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In trying to solve another the problem posed in the question https://www.mathoverflow.net/q/385777/78539, I'm led to consider the following problem.

Let $\mu_\gamma$ be the Marchenko-Pastur distribution with parameter $\gamma \in (0,1)$. Note that $\mu_\gamma$ is supported on $[t_-,t_+]$, where $t_{\pm} = (1\pm \sqrt{\gamma})^2$. For $\lambda \ge 0$, define $$ I_\gamma(\lambda):=\int_{t_-}^{t_+} \dfrac{t}{(t + \lambda)^2}d\mu_\gamma(t). $$ It is clear that $I_\gamma$ is a decreasing function on $[0,\infty)$.

Question 1. What is an analytic formula for $I_\gamma(\lambda)$ ?

Note. I'm fine with good lower-bounds on $I_\gamma(\lambda)$

Solution for the extreme cases: $\lambda \to 0^+$ and $\lambda \to \infty$

For any $t > 0$, one may write $$ \frac{t}{(t + \lambda)^2} = \frac{1}{t+\lambda} - \frac{\lambda}{(t+\lambda)^2} = \begin{cases}\dfrac{1}{t},&\mbox{ if }\lambda \to 0^+,\\ \dfrac{1}{t+\lambda}+\mathcal O(\dfrac{1}{\lambda}),&\mbox{ if }\lambda \to \infty.\end{cases} $$

Let $m_\gamma(z) := \int \dfrac{1}{z-t}d\mu_\gamma(t)$ be the Stieltjes transform of $\mu_\gamma$. It is well-known that for all real $z>0$, $$ m_\gamma(-z)=\frac{-(1-\gamma + z) + \sqrt{(1-\gamma+z)^2 + 4\gamma z}}{2\gamma z} $$ be the Cauchy transform of $\mu_\gamma$.

  • Case 1: $\lambda \to 0^+$. One recognizes $ I_\gamma(0^+) = \lim_{\lambda \to 0^+}\int_{t_-}^{t_+}\dfrac{1}{t + \lambda}d\mu_\gamma(t) = \lim_{\lambda \to 0^+}m_\gamma(-\lambda)=\dfrac{1}{1-\gamma}. $

  • Case 2: $\lambda \to \infty$. Likewise, one recognizes $I_\gamma(\infty) = \lim_{\lambda \to \infty}\int_{t_-}^{t_+}\dfrac{1}{t + \lambda}d\mu_\gamma(t) = \lim_{\lambda \to \infty}m_\gamma(-\lambda) = 0$.

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  • $\begingroup$ I think case 1 should be $1/(1-\gamma)$ (it should not vanish for $\gamma=0$). $\endgroup$
    – Carlo Beenakker
    Commented Mar 7, 2021 at 19:28
  • $\begingroup$ @CarloBeenakker Indeed, thanks again. $\endgroup$
    – dohmatob
    Commented Mar 7, 2021 at 19:35

1 Answer 1

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$$I(\lambda)=\int_{t_-}^{t_+}\frac{\sqrt{\left(t_+-t\right) \left(t-t_-\right)}}{2 \pi {\gamma} (t+\lambda)^2}\,dt =\frac{-\sqrt{ {\gamma}^2+2 {\gamma} ( {\lambda}-1)+( {\lambda}+1)^2}+ {\gamma}+ {\lambda}+1}{2 {\gamma} \sqrt{ {\gamma}^2+2 {\gamma} ( {\lambda}-1)+( {\lambda}+1)^2}}.$$

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  • $\begingroup$ Thanks for the nice analytic answer (upvoted). Please could kindly add a short sentence explaining how you go the answer ? Is it via some kind of symbolic integration (wolfram, sympy, mathematica, etc.) or is there a secret table of MP-integrals hidden somewhere of which I'm not aware :) $\endgroup$
    – dohmatob
    Commented Mar 7, 2021 at 18:59
  • $\begingroup$ Mathematica output $\endgroup$
    – Carlo Beenakker
    Commented Mar 7, 2021 at 18:59
  • $\begingroup$ The integral is "elementary", via change of variable to a rational coordinate on $u^2 = (t_+^2-t) (t - t_-^2)$, but cumbersome to work out by hand. Using mathematica or similar packages such as maxima is probably the best way to proceed, as long as you check that the formula matches the output of numerical integration for some representative values of $t_\pm$ and $\lambda$. $\endgroup$
    – Noam D. Elkies
    Commented Mar 7, 2021 at 19:03
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    $\begingroup$ @CarloBeenakker Thanks for the clarification. Time to start using mathematica. $\endgroup$
    – dohmatob
    Commented Mar 7, 2021 at 19:04
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    $\begingroup$ @NoamD.Elkies Thanks for mentioning the opensource tool "Maxima". Just tried it at the moment, and it seems really easy to use. $\endgroup$
    – dohmatob
    Commented Mar 7, 2021 at 19:26

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