).$

(18)

Hence, we have an another expression

$\displaystyle P(x)=-\frac{1}{8}\left(2+\pi J_{0}(x)Y_{0}(x)+\frac{\sqrt{\pi}x}{2}G^{20}_{13}\left(x^{2}\middle|\begin{array}[]{c}1\\ -\frac{1}{2},-\frac{1}{2},-\frac{1}{2}\end{array}\right)\right).$

(21)

We now obtain the asymptotic approximation of $P(x)$ at large $x$. The asymptotic behavior can be helpful to analyze the behavior of the combination of Green’s functions $\left.\partial G_{n}(r,\theta;m^{{}^{\prime}2})/\partial n\right|_{n=1}-G_{1}(r,\theta;m^{{}^{\prime}2})$, especially the divergence and the regulariztion of it.

Obviously, when $x\to\infty$, the last two terms in (15), involving $J_{0}$ and $Y_{0}$, go to zero fast (at the order of $O(x^{-1})$). So we have

$\displaystyle\lim_{x\to\infty}P(x)=-\frac{1}{4}.$

(22)

This is not a good news for us because $P(x)$ as a part of the integrand in $\left.\partial G_{n}(r,\theta;m^{{}^{\prime}2})/\partial n\right|_{n=1}$, the non-zero limits highly increases the probability that we have a divergent integral. Indeed, the integral is actually divergent.

However, things are not so bad. We define

	$\displaystyle Q(x)\equiv$	$\displaystyle-4P(x)-\sum_{k=0}^{\infty}J_{k}^{2}(x)=-4P(x)-1$
		$\displaystyle=\frac{1}{2}\left(\pi J_{0}(x)Y_{0}(x)-\pi x\int^{\infty}_{x}\frac{J_{0}(t)Y_{0}(t)}{t^{2}}dt\right).$		(23)

We can rewrite the combination of Green’s functions in terms of $Q(x)$:

$\displaystyle\left.\frac{\partial}{\partial n}G_{n}(r,\theta;m^{{}^{\prime}2})\right|_{n=1}-G_{1}(r,\theta;m^{{}^{\prime}2})=\frac{1}{2\pi}\int_{0}^{\infty}\frac{Q(\lambda r)}{\lambda^{2}+m^{{}^{\prime}2}}\lambda d\lambda.$

(24)

It is easy to check that $Q(x)$ vanishes when $x$ approaches to infinity. By introducing asymptotic approximations for $J_{0}(x)$ and $Y_{0}(x)$ at large $x$

	$\displaystyle J_{n}(x)\sim\sqrt{\frac{2}{\pi x}}\cos\left(x-\frac{n\pi}{2}-\frac{\pi}{4}\right),$
	$\displaystyle Y_{n}(x)\sim\sqrt{\frac{2}{\pi x}}\sin\left(x-\frac{n\pi}{2}-\frac{\pi}{4}\right),$		(25)

we have $\pi J_{0}(x)Y_{0}(x)\sim-\cos(2x)/x$. Substituting this approximation into the integral and integrating by part, we derive the asymptotic approximation for the second term in $Q(x)$, which gives

	$\displaystyle\int_{x}^{\infty}\frac{J_{0}(t)Y_{0}(t)}{t^{2}}dt\sim-\int_{x}^{\infty}\frac{\cos(2t)}{t^{3}}dt=\frac{\sin(x)}{2x^{3}}-\frac{3}{2}\int_{x}^{\infty}\frac{\sin(2t)}{t^{4}}dt$
	$\displaystyle\sim O(x^{-3}).$		(26)

Comparing to the first term, we can neglect the second term when $x$ is sufficiently large. Thus, the asymptotic approximation of $Q(x)$ turns

$\displaystyle Q(x)\sim-\frac{\cos(2x)}{x}.$

(27)

Also, we can evaluate the value at $x=0$. By using L’Hopital rule, we find

	$\displaystyle\lim_{x\to 0}$	$\displaystyle\frac{1}{1/x}\left(\frac{J_{0}(x)Y_{0}(x)}{x}-\int^{\infty}_{x}\frac{J_{0}(t)Y_{0}(t)}{t^{2}}dt\right)$
		$\displaystyle=\lim_{x\to 0}\frac{-1}{1/x^{2}}\left(\frac{x(J_{0}(x)Y_{0}(x))^{\prime}-J_{0}(x)Y_{0}(x)}{x^{2}}+\frac{J_{0}(x)Y_{0}(x)}{x^{2}}\right)$
		$\displaystyle=-\lim_{x\to 0}x(J_{0}(x)Y_{0}(x))^{\prime}.$		(28)

Using identities (8.471 and 8.477 in GR )

	$\displaystyle Z^{\prime}_{n}(x)=\frac{1}{2}(Z_{n-1}(x)-Z_{n+1}(x)),$
	$\displaystyle J_{n}(x)Y_{n+1}(x)-J_{n+1}(x)Y_{n}(x)=-\frac{2}{\pi x},$

where $Z$ denotes $J$ or $Y$, we have

	$\displaystyle-\lim_{x\to 0}x(J_{0}(x)Y_{0}(x))^{\prime}$	$\displaystyle=\lim_{x\to 0}x(J_{1}(x)Y_{0}(x)+J_{0}(x)Y_{1}(x))$
		$\displaystyle=\lim_{x\to 0}x\left(2J_{1}(x)Y_{0}(x)-\frac{2}{\pi x}\right)$
		$\displaystyle=-\frac{2}{\pi}.$		(29)

Thus, $Q(0)=-1$. Moreover, no singularity at the origin (of course, no singularity at other place as well) and the simple asymptotic form of $Q(x)$ tell us the combination in (24) is convergent. Also, by using the identity (7.811 5) in GR1 and the relation (9.31 1) in GR to deal with the Meijer G function, one can evaluate the combination exactly

$\displaystyle\left.\frac{\partial}{\partial n}G_{n}(r,\theta;m^{{}^{\prime}2})\right|_{n=1}-G_{1}(r,\theta;m^{{}^{\prime}2})=\frac{1}{16\pi}\left(\sqrt{\pi}m^{\prime}rG_{13}^{30}\left(m^{{}^{\prime}2}r^{2}\middle|\begin{array}[]{c}1\\ -\frac{1}{2},-\frac{1}{2},-\frac{1}{2}\\ \end{array}\right)-4K^{2}_{0}\left(m^{\prime}r\right)\right).$

(32)

Furthermore, we can also obtain

$\displaystyle\int d^{2}r\left(\left.\frac{\partial}{\partial n}G_{n}(r,\theta;m^{{}^{\prime}2})\right|_{n=1}-G_{1}(r,\theta;m^{{}^{\prime}2})\right)=-\frac{1}{6m^{{}^{\prime}2}},$

(33)

which agrees the result in Calabrese_2004 . Because the radial integral is definitely a Mellin transform so this result can be found in Mellin transform table in tables .

We study a summation of series involving Bessel functions and the order derivatives of Bessel functions, which arises in the study of entanglement entropy in two dimensional bosonic free field. We find a closed-form expression of such summation which can be expressed simply in terms of $J_{0}$, $Y_{0}$ and a Meijer G function. Further, we study the asymptotic behavior of this summation of series and finally we find a simple expression for the combination of Green’s functions in (32).

The expression in (32) and also $P(x)$, $Q(x)$ might be useful for calculating entanglement entropies in different fields. For instance, if the field is coupled with a potential such that the equation of motion is no longer a homogeneous Helmholtz equation and the Green’s equation becomes

$\displaystyle(-\nabla^{2}_{\mathbf{r}}+m^{2}-f(\mathbf{r}))G_{n}(\mathbf{r},\mathbf{r^{\prime}})=\delta(\mathbf{r}-\mathbf{r^{\prime}}).$

(34)

Generally, the exact solution would not be found easily. However, the equation can be solved perturbatively by applying the resolvent formalism

$\displaystyle G_{n}(\mathbf{r},\mathbf{r^{\prime}})=\tilde{G}_{0}(\mathbf{r},\mathbf{r^{\prime}})+\tilde{G}_{1}(\mathbf{r},\mathbf{r^{\prime}})+...,$

(35)

where

		$\displaystyle\tilde{G}_{0}(\mathbf{r},\mathbf{r^{\prime}})=\tilde{G}(\mathbf{r},\mathbf{r^{\prime}}),$
		$\displaystyle\tilde{G}_{1}(\mathbf{r},\mathbf{r^{\prime}})=(\tilde{G}\circ f\circ\tilde{G})(\mathbf{r},\mathbf{r^{\prime}}),$
		$\displaystyle...$
		$\displaystyle\tilde{G}_{i}(\mathbf{r},\mathbf{r^{\prime}})=(\tilde{G}\circ f\circ\tilde{G}\circ...\circ f\circ\tilde{G})(\mathbf{r},\mathbf{r^{\prime}}),$		(36)

where $\tilde{G}$ is the Green’s function with $f=0$ and $(G_{n}\circ f)(\mathbf{r})=\int G(\mathbf{r},\mathbf{r^{\prime}})f(\mathbf{r^{\prime}})d^{2}r$. In this method, the exact Green’s function can be also written as (see Appendix B)

$\displaystyle G_{n}(\mathbf{r},\mathbf{r^{\prime}})=\sum_{l=0}^{\infty}\sum_{i+j=0}^{l}B_{i,j,l}[f](2\partial_{\omega})^{i}(2\partial_{\bar{\omega}})^{j}\tilde{G}_{n}^{\left(1+\frac{i+j+l}{2}\right)}(\mathbf{r},\mathbf{r^{\prime}}).$

(37)

This method is well developed in dikii1955zeta ; dikii1958trace for Sturm-Liouville problems. More details can be found in gelfand ; SEELEY . Furthermore, we can apply the Green’s function to calculate the entanglement entropy

	$\displaystyle\left.\frac{\partial}{\partial n}G_{n}(r,\theta;m^{{}^{\prime}2})\right|_{n=1}-G_{1}(r,\theta;m^{{}^{\prime}2})$
	$\displaystyle=\sum_{l=0}^{\infty}\sum_{i+j=0}^{l}\left\{B_{i,j,l}[f](2\partial_{\omega})^{i}(2\partial_{\bar{\omega}})^{j}\left[\left.\frac{\partial}{\partial n}\tilde{G}_{n}^{\left(1+\frac{i+j+l}{2}\right)}(\mathbf{r},\mathbf{r^{\prime}})\right|_{n=1}-\tilde{G}_{1}^{\left(1+\frac{i+j+l}{2}\right)}(\mathbf{r},\mathbf{r^{\prime}})\right]\right\}_{\mathbf{r}=\mathbf{r^{\prime}}}.$		(38)

For those terms with $i=j$ in the summation, it is easy to figure out by using $Q(x)$, i.e.

$\displaystyle(4\partial_{\omega}\partial_{\bar{\omega}})^{i}\left[\left.\frac{\partial}{\partial n}\tilde{G}_{n}^{\left(1+i+l/2\right)}(\mathbf{r},\mathbf{r^{\prime}})\right|_{n=1}-\tilde{G}_{1}^{\left(1+i+l/2\right)}(\mathbf{r},\mathbf{r^{\prime}})\right]_{\mathbf{r}=\mathbf{r^{\prime}}}=\frac{1}{2\pi}\int_{0}^{\infty}\frac{-(-\lambda)^{2i+1}}{\left(\lambda^{2}+m^{{}^{\prime}2}\right)^{1+i+l/2}}Q(\lambda r)d\lambda.$

(39)

However, the remaining terms where $i\neq j$ are still a problem that we can not solve easily. So how to deal with other terms can be a topic for further study.