Integral involving a generalized Laguerre Polynomial

Question

I want to evaluate the following integral:

$$ \int_0^\infty z^{1/2}e^{-a\space z}L_{m}^{1/2}(z)\space dz, $$

where $L_{m}^{1/2}$ is a generalized Laguerre polynomial. I found a certain indefinite integral here, but im not sure how to evaluate it for $z=\infty$. I calculated the integral for the first few m with a software and it appears there is a simple closed form for it. The results are: $$ \frac{3 \sqrt{\pi } (a-1)}{4 a^{5/2}},\frac{15 \sqrt{\pi } (a-1)^2}{16 a^{7/2}},\frac{35 \sqrt{\pi } (a-1)^3}{32 a^{9/2}},\frac{315 \sqrt{\pi } (a-1)^4}{256 a^{11/2}}, $$ for $m=1,2,3,4$.

Can you help me to find a closed form for any $m\in \mathbb{N}$? I'm more interested in the result than the derivation.

Answer 1

Using the substitution $u=az\to dz=\frac1adu$ gives the integral $$\int_0^\infty (\frac{u}a)^\frac12e^{-u}L_m^\frac12(\frac{u}a)(\frac1a)du=\frac{1}{a\sqrt{a}}\int_0^\infty e^{-u}u^\frac12L_m^\frac12(\frac{u}a)du$$

Then by using the generating function $$\sum_{k=0}^\infty t^ke^{-u}u^\frac12L_k^\frac12(\frac{u}{a})=\frac{u^\frac12}{(1-t)^\frac32}e^{-\frac{u(t-at+a)}{a(1-t)}}$$ we can integrate every polynomial simultaneously giving $$\int_0^\infty\sum_{k=0}^\infty t^ke^{-u}u^\frac12L_k^\frac12(\frac{u}{a})du=\int_0^\infty\frac{u^\frac12}{(1-t)^\frac32}e^{-\frac{u(t-at+a)}{a(1-t)}}du$$ $$=\frac1{(1-t)^\frac32}\int_0^\infty u^\frac12e^{-\frac{u(t-at+a)}{a(1-t)}}du$$ Then we can use the substitution $v=\frac{u(t-at+a)}{a(1-t)}\to du=\frac{a(1-t)}{t-at+a}dv$ $$=\frac1{(1-t)^\frac32}\int_0^\infty \Big(\frac{a(1-t)}{t-at+a}v\Big)^\frac12e^{-v}\Big(\frac{a(1-t)}{t-at+a}\Big)dv$$ $$=\frac1{(1-t)^\frac32}\Big(\frac{a(1-t)}{t-at+a}\Big)^\frac32\int_0^\infty v^\frac12e^{-v}dv$$ $$=\frac1{(1-t)^\frac32}\Big(\frac{a(1-t)}{t-at+a}\Big)^\frac32\Gamma(\frac32)$$ $$=\Big(\frac{a}{t-at+a}\Big)^\frac32\frac{\sqrt{\pi}}{2}$$ $$=(t(1-a)+a)^{-\frac32}\frac{a\sqrt{a\pi}}{2}$$ $$=(1+\frac{t(1-a)}{a})^{-\frac32}\frac{\sqrt{\pi}}{2}$$ $$=\frac{\sqrt{\pi}}{2}\sum_{k=0}^\infty\binom{-\frac32}{k}\Big(\frac{t(1-a)}{a}\Big)^k$$ $$=\sum_{k=0}^\infty\frac{\sqrt{\pi}}{2}\binom{-\frac32}{k}\Big(\frac{1-a}{a}\Big)^kt^k$$

So the value of the integral for a given value of $k$ (or $m$ in your case) is $$\int_0^\infty z^{1/2}e^{-a\space z}L_k^{1/2}(z)\space dz = \frac{1}{a\sqrt{a}}\Big(\frac{\sqrt{\pi}}{2}\binom{-\frac32}{k}\Big(\frac{1-a}{a}\Big)^k\Big)=\boxed{\frac1{2a}\sqrt{\frac{\pi}{a}}\binom{-\frac32}{k}\Big(\frac{1-a}{a}\Big)^k}$$

Answer 2

Using the generating function $$\sum_{n=0}^{\infty} L_{n}^{(\alpha)}(x) \, t^{n} = (1-t)^{-\alpha -1} \, e^{-xt/(1-t)}$$ then, where the integral is defined by $$I_{n}(a) = \int_{0}^{\infty} e^{-a x} \, L_{n}^{(1/2)}(x) \, \sqrt{x} \, dx$$, leads to \begin{align} \sum_{n=0}^{\infty} I_{n}(a) \, t^{n} &= \int_{0}^{\infty} e^{-a x} \, \sqrt{x} \, (1-t)^{-3/2} \, e^{-x t/(1-t)} \, dx \\ &= (1-t)^{-3/2} \, \int_{0}^{\infty} e^{-(a + t/(1-t)) x} \, \sqrt{x} \, dx \\ &= (1-t)^{-3/2} \, \int_{0}^{\infty} e^{- p x} \, x^{3/2 -1} \, dx, \hspace{5mm} \text{with} \hspace{5mm}p = \frac{a - (a-1) t}{1-t} \\ &= \frac{\Gamma(3/2)}{p^{3/2}} \, (1-t)^{-3/2} \\ &= \frac{\Gamma(3/2)}{a^{3/2}} \, \left(1 - \frac{a-1}{a} \, t\right)^{-3/2} \\ &= \frac{\Gamma(3/2)}{a^{3/2}} \, \sum_{n=0}^{\infty} \frac{(3/2)_{n}}{n!} \, \left(\frac{a-1}{a}\right)^{n} \, t^{n}. \end{align}

This leads to $$ I_{n}(a) = \frac{\Gamma\left(n + \frac{3}{2}\right) \, (a-1)^{n}}{n! \, a^{n+3/2}}. $$