1
$\begingroup$

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.

$\endgroup$
9
  • $\begingroup$ 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 $\endgroup$ 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
    Aug 7, 2022 at 12:26
  • $\begingroup$ Not to my knowledge. $\endgroup$ Aug 7, 2022 at 13:07
  • $\begingroup$ Thank you a lot $\endgroup$
    – Ryo Ken
    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) $\endgroup$ Aug 8, 2022 at 10:25

1 Answer 1

2
$\begingroup$

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

$\endgroup$
1
  • $\begingroup$ Thank you a lot $\endgroup$
    – Ryo Ken
    Dec 31, 2022 at 10:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.