17
$\begingroup$

On stackoverflow, a question was asked about getting Mathematica to evaluate the integral,

$\int^\infty_0 \frac{e^{-x}}{\sin x} \, \mathrm{d}x$

which we know is divergent. In one of the answers, the integrand is replaced with its Taylor expansion, and integrated term by term, and in physics, it is often taken for granted that this works. But, under what circumstances is this valid for improper integrals, in general? More precisely, what must be done to properly interchange the two limiting processes?

  • 0
    It's not true. In Physics, it's not taken for granted that it works. Those types of integrals involves 'energy denominators' and the divergence appears since some interaction is neglected. Once that interaction is recovered the energy poles becomes complex and anything works fine. Sometimes, it's quite difficult to include the neglected interaction and we can switch to some kind of 'disorder average' which is pretty justified since the original model was 'too much' an ideal one.2014-06-18

3 Answers 3

9

Uniform convergence is often both too strong and too weak (it isn't sufficient for an improper integral over an infinite interval). Better ones are dominated convergence and monotone convergence (see the Lebesgue dominated convergence theorem and the Lebesgue monotone convergence theorem).

  • 0
    @rcollyer: I've fixed the links; they contained an excess slash in the end.2011-04-08
3

Here's a nice general treatment of the interchange of limiting processes using uniform convergence. It applies to integration, differentiation and series alike.

  • 0
    @shamovic: I think you're right, but how is that relevant? Those conditions are fulfilled here.2015-08-29
3

In some cases, a class of theorems called Tauberian theorems can help you to justify interchanging the order of two limiting operators. For example, if the improper integral $\int_{0}^{\infty} f(x) \, dx$ exists, then $\int_{0}^{\infty} \int_{0}^{\infty} x^{n} s^{n-1} f(x) \; e^{-xs} \, ds dx = \int_{0}^{\infty} \int_{0}^{\infty} x^{n} s^{n-1} f(x) \; e^{-xs} \, dx ds$ holds for all $n$. (Of course, existence of both iterated integrals are also guaranteed.) Originally, Tauberian theorems are answers to the following question: For what condition (Tauberian condition) ensures that a stronger summability method implies a weaker summability? Since a stronger summability method often exploits a good approximation to the identity, these theorems may be regarded as a special kind of interchanging the order of limiting operators. For example, a function $f(x)$ is Abel-summable to $I$ if $\lim_{\delta \to 0+} \int_{0}^{\infty} f(x) e^{-\delta x} \, dx$ exists with the value $I$. Then this reduces to the ordinary summability if we have $\lim_{\delta \to 0+} \int_{0}^{\infty} f(x) e^{-\delta x} \, dx = \int_{0}^{\infty} f(x) \, dx = \int_{0}^{\infty} \lim_{\delta \to 0+} f(x) e^{-\delta x} \, dx.$

For the integral in question, we may understand it as the Cauchy principal value. That is, we identify this integral with $ \lim_{\epsilon \to 0+} \sum_{n=1}^{\infty} \int_{\pi(n-1) + \epsilon}^{ \pi n - \epsilon} \frac{e^{-x}}{\sin x} \, dx.$ By circumventing poles, this integral can be managed by several techniques.

  • 0
    We determined that the pole at the origin was the cause of the problem. But, I also found that if the integration limits each corresponded to a pole, i.e. $x \in [\pi(n-1), \pi n]$, you are unable to compensate for the pole. But, if you shifted the integral to $x \in [\pi(n - 1/2), \pi (n + 1/2)]$ the poles can be compensated for, and the series of integrals is an alternating series, $i \sum (-1/e)^n$, which converges to $-i/(1-e)$.2011-04-10