# Thanks for helping

Posted 3 months ago
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 Thanks for all your help
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Posted 3 months ago
 If r and z are the coordinates, why aren't you integrating over them? What's x?
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Posted 3 months ago
 Hi, We want to "remove x" to get a function depending on r and z. I get what you mean but it is an assignment for a course.
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Posted 3 months ago
 HelloI found formula on this book (my book have Date - Edition 1986 year) on page: 218 example: 1. $$\int_0^{\infty } \exp (-x z) J_0(R x) J_0(r x) \, dx=\Re\left(\frac{Q_{-\frac{1}{2}}\left(\frac{r^2+R^2+z^2}{2 r R}\right)}{\pi \sqrt{r R}}\right)$$for: Re[z] > Abs[Im[r]] + Abs[Im[R]]Numeric check for forumla: f[R_?NumericQ, r_?NumericQ, z_?NumericQ] := NIntegrate[ Exp[-x z] BesselJ[0, R x] BesselJ[0, r x], {x, 0, Infinity}]; f[20, 2, 1] (* 0.0500619 *) g[R_?NumericQ, r_?NumericQ, z_?NumericQ] := Re[LegendreQ[-(1/2), (r^2 + R^2 + z^2)/(2 r R)]/(π Sqrt[r R])] // N; g[20, 2, 1] (* 0.0500619 *) Regards MI.
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Posted 3 months ago
 Hi! Thanks for your answer. I am not sure what ℜ and Q-1/2 stands for. Regards, Lina
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Posted 3 months ago
 Q-1/2 is LegendreQ function.$\Re(z)$ is Re[z].
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