Message Boards Message Boards

GROUPS:

Thanks for helping

Posted 3 months ago
422 Views
|
5 Replies
|
1 Total Likes
|

Thanks for all your help

5 Replies

If r and z are the coordinates, why aren't you integrating over them? What's x?

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.

Hello

I 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.

Posted 3 months ago

Hi! Thanks for your answer. I am not sure what ℜ and Q-1/2 stands for.

Regards, Lina

Q-1/2 is LegendreQ function.$\Re(z)$ is Re[z].

Reply to this discussion
Community posts can be styled and formatted using the Markdown syntax.
Reply Preview
Attachments
Remove
or Discard

Group Abstract Group Abstract