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There exists such an equality which converts the product of Bessel functions of the first and second kind into a Meijer G-funtion, $$x^\mu J_\nu(x)Y_\nu(x)=-\pi^{-1/2}G_{13}^{20}\left(x^2\left| \begin{matrix}(1+\mu)/2\\ \mu/2+\nu, \mu/2,\mu/2-\nu\end{matrix}\right. \right) $$ which is equation (60) in page 220 in book, Higer transcendental functions [1]. However, there was no proof in this book.

Could anyone offer me a proof? A further question is: can we convert $\frac{\sin(t)Y_0(t)}{t}$ into a certain Meijer G-function? Thanks a lot.

[1] Bateman H, Erdelyi A (1953). Higher transcendental functions [M]. McGraw-Hill; New York.

mywyk5522
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  • $$\frac {Y_0(x) \sin x} x = \frac 1 {\sqrt 2} G_{3, 5}^{2, 2} \left( x^2 \middle| {-\frac 1 4, \frac 1 4, -\frac 1 2 \atop 0, 0, -\frac 1 2, -\frac 1 2, - \frac 1 2} \right).$$ Note that the rhs is an even function of $x$; neither formula holds for $\operatorname{Re} x < 0$. – Maxim Aug 24 '18 at 18:29
  • Thank you very much. Could you show me how to get the equation for $\frac{Y_0(x)\sin(x)}{x}$? How about $ \frac{Y_0(x)(1-\cos(x))}{x}$? Or even more complicate case: can the functions $\frac{J_\mu(x) J_\nu(x)(1-\cos(x))}{x}$ and $\frac{J_\mu(x)J_\nu\sin(x)}{x}$ be converted to the Meijer G-function? – mywyk5522 Aug 25 '18 at 00:21
  • $Y_0(x) \sin x$ is this formula with $\nu = 0$. Multiplication by $(x^2)^{-1/2}$ shifts all coefficients by $-1/2$. – Maxim Aug 25 '18 at 01:31
  • Thanks a lot for bringing the formula for me. Where can I find the proof for this formula? – mywyk5522 Aug 25 '18 at 02:00
  • I don't know. For $J_\nu(x) Y_\nu(x)$ (you need only the case $\mu = 0$), a standard method is to write the G-function as its defining integral and evaluate it as a sum of residues. You get a sum of two ${_1F_2}$ terms. $Y_0(x) \sin x$ seems more difficult because you would have to compute residues at double poles. – Maxim Aug 25 '18 at 03:27

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