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Is there a formulas for the following integral $$\int^\infty_0 e^{-ar^2}L^1_k(b r^2)J_1(cr)r^d dr $$ where $L^1_k$ is the Laguerre polynomials of type 1 and $J_1$ is the Bessel function with $a,b,c,d\geq 0$ Thank you in advance.

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  • $\begingroup$ 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 $\endgroup$ Commented Aug 7, 2022 at 12:06
  • $\begingroup$ 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 $\endgroup$
    – Ryo Ken
    Commented Aug 7, 2022 at 12:26
  • $\begingroup$ Not to my knowledge. $\endgroup$ Commented Aug 7, 2022 at 13:07
  • $\begingroup$ Thank you a lot $\endgroup$
    – Ryo Ken
    Commented Aug 7, 2022 at 13:15
  • $\begingroup$ 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) $\endgroup$ Commented Aug 8, 2022 at 10:25

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To wrap up here is the result for the OP's integral: $$ \int_{0}^{\infty} dr \ e^{-a r^2}\ L_{k}^{1}(b \ r^2) \ J_{1}( c\ r)\ r^d = \\ \frac{(k+1)! \ c}{4 \ a^{\frac{d}{2}+1}}\sum_{n=0}^{k}\left(-\frac{b}{a}\right)^n \ \frac{\Gamma(n+\frac{d}{2}+1)}{(k-n)!\ n!\ (n+1)!} \ _{1}F_{1}\left(n+\frac{d}{2}+1,2;-\frac{c^2}{4\ a} \right) $$

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  • $\begingroup$ Thank you a lot $\endgroup$
    – Ryo Ken
    Commented Dec 31, 2022 at 10:47

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