Infinite sum of even Bessel functions - Identities

Question

Recently, I came across the following identities among first-kind Bessel functions, namely $$ 2\sum_{k=1}^{\infty}(-1)^k\,k^5\,J_{2k}(x) = \frac{x^2}{4}\left[x\,J_1(x)-J_0(x)\right] \label{1}\tag{1} $$ and $$ 2\sum_{k=1}^{\infty}\,(-1)^k\,k^3\,J_{2k}(x) = -\frac{x^2}{4}J_0(x) \label{2} \tag{2}. $$ It's straightforward to verify that the identities hold perturbatively up to very high orders in $x$. However, I'm curious if anyone knows how to formally prove them analytically. I've searched the mathematical literature, but I couldn't find any identities involving the multiplication of even Bessel functions by a monomial of power 5 or 3, as in \eqref{1} and \eqref{2}. Any hints or suggestions would be greatly appreciate.

Many thanks in advance

Answer 1

Let $L$ denote the left-hand side of your identity \eqref{2}. Then, using the identity $$J_a(x)=\sum_{m\ge0}\frac{(-1)^m}{m!(m+a)!}(x/2)^{2m+a} \tag{$\dagger$}\label{3},$$ we get $$ \begin{split} L&=2\sum_{k\ge1}(-1)^k k^3 J_{2k}(x) \\ &=2\sum_{k\ge1}(-1)^k k^3 \sum_{m\ge0}\frac{(-1)^m}{m!(m+2k)!}(x/2)^{2m+2k} \\ &=2\sum_{n\ge1}(x/2)^{2n}\sum_{k=1}^n(-1)^k k^3 \frac{(-1)^{n-k}}{(n-k)! (n+k)!} \\ &=2\sum_{n\ge1}(-1)^n(x/2)^{2n}\sum_{k=1}^n\frac{k^3}{(n-k)! (n+k)!} \\ &=2\sum_{n\ge1}(-1)^n(x/2)^{2n}\frac1{2(n-1)! (n-1)!} \\ &=-\frac{x^2}4\,\sum_{m\ge0}(-1)^m(x/2)^{2m}\frac1{m! m!} =-\frac{x^2}4\,J_0(x), \end{split} $$ by \eqref{3}. This proves your identity \eqref{2}.

Your identity \eqref{1} can be proved similarly. (According to MathOverflow guidelines, there should be only one question in one post.)

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To verify the identity $$\sum_{k=1}^n\frac{k^3}{(n-k)! (n+k)!}=\frac1{2(n-1)! (n-1)!},$$ used in the multi-line display above, one can use the identities $$ \begin{split} 2k^3 &=((n+k)^{(3)}-(n-k)^{(3)}) \\ &\qquad -(2 - 6 n + 3 n^2)((n+k)^{(1)}-(n-k)^{(1)}) \end{split} $$ and $(n\pm k)^{(m)}/(n\pm k)!=1/(n-m\pm k)!$, assuming that $\frac{1}{p!}=0$ if $p\in\{-1,-2,\dots\}$, where $a^{(m)}:=\frac{\Gamma(a+1)}{\Gamma(a-m+1)}=a(a-1)\cdots(a-(m-1))$ for nonnegative integers $m$.

Using the similar identity for $2k^5$ instead of $2k^3$, we can check that $$ \begin{split} \sum_{k=1}^n\frac{k^5}{(n-k)! (n+k)!} &=\frac{2n-1}{2(n-1)! (n-1)!} \\ &=\frac{1}{(n-2)! (n-1)!}+\frac{1}{2(n-1)! (n-1)!} \end{split}.$$

Answer 2

It appears to me that these identities can be interpreted in terms of Neumann series. To briefly sketch the idea, suppose that $f(z)$ is analytic in some disc centered at the origin. Then $f(z) = \sum_{n=0}^{\infty} a_n J_n(z)$ is called the Neumann series for $f$, and the coefficients $a_n$ can be determined in various ways. For instance, Chapter 16.11 of Watson's book gives that if $f(z) = \sum_{n=0}^{\infty} b_n z^n$ is the power series expansion of $f$ at the origin, then $a_0 = b_0$ and $$a_n = n \sum_{0 \leq m \leq n/2} 2^{n-2m} \frac{(n-m-1)!}{m!} b_{n-2m}.$$

Now let's derive the Neumann series expansion of $-\frac{x^2}{4} J_0(x)$. A little calculation with the power series definition of $J_0(x)$ gives that $b_0 = 0$, $b_n = 0$ for $n$ odd, and for $n=2k$ with $k \geq 1$ then $$b_{2k} = \frac{(-1)^k}{2^{2k}(k-1)!^2}.$$ Hence $a_0 = 0$, $a_n = 0$ for $n$ odd, and for $k \geq 1$ we have $$a_{2k} = 2k \sum_{m=0}^{k} 2^{2k-2m} \frac{(2k-m-1)!}{m!} \frac{(-1)^{k-m}}{2^{2k-2m}(k-m-1)!^2}.$$ So now to verify the proposed identity we need to show that $$ \sum_{m=0}^{k} \frac{(2k-m-1)!}{m!} \frac{(-1)^{m}}{(k-m-1)!^2} = k^2.$$ Out of laziness, I didn't bother to verify this identity, but I plugged the left hand side into Wolfram Alpha and it was able to evaluate it as $k^2$.

I imagine the other identity can be worked out similarly.

Perhaps this line of thinking can help see the more general pattern and derive additional identities of the type given in the OP. Presumably one can invert the relationship between $a_n$ and $b_n$ and from a desired Neumann series, one can derive the power series.