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How to prove this:

$$2\sum_{n=1}^{\infty}(-1)^n\frac{J_{2n}(z)}{n}=\sum_{m=1}^{\infty}\left\{\frac{(-1)^m(\frac{z}{2})^{2m}}{(m!)^2}\sum_{k=1}^{m}\frac{1}{k}\right\}$$

This equation occurs in converting standard BesselY function to Neumann's representation.

You can find this problem in the book 'A Treatise on the Theory of Bessel Functions' by G. N. Watson, 1944, page 67. screenshot of page 67

Thank you very much.

Ying Tang
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1 Answers1

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It is well-known that $$ J_{2n}(z)=\sum_{l\geq 0}\frac{(-1)^l z^{2l+2n}}{2^{2l+2n}l!(2n+l)!} \tag{1}$$ and that $J_n(z)$ has a very fast decay to zero as $n\to +\infty$ for a fixed $z$. It follows that:

$$\begin{eqnarray*} \sum_{n\geq 1}\frac{(-1)^n}{n}J_{2n}(z) &=& \sum_{n\geq 1}\sum_{l\geq 0}\frac{(-1)^{l+n}(z/2)^{2l+2n}}{nl!(2n+l)!}\\&=&\sum_{h\geq 1}(-1)^h (z/2)^{2h}\sum_{l=0}^{h-1}\frac{1}{(2h-l)!l!(h-l)}\end{eqnarray*}\tag{2} $$ and it is enough to find a closed form for the very last sum: $$\begin{eqnarray*} \sum_{l=0}^{h-1}\frac{1}{(2h-l)!l!(h-l)} &=& \sum_{l=1}^{h}\frac{1}{l(h-l)!(h+l)!}\\&=&\color{blue}{\frac{1}{h!^2}}\sum_{l=1}^{h}\binom{h}{l}\frac{ h!(l-1)!}{(h+l)!}\end{eqnarray*} \tag{3}$$ Here comes the magic of Euler's Beta function: $$\begin{eqnarray*} \sum_{l=1}^{h}\binom{h}{l}\frac{\Gamma(h+1)\,\Gamma(l)}{\Gamma(h+l+1)}&=&\int_{0}^{1}\sum_{l=1}^{h}\binom{h}{l}z^{l-1}(1-z)^h\,dz\\&=&\int_{0}^{1}\frac{(1+z)^h-1}{z}(1-z)^h\,dz\tag{4}\end{eqnarray*}$$ leading to: $$\begin{eqnarray*} \sum_{l=1}^{h}\binom{h}{l}\frac{ h!(l-1)!}{(h+l)!} &=& \int_{0}^{1}\frac{(1-z^2)^h-1}{z}\,dz -\int_{0}^{1}\frac{(1-z)^h-1}{z}\,dz\\&=&\frac{1}{2}\int_{0}^{1}\frac{(1-z)^h-1}{z}\,dz-\int_{0}^{1}\frac{(1-z)^h-1}{z}\,dz\\&=&\frac{1}{2}\int_{0}^{1}\frac{1-z^h}{1-z}\,dz = \color{red}{\frac{H_h}{2}}\tag{5}\end{eqnarray*} $$ and finally to:

$$ \color{red}{2}\sum_{n\geq 1}\frac{(-1)^n}{n}J_{2n}(z) = \sum_{h\geq 1}\frac{(-1)^h (z/2)^{2h}}{\color{blue}{h!^2}}\color{red}{\sum_{k=1}^{h}\frac{1}{k}}.\tag{6}$$

Jack D'Aurizio
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  • In the equation 2, how could you make the upper limit of the inner summation to 'h-1'? how is the term '$(z/2)^{2l}$' disappeared?. In the equation 5, how is the formula derivated? Thank you very much for the answer. – Ying Tang May 01 '17 at 22:19
  • @YingTang: in $(2)$ at some point I perform the substitution $l\to h-l$ bringing the old summation indices ($1..h$) into the new ones ($0..h-1$). After $(2)$, I care just about the coefficient of $z^{2h}$. – Jack D'Aurizio May 01 '17 at 22:23
  • In $(5)$ an integral is simplified through the substitution $z\to\sqrt{z}$, then another substitution is performed, $z\to (1-z)$. Everything pretty standard. – Jack D'Aurizio May 01 '17 at 22:24
  • Thank you for the reply. I think the substitution $l=> h-l$ is done in step 3. The problem I don't understand is one step earlier. – Ying Tang May 01 '17 at 22:30
  • @YingTang: oh, sorry, you are right. In $(2)$ I used $h=n+l$ as a new summation index. So $h$ ranges from $1$ to infinity and $l\leq h-1$ since $n\geq 1$. – Jack D'Aurizio May 01 '17 at 22:35