1. 2012

    A simple regularization example

    Earlier this month I promised to show you a simple example of regularization. Well, here it is. This is a particular combination of integrals I’ve been working with a lot lately:

    $$\iint_{\mathbb{R}^2} \uddc\mathbf{x}\frac{e^{-x^2}}{x^2}e^{-i\mathbf{k}\cdot\mathbf{x}} - \iint_{\mathbb{R}^2} \frac{\uddc\mathbf{y}}{y^2}e^{-i\xi\mathbf{k}\cdot\mathbf{y}}$$

    A quick look at the formulas shows you that both integrands have singularities at \(\mathbf{x} = 0\) and \(\mathbf{y} = 0\).

    OK, well, that’s why we have two integrands. We can change variables from \(\xi\mathbf{y}\to\mathbf{x}\), subtract them, and the singularities will cancel out, right? You can do this integral by hand in polar coordinates, or just pop it into Mathematica:

    $$\iint_{\mathbb{R}^2} \uddc\mathbf{x}\frac{e^{-x^2} - 1}{x^2}e^{-i\mathbf{k}\cdot\mathbf{x}} = -\pi\Gamma\biggl(0, \frac{k^2}{4}\biggr)$$

    Just one problem: that’s not the right answer! You can tell because the value of this integral had better depend on \(\xi\), but this expression doesn’t. This isn’t anything so mundane …

  2. 2012

    Not really a simple regularization analogy

    Last year I posted about an infinite sum involving the mean of the harmonic numbers,

    $$\lim_{n\to\infty}\frac{1}{n}\sum_{k=1}^{n}\frac{1}{k} = \lim_{n\to\infty}\frac{\ln n}{n}$$

    The method I used to evaluate this was to approximate the sum by an integral. This is a technique that is used in various places in physics, such as in the computation of the Fermi surface in metals.

    In this particular case, we only cared about the large-\(n\) limiting behavior of the sum, namely that it grows sublinearly for large \(n\). But suppose you wanted to know whether there was, say, a constant term as well. Here’s one way to figure that out. Instead of converting the entire sum to an integral, you choose some cutoff value \(a\), and keep the first \(a\) terms in the sum explicitly.

    $$\sum_{k=1}^{n}\frac{1}{k} \approx \sum_{k=1}^{a}\frac{1}{k} + \sum_{k=a+1}^{n}\frac{1}{k} = \sum_{k=1}^{a}\frac{1}{k} + \ln\frac{n}{a+1}$$

    Now you have a value, \(a\), that represents a “break” in your sequence, but it’s an arbitrary …