I want to lower bound the expected value of the square root of a randomly chosen eigenvalue of a Wishart matrix.


To get the bound I want I need a lower bound on


$$T\_n = \int\_0^\infty\sqrt{x}e^{-x}L\_n(x)^2dx,$$


for $n>1$ where $L\_n(x)$ is the $n$-th Laguerre polynomial.


Mathematica convinced me that $T\_n > \sqrt{n+1}$ but I wasn't able to find a proof for this.


Does anyone have an idea of how to estimate this integral?


Note: if the $\sqrt{x}$ is replaced by $x$ then Wikipedia has a closed form solution for it.


Thanks!



Lower bound of integral involving Laguerre polynomials
Here is a proof.


First evaluate the integral. I use the generating function for squares of Laguerre polynomials ($I\_0$ the modified Bessel function):


$$
\sum\_{n=0}^{\infty}L\_{n}(x)^2 z^n = \frac{1}{1-z} \exp\left(-\frac{2 x z}{1-z}\right) I\_{0}\left(\frac{2 x \sqrt{z}}{1-z}\right)
$$


for $|z|<1$ (All the formulas I use, one can find it in e.g. Gradshteyn and Ryzhik.) 


This gives for the integral


$$
T\_{n} = \int\_{0}^{\infty} dx \sqrt{x} e^{-x} L\_{n}(x)^2 = [z^n] \frac{1}{1-z} \int\_{0}^{\infty} dx \sqrt{x} \exp\left(-x \frac{1+z}{1-z}\right) I\_{0}\left(x \frac{2 \sqrt{z}}{1-z}\right)
$$


$[z^n]$ means, as usual, "take the coefficient of $z^n$ in the (formal) power expansion in $z$ of the following expression".


The integral can be evaluated (it is in Gradshteyn and Ryzhik) and the result is


$$
T\_{n} = [z^n] \frac{\sqrt{\pi}}{2} (1-z)^{-1} P\_{1/2}\left(\frac{1+z}{1-z}\right)
$$


with $P\_{1/2}(x)$ the Legendre Function.


Expanding the Legendre function and the prefactor in powers of $z$ gives for the coefficient of $[z^n]$


$$
T\_{n} = \frac{1}{8\sqrt{\pi}}\sum\_{m=0}^{n}\frac{\Gamma\left(m-\frac{1}{2}\right) (2(n-m)+1)!}{4^{n-m} m!^2 (n-m)!^2}
$$


Now I was lazy and asked Mathematica to simplify, which gives


$$
T\_{n}= \frac{\Gamma(n + \frac{3}{2})}{\Gamma(n+1)} {\_3}F\_{2}\left(-\frac{1}{2},-\frac{1}{2},-n;1,-\frac{1}{2}-n;1\right)
$$


with the generalized hypergeometric function ${\_3}F\_{2}$ evaluated at $1$.


With the exact solution at hands we are almost done. We observe that the sum defining the hypergeometric function 


$$
{\_3}F\_{2}\left(-\frac{1}{2},-\frac{1}{2},-n;1,-\frac{1}{2}-n;1\right)
=\sum\_{m=0}^{\infty} \frac{\left(-\frac{1}{2}\right)\_{m}^{2}(-n)\_m}{m!^2 \left(-n-\frac{1}{2}\right)\_{m} }
$$


has only positive terms (the minus signs in the Pochhammer symbols $(-n)\_m$ and $(-n-\frac{1}{2})\_m$ cancel.). We get a lower bound by taking only the first two expansion terms of ${\_3}F\_{2}$ and the well known estimate for the Gamma functions in front


$$
\frac{\Gamma(n + \frac{3}{2})}{\Gamma(n+1)} \geq n^{1/2}
$$


and find


$$
T\_{n} \geq n^{1/2} \left(1+\frac{n}{2(n+1)}\right)
$$


which is larger than $\sqrt{n+1}$ for $n\geq 2$.


Btw, the asymptotic ($n\rightarrow \infty$) value of $T\_n$ is


$$
T\_{n}\sim \frac{4}{\pi} n^{1/2}
$$


I am pretty sure that the calculation can be done more elegantly, but had not the time to dig deeper.


**Edit:**
With the same trick one finds more generally:


$$
 \int\_{0}^{\infty} dx \, x^{\nu} e^{-x} L\_{n}(x)^2 = \frac{\Gamma(n+\nu+1)}{\Gamma(n+1)} {\_3}F\_{2}\left(-n,-\nu,-\nu;1,-n-\nu;1\right) = n^{\nu} {\_2}F\_{1}(-\nu,-\nu;1;1)) (1+ O(n^{-1})) 
$$
The last term simplifies to:
$$
\frac{\Gamma(n+\nu+1)}{\Gamma(n+1)} \frac{\Gamma(2 \nu +1)}{\Gamma(\nu+1)^2} (1+ O(n^{-1})) 
$$


for $\nu > 0$.


**Edit:**
After some massaging with Mathematica I could determine the coefficient of $n^{-1}$ in the above expansion:
$$
 \frac{\Gamma(n+\nu+1)}{\Gamma(n+1)} {\_3}F\_{2}\left(-n,-\nu,-\nu;1,-n-\nu;1\right) = n^{\nu} {\_2}F\_{1}(-\nu,-\nu;1;1)) (1 + \frac{\nu}{2 n} + O(n^{-1-2 \nu}))
$$
The determination of the next order coefficient as well as the exact exponent is rather cumbersome. One could do better than $O(n^{-1-2 \nu})$.


