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Forums
Mathematics
Calculus
Orthogonality of spherical Bessel functions
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[QUOTE="sunrah, post: 5498066, member: 365167"] Thanks for replying. If I understand correctly, I should try integrating by parts. So doing this I find my function can be approximated as [itex] I_{\ell}(k,k_{i}) \approx \frac{k_{i}}{k^{2}}\int^{\infty}_{0} y^{2}(\ln{y} - 1)j_{\ell}(k_{i}y)j_{\ell}(ky)dy [/itex] to do this I'v used [itex]j_{\ell}(0) = 0 [/itex] for [itex]\ell > 0[/itex] (I'm not interested in monopols) and that the value of higher-order integrals, e.g. [itex] \int^{\infty}_{0}dz (\int^{z}_{0}dy\int^{y}_{0}j_{\ell}(x)dx) [/itex] are progressively much smaller than first-order integrals (if that's clear??). My opinion is that for this function [itex]I_{\ell}[/itex] to peak at k=k[SUB]i[/SUB] then the approximation shown featuring the two un-integrated sph. Bessel functions must dominate. Also just plugging some values into python seems show that there are orders of magnitude difference between higher-order integrals of j[SUB]l[/SUB]. Is this what you meant? [/QUOTE]
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Mathematics
Calculus
Orthogonality of spherical Bessel functions
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