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Mathematics
Calculus
Orthogonality of spherical Bessel functions
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[QUOTE="Twigg, post: 5498598, member: 572426"] That is what I had in mind, but to be honest I hadn't examined the remainders when I first replied. It might not be the best method, it's just what jumped out at me. After working it through on paper, I got the same approximation as you when I expanded the inner integral (the dx integral) to first order using integration by parts. When I did the second-order correction, I found that the next term in the part of the inner (dx) integral that is a multiple of the ##\ell##-th spherical bessel depends on ##\ell##, due to the recursion formulas. So when you run your Python calculations, you might want to examine those higher-order integrals for some high-order spherical bessels, like ##j_{10}(x)## and ##j_{100}(x)##, just to be on the safe side. Here's what I got for the second-order integration by parts expansion: ##\int_{0}^{y} \frac{y-x}{x} j_{\ell}(kx)dx = [ y (\ln y - 1) - \ell y (\ln y - \frac{3}{2}) ]j_{\ell}(kx) - ky^{2} (\ln y - \frac{3}{2}) j_{\ell + 1}(kx) + ...## By orthogonality, you can ignore the ##\ell + 1## term and higher in ##I_{\ell}##, but the coefficient on the first term will probably be an infinite series in ##\ell##. You might even be able to extrapolate the series. To get the above result, I used the following recursion relation after the first integration by parts: ##j'_{\ell}(z) = \frac{\ell}{z} j_{\ell}(z) + j_{\ell + 1} (z)## This is where I'm getting my series for the coefficient on the ##\ell##-th spherical bessel. Hope this helps! [/QUOTE]
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Orthogonality of spherical Bessel functions
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