An Integral Involving the Product of Bessel Function and Associated Laguerre Polynomial

Question:

Try to prove the following integral:

\[\int_{0}^{\infty} z^{\mu/2} \mathrm{e}^{-z} L_{n}^{\mu}(z)J_{\mu}\left(2\sqrt{xz}\right) \mathrm{d}z = \frac{1}{n!} x^{n+\mu/2}\mathrm{e}^{-x},\]

where $x>0,n\in\mathbb{N}$ and $L_{n}^{\mu}(z), J_{\mu}(z)$ denote associated Laguerre polynomial and Bessel function respectively.

This question was asked on Zhihu recently, and you can find my solution in it. You can see that this integral looks awful, but has a very concise result. That’s why I post this question in my blog.

Solution:

Train of Thought: Recall the orthogonality of the associated Laguerre polynomials:

\[\int_0^{\infty}z^{\mu} \mathrm{e}^{-z} L_{n}^{\mu}(z) L_{n'}^{\mu}(z) \ \mathrm{d}z = \frac{\Gamma(\mu+n+1)}{n!}\delta_{nn'}.\]

So if $z^{-\mu/2}J_{\mu}(2\sqrt{xz})$ can be expressed as the linear combination of associated Laguerre polynomials $L_{m}^{\mu}(z)$ with different $m$:

\[z^{-\mu/2}J_{\mu}(2\sqrt{xz}) \sim \sum_{m=0}^{\infty} a_m L_{m}^{\mu}(z) ,\]

then we just need to use the orthogonality to find answer. You can see the following steps for detailed process.

Lemma 1

\[\color{Green}{\text{Lemma 1:}\quad\displaystyle{ \int_0^{\infty}z^{\mu} \mathrm{e}^{-z} L_{n}^{\mu}(z) L_{n'}^{\mu}(z) \ \mathrm{d}z = \frac{\Gamma(\mu+n+1)}{n!}\delta_{nn'}}} .\]

Recall the generating function of the associated Laguerre polynomials:

\[\sum_{n=0}^{\infty}L_{n}^{\mu}(z)t^{n} = \frac{\mathrm{e}^{-zt/(1-t)}}{(1-t)^{\mu+1}} ,\]

then we can calculate the product of two different generating functions,

\[\begin{aligned} &\sum_{n=0}^{\infty}t^{n}L_{n}^{\mu}(z) \cdot \sum_{n'=0}^{\infty} s^{n'}L_{n'}^{\mu'}(z)\\[.5cm] ={}&\sum_{n=0}^{\infty}\sum_{n'=0}^{\infty} t^{n}s^{n'} L_{n}^{\mu}(z)L_{n'}^{\mu'}(z) \\[.5cm]={}& \frac{1}{(1-t)^{\mu+1}(1-s)^{\mu'+1}}\, \mathrm{exp}\left[-z \left(\frac{t}{1-t}+ \frac{s}{1-s}\right) \right]. \end{aligned}\]

Multiplying both sides by $z^{\lambda}\mathrm{e}^{-z}$ and integrating,

\[\begin{aligned} &\sum_{n=0}^{\infty}\sum_{n'=0}^{\infty} t^{n}s^{n'}\int_{0}^{\infty} z^{\lambda} \mathrm{e}^{-z} L_{n}^{\mu}(z) L_{n'}^{\mu'}(z) \ \mathrm{d}z\\[.5cm] ={}&\int_0^{\infty} \frac{z^{\lambda}}{(1-t)^{\mu+1}(1-s)^{\mu'+1}} \, \exp\left[-z \frac{1-ts}{(1-t)(1-s)}\right] \ \mathrm{d}z\\[.5cm] ={}&\frac{(1-t)^{\lambda-\mu}(1-s)^{\lambda -\mu'}}{(1-ts)^{\lambda+1}}\int_0^{\infty} \mathrm{e}^{-x}x^{\lambda}\ \mathrm{d}x\\[.5cm] ={}&\Gamma(\lambda+1)\sum_{l=0}^{\infty}\binom{\lambda-\mu}{l}(-t)^{l} \sum_{l'=0}^{\infty} \binom{\lambda-\mu'}{l}(-s)^{l'}\sum_{k=0}^{\infty}\binom{-\lambda-1}{k}(-ts)^{k} \\[.5cm] ={}&\Gamma(\lambda+1)\sum_{n=0}^{\infty}\sum_{n=0}^{\infty}t^{n}s^{n'} \sum_{k=0}^{\infty}(-)^{n+n'+k} \binom{\lambda-\mu}{n-k}\binom{\lambda-\mu'}{n'-k}\binom{-\lambda-1}{k}. \end{aligned}\]

Comparing the coefficients of $t^{n}s^{n’}$ on both sides, we can find:

\[\int_0^{\infty} z^{\lambda} \mathrm{e}^{-z} L_{n}^{\mu}(z) L_{n'}^{\mu'}(z) \ \mathrm{d}z = (-)^{n+n'}\Gamma(\lambda+1) \sum_{k=0}^{\infty} \binom{\lambda-\mu}{n-k}\binom{\lambda-\mu'}{n'-k}\binom{\lambda+k}{k} .\]

Let $\lambda = \mu = \mu’$, the above equation becomes

\[\int_0^{\infty}z^{\mu} \mathrm{e}^{-z} L_{n}^{\mu}(z) L_{n'}^{\mu}(z) \ \mathrm{d}z = \frac{\Gamma(\mu+n+1)}{n!}\delta_{nn'}.\]

Thus, we have proved the lemma 1.

Lemma 2

\[\color{Green}{\text{Lemma 2:}\quad\displaystyle{J_{\mu}(2\sqrt{xz})\mathrm{e}^{x}(xz)^{-\mu/2} = \sum_{n=0}^{\infty}\frac{x^{n}}{\Gamma(n+\mu+1)}L_{n}^{\mu}(z)}} .\]

From the definition of the Bessel functions:

\[J_{\nu}(z) =\left(\frac{z}{2}\right)^{\nu}\sum_{k=0}^{\infty} \frac{(-)^{k}}{k!} \frac{1}{\Gamma(\nu+k+1)}\left(\frac{z}{2}\right)^{2k},\]

we can find

\[J_{\mu}(2\sqrt{xz})\mathrm{e}^{x} (xz)^{-\mu/2} = \sum_{k=0}^{\infty}\frac{(-)^{k}}{k!} \frac{(xz)^{k}}{\Gamma(\mu+k+1)}\mathrm{e}^{x} .\]

On the other hand, from the definition of the associated Laguerre polynomials:

\[L_{n}^{\mu}(z) = \sum_{m=0}^{n} (-)^{m}\binom{n+\mu}{n-m} \frac{z^{m}}{m!},\]

we can calculate the right hand side of the lemma 2,

\[\begin{aligned} &\sum_{n=0}^{\infty}\frac{x^n}{\Gamma(n+\mu+1)}L_{n}^{\mu}(z) \\[.5cm] ={}& \sum_{n=0}^{\infty}\sum_{m=0}^{n}\frac{(-)^{m}}{m!} \frac{z^{m}x^{n}}{\Gamma(n-m+1)\Gamma(\mu+m+1)}\\[.5cm] ={}&\sum_{n=0}^{\infty}\sum_{m=0}^{n}\frac{x^{n-m}}{\Gamma(n-m+1)}\frac{(-xz)^{m}}{m!\Gamma(\mu+m+1)} \quad \text{Cauchy product}\\[.5cm] ={}&\sum_{k=0}^{\infty}\frac{x^{k}}{\Gamma(k+1)}\cdot \sum_{k=0}^{\infty}\frac{(-xz)^{k}}{k! \Gamma(\mu+k+1)}\\[.5cm] ={}&\sum_{k=0}^{\infty}\frac{(-)^{k}}{k!} \frac{(xz)^{k}}{\Gamma(\mu+k+1)}\mathrm{e}^{x} .\end{aligned}\]

It coincides exactly with the result of the Bessel function expansion, so we have proved the lemma 2.

Answer

Back to our original question, using lemma 2, we can find that

\[J_{\mu}(2\sqrt{xz})z^{-\mu/2} = \mathrm{e}^{-x}x^{\mu/2}\sum_{n=0}^{\infty}\frac{x^{n}}{\Gamma(n+\mu+1)}L_{n}^{\mu}(z) .\]

So we can calculate the original integral directly,

\[\begin{aligned} &\int_{0}^{\infty} z^{\mu/2} \mathrm{e}^{-z} L_{n}^{\mu}(z)J_{\mu}\left(2\sqrt{xz}\right) \mathrm{d}z\\[.5cm] ={}&\int_{0}^{\infty} z^{\mu} \mathrm{e}^{-z} L_{n}^{\mu}(z) z^{-\mu/2}J_{\mu}\left(2\sqrt{xz}\right) \mathrm{d}z\\[.5cm] ={}&\sum_{n'=0}^{\infty}\mathrm{e}^{-x}x^{\mu/2} \cdot \frac{x^{n'}}{\Gamma(n'+\mu+1)}\int_0^{\infty}z^{\mu} \mathrm{e}^{-z} L_{n}^{\mu}(z)L_{n'}^{\mu}(z)\ \mathrm{d}z,\quad \color{Green}\text{Lemma 1}\\[.5cm] ={}&\sum_{n'=0}^{\infty}\mathrm{e}^{-x} \frac{x^{n'+\mu/2}}{\Gamma(n'+\mu+1)}\frac{\Gamma(\mu+n+1)}{n!}\delta_{nn'}\\[.5cm] ={}&\frac{1}{n!} x^{n+\mu/2}\mathrm{e}^{-x}. \end{aligned}\]

Thus we have proved the identity,

\[\color{red}{\int_{0}^{\infty} z^{\mu/2} \mathrm{e}^{-z} L_{n}^{\mu}(z)J_{\mu}\left(2\sqrt{xz}\right) \mathrm{d}z = \frac{1}{n!} x^{n+\mu/2}\mathrm{e}^{-x}.}\]