Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
    
It should be at least convergent for $x$ in the integration range (which in this case doesn't seem to be given). –  Fabian Apr 7 '11 at 17:54
    
A sufficient criterion is that the sum converges uniformly. –  Fabian Apr 7 '11 at 17:57
2  
The integrand has a nasty fence of poles within the interval of integration... why would somebody expect the whole mess to converge? –  J. M. Apr 10 '11 at 5:37
    
@JM, I have no idea, but it did bring up some questions on how to proceed. –  rcollyer Apr 10 '11 at 14:27
    
@J.M.: because in physics all functions are well-behaved. (Whatever well-behaved means in context). They are only outside $C^{\infty}$ when it is convenient. I still (many years later) remember the revolt when the professor talked about the derivative of the delta function... –  Ross Millikan Apr 12 '11 at 4:42
show 2 more comments

3 Answers

up vote 8 down vote accepted

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).

share|improve this answer
    
Could you say more specifically for which case uniform convergence isn't sufficient? The proof I linked to shows that in a certain sense it's always sufficient (but that sense might not apply to the situation you have in mind). –  joriki Apr 7 '11 at 18:31
3  
@Joriki: the functions $f_n(x)=\tfrac1n$ converge uniformly to the zero function on $\mathbb R$, yet $\int_{\mathbb R}f_n\not\to0$! This is a silly example, but shows you where the problem lies. –  Mariano Suárez-Alvarez Apr 7 '11 at 18:34
    
@Fabian the functions are constant so the convergence is independent of $x$ and thus uniform. –  shamovic Apr 7 '11 at 19:12
    
the wikipedia page with Lebesque's theorem doesn't exist as such, did you mean? –  rcollyer Apr 8 '11 at 17:19
    
@rcollyer: I've fixed the links; they contained an excess slash in the end. –  joriki Apr 8 '11 at 17:27
add comment

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

share|improve this answer
    
Maybe I'm wrong, but it seems that the author assumes that the limits exist and are finite. Is it true or am I missing something? –  shamovic Apr 7 '11 at 19:12
add comment

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.

share|improve this answer
    
I don't see how to get the principal value of the integral from 0 to any number less than $\pi$... on the other hand, the principal value integrals with respect to the other poles are expressible in terms of digamma functions. –  J. M. Apr 10 '11 at 17:36
    
Oh, you're right. It also seems to me that there is no way to cancel out the pole at the origin, causing the value to explode to $+\infty$. I shoud have checked its existence first... –  sos440 Apr 10 '11 at 18:46
    
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 $-i/(1-e)$. –  rcollyer Apr 10 '11 at 19:06
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.