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I'd like to proove the following integral relation

$$ \frac{1}{2^m m!} \frac{1}{\sqrt{\pi}} \int_{-\infty}^{\infty}\,\mathrm{d}\zeta \, e^{-\zeta^2} H_m(\zeta+\zeta_1)H_m(\zeta+\zeta_2) = L_m(-2\zeta_1\zeta_2)$$

where $H_m(x)$ is the $m$th Hermite polynomial and $L_m(x)$ is the $m$th Laguerre polynomial. I tried proving it with these realtions I found on Wikipedia

$$ H_m(x+y)=\sum_{k=0}^m \binom{m}{k}H_k(x)(2 y)^{(n-k)}$$

and

$$ L_m(x)=\sum_{k=0}^m \binom{m}{k}\frac{(-x)^k}{k!}$$

by using the othonormality of the Hermite polynoms, but I got stuck at some point. Can you help me here?

GNUSupporter 8964民主女神 地下教會
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Marie
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1 Answers1

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The solution is most easily constructed if you use the Hardy-Hille formula for Hermite polynomials. (I gave a different kind of proof in MSE 3694450.) To do so, we'll make a generating function from the left hand side of the equation: $$ I_m(x,y):=\frac{1}{\sqrt{\pi}} \frac{2^{-m}}{m!} \int_{-\infty}^\infty dt\ e^{-t^2} H_m(t+x) \ H_m(t+y) $$ $$ F(x,y,z)=\sum_{m=0}^\infty z^m I_m(x,y) = \frac{1}{\sqrt{\pi}} \int_{-\infty}^\infty dt\ e^{-t^2} \sum_{m=0}^\infty \frac{z^m}{m!2^m}H_m(t+x) \ H_m(t+y) $$ $$ = \frac{1}{\sqrt{\pi(1-z^2)}} \int_{-\infty}^\infty dt\ e^{-t^2} \exp{\big( \frac{2z(t+x)(t+y) - ((t+x)^2 + (t+y)^2)z^2}{1-z^2} \big) } $$ where in the last step the Hardy-Hille formula was used. Do the Gaussian integral and algebra. You'll find $$ F(x,y,z)=\frac{1}{1-z} \exp{\big(\frac{2\ xy \ z}{1-z} \big)} $$ However, it is well known that the Laguerre polynomials have the generating function

$$\frac{1}{1-z} \exp{\big(\frac{-q \ z}{1-z} \big)}= \sum_{n=0}^\infty L_n(q) z^n $$

Equate coefficients of z to complete the proof.

user321120
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