Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

For continuous indices 'n' and 'm' is it possible to have

$$ \int_{0}^{\infty}dx e^{-x}L_{n}(x)L_{m}(x) = \delta (n-m)? $$

Another question: let $g(m)$ be the function defined via the transform

$$ \int_{0}^{\infty}dx e^{-x}f(x)L_{m}(x)=g(m). $$

Is its inverse given by $$ \int_{0}^{\infty}dm e^{-m}g(m)L_{m}(x)=f(x), $$

so that $$ \int_{0}^{\infty}e^{-m}dm\int_{0}^{\infty}dx e^{-x}L_{n}(x)L_{m}(x) =1? $$

to be more explicit , let be the integral transform

$$ g(m)= \int_{0}^{\infty}dxe^{-x}f(x)L_{m}(x) $$ , then how could we obtain $ f(x) $ from $ g(m) $ ??

share|cite|improve this question
By $\delta(n-m)$ do you mean a Dirac delta? If so that isn't true since we know that the integral is equal to a finite value when $n=m$. That aside this is an interesting idea, do you have a definition for what $L_{m}(x)$ means for non-integer values? Maybe we can modify the recurrence realtions to make this work. – Spencer Jul 18 '13 at 19:07
Non integer index reminds me of scattering states of a quantum mechanical system.. – Torsten Hĕrculĕ Cärlemän Jul 18 '13 at 19:11
By any chance, were you talking about the q-Laguerre polynomials? – Torsten Hĕrculĕ Cärlemän Jul 18 '13 at 19:39
@Spencer Yeah.We must also keep measures in our mind. But I think the OP is actually talking about the quantum-laguerre polynomials here – Torsten Hĕrculĕ Cärlemän Jul 18 '13 at 19:43
for non integer 'm' we can use fractional derivative to define $ L_{m}(x)= \frac{e^{x}}{\Gamma (m+1)}D_{x}^{n}(x^{n}e^{-x}) $ – Jose Garcia Jul 18 '13 at 21:22

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.