# Stieltjes Transform

Stieltjes transform for a distribution $F(x)$ is defined as

$$m(z)=\int \frac{dF(x)}{x−z}$$ where z is complex with positive imaginary parts and $F(x)$ is a distribution function.

Basically, I am stuck with evaluating Stieltjes transforms for given PDFs.

I referred to "Random matrix theory and wireless communications By Antonia M. Tulino, Sergio Verdú" and on page 37 http://bit.ly/qnf39U, they give an example

$$S(z)=\frac{1}{2π}\int_{-2}^2 \frac{\sqrt{4−x^2}}{(x−z)}\;dx = \frac{1}{2}\left[−z\pm \sqrt{z^2−4}\right].$$ I can't see how they came to this result. Any leads?

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This answer is late, but anyway

One can calculate the integral in the question by integrating over a complex contour around the branch cut [-2, 2]. This is standard technique in complex integration.

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I guess you can prove that this identity holds by using a Riemann-Hilbert problem.

Consider the follwing problem:

(RHP) Seek an analytic function $f:\mathbb{C}\setminus [-2,2]\to\mathbb{C}$ such that:

(i) $f_{+}(s)=f_{-}(s)-\sqrt{4-s^2}$ for $s\in [-2,2]$
(ii) $f(z)\to 0$, when $|z|\to\infty$.

Here $f_{+}(s)$ denotes the boundary value of $f(z)$ when $z$ tends to $s\in [-2,2]$ from the upper-half plane. Similarly we define $f_{-}(s)$ but now the point $z$ approaches to $s$ from the lower-half plane. Note that we are considering the interval oriented from $-2$ to $2$.

Using the Morera's and Liouville's Theorem you can show that the solution is unique (nice exercise). To proceed we remark that the Plemelj's formula give us the solution for this scalar Riemann-Hilbert problem, that is, $$f(z)=\frac{1}{2\pi i}\int_{[-2,2]}\frac{f_{+}(s)-f_{-}(s)}{s-z}\ ds$$ So your task now is only compute the jumps for $\frac{1}{2}[z+\sqrt{z^2-4}]$ in the contour $[-2,2]$. Replacing the orientation of the interval you will get the second value $\frac{1}{2}[z-\sqrt{z^2-4}]$. I will leave this final computation for you, if you get any troubles come back.

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Thanks. But I'm not looking to prove the identity. I am trying to get the result on the right by solving the integral. – sauravrt Oct 21 '11 at 13:06
@sauravrt, your comment is a bit confusing. Leandro is giving an outline for how you can prove that the integral on the left hand side equals the expression on the right-hand side. What more did you desire? – user16299 Oct 21 '11 at 19:04
@Leandro: $z \in \mathbb{C}^{+} \backslash \mathbb{R}$. Given this domain of z, I don't understand how to compute the jumps for $\frac{1}{2}[z + \sqrt{(z^2 - 4)}]$ in the contour [-2,2] – sauravrt Oct 22 '11 at 1:48
I was able to get the result with help from Mathematica. I think my confusion is that since the integrand has a complex variable $z$, should I treat it as a constant when solving for the integral? – sauravrt Oct 22 '11 at 2:42
@sauravrt: Yes; in the integral, $x$ is the variable, $z$ is a complex parameter – user16299 Oct 22 '11 at 16:26