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Can we show that the series given below,

$$ \begin{align*} 2\pi e^{-1} \Biggl[ \delta(\xi) + e^{-|a\xi|} \sum_{n=0}^{\infty} \sum_{j=0}^{\infty} \frac{(n+2j)!}{(n+j+1)!(n+j)!j!n!} \frac{|a|^{n+1} |\xi|^n}{2^{n+2j+1}} \Biggr] \end{align*} $$

be given terms of a single summation of Modified Bessel function of the following form

$$ \begin{align*} \mathscr{F}\{f\}(x)=2\pi e^{-1}\delta(x)+|a|e^{-1}\sqrt{\pi}\sum_{n=1}^\infty \frac{K_{n-1/2}(|ax|)|ax|^{n-1/2}}{2^{n-5/2}n!(n-1)!}\,? \end{align*} $$

This question is motivated by the answers given for a different problem that is discussed here.

Gary
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user35952
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1 Answers1

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Basically, this asks for an alternative expression for $$f(z)=\sum_{n=1}^\infty\left(\frac z2\right)^{n-1/2}\frac{K_{n-1/2}(z)}{n!(n-1)!}=\sum_{n=0}^\infty\left(\frac z2\right)^{n+1/2}\frac{K_{n+1/2}(z)}{n!(n+1)!}$$ (at $z=|ax|$ with some irrelevant factors and terms). Just use $$K_{n+1/2}(z)=e^{-z}\sqrt{\frac{\pi}{2z}}\sum_{k=0}^n\frac{(n+k)!}{k!(n-k)!}(2z)^{-k},$$ a known closed form; there are many proofs, say using induction on $n$ and the recurrences for $K_\nu(z)$, or the integral representation $K_{n+1/2}(z)=\left(\frac z2\right)^{n+1/2}\frac{\sqrt\pi}{n!}\int_1^\infty(t^2-1)^n e^{-zt}\,dt$. Thus, \begin{align} \frac{2e^z}{\sqrt\pi}f(z)&=\sum_{n=0}^\infty\frac{(z/2)^n}{n!(n+1)!}\sum_{k=0}^n\frac{(n+k)!}{k!(n-k)!}(2z)^{-k} \\&=\sum_{k=0}^\infty\sum_{n=k}^\infty\frac{2^{-n-k}z^{n-k}(n+k)!}{n!(n+1)!k!(n-k)!} \\\color{gray}{[n=k+j]}&=\sum_{k=0}^\infty\sum_{j=0}^\infty\frac{2^{-2k-j}(2k+j)!\ z^j}{(k+j)!(k+j+1)!k!j!}, \end{align} equivalent to the equality being asked for.

metamorphy
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