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Oct 10, 2022 at 11:58 answer added Johannes Trost timeline score: 2
Aug 9, 2022 at 11:24 comment added Johannes Trost That would mean that the integral is independent of the exponent $d$ of $r$ under the integral.
Aug 9, 2022 at 11:11 comment added Ryo Ken I m sure of it@ Johannes Trost
Aug 8, 2022 at 15:04 comment added Johannes Trost I try to check your result for the integral that you stated in the comments. I can not reproduce it, yet. Are you sure, that it is correct ?
Aug 8, 2022 at 11:41 comment added Ryo Ken Thank you a lot@ Johannes Trost. I will use it.
Aug 8, 2022 at 10:25 comment added Johannes Trost There IS in fact a simplification: for integer parameters the confluent hypergeometric function reduces of course considerably : $ _1F_1(n+m,n,z)=\exp(z) \sum_{j=0}^{m} {m \choose j } z^{j} / (n)_{j}$ (I have no link at hands for that formula, but should be online somewhere)
Aug 7, 2022 at 13:15 comment added Ryo Ken Thank you a lot
Aug 7, 2022 at 13:07 comment added Johannes Trost Not to my knowledge.
Aug 7, 2022 at 12:26 comment added Ryo Ken Thnak you for your help. If we use your relations then the integral is equivalent to the sum $\sum^k_{j=0} \binom{k+1}{k-j}(-2)^j (j+1)_{1}F_{1}\left(j+2,2,-\frac{c^{2}}{4\ a}\right)$. Is there a closed formula for this
Aug 7, 2022 at 12:06 comment added Johannes Trost The Laguerre polynomial can be found in NIST Special functions: dlmf.nist.gov/18.5.E12 . And use $\int^\infty_0 e^{-ar^2}J_1(cr)r^\delta dr =$ $\frac{1}{4} c a^{-1-\frac{\delta}{2}} \Gamma(1+\frac{\delta}{2}) _{1}F_{1}\left(1+\frac{\delta}{2},2,-\frac{c^{2}}{4\ a}\right)$. (Mathematica result) Could not find a reference for the integral, but maybe later. So called Confluent Hypergeometric function $_{1}F_{1}$ is here functions.wolfram.com/HypergeometricFunctions/Hypergeometric1F1
S Aug 7, 2022 at 11:25 review First questions
Aug 7, 2022 at 17:29
S Aug 7, 2022 at 11:25 history asked Ryo Ken CC BY-SA 4.0