5
$\begingroup$

I ask this question because similar questions don't address contour integrals and infinite series specifically, only evaluating a contour integral by converting it into an infinite series.

I'm trying to take the contour integral of an infinite series. My understanding is that under Fubini's theorem that under certain conditions, integrals can commute with each other, summations can commute with each other, and integrals can commute with summations.

However, the requirement (as I understand it) is that the summation must converge for the summation and integral to commute; but if I knew whether or not the summation converges, I would not need to integrate first.

Here's an example:

$$\oint_C \sum_{n=0}^ \infty a_{n} \cdot f_{n} \big(x\big) \stackrel{?}{=} \sum_{n=0}^ \infty \oint_C a_{n} \cdot f_{n} \big(x\big)$$

EDIT: I'd like to further explain my question and perhaps why it's confusing. I want to find an analytical solution to the area of a simply-connected space. I found that I could conveniently describe this space using parametric equations, and therefore could use Green's Theorem and integrate around the boundary to obtain the area. The parametric equations are:

$$x[\theta] = a \cdot Cos[\theta]$$

$$y[\theta] = \sum_{n=0}^ \infty \frac{b_{n}}{n} \cdot Sin[(2n+1) \cdot \theta] + \frac{c_{n}}{n} \cdot Cos[(2n+1) \cdot \theta ]$$

From there you can calculate $dx/d\theta$ and $dy/d\theta$ and from there $ A = \frac{1}{2} \oint_C x \cdot dy-y \cdot dx$.

However, summing the integral means summing the individual areas of each term's space, while integrating the sum means integrating the area from the sum of the contribution of each term to the coordinates at a given point on the contour.

These seem to be different operations to me and therefore I'm confused as to under what circumstances contour integral's and infinite series commute.

EDIT 2: I'm asking this question because I can do the integral on the right hand side (the sum of the contour integrals) in the example, but not the other (the contour integral of the sum), and need to know (1) how to test to see if both expressions are equivalent. The suggestions so far have been to integrate both sides and see if they are equivalent, but I can't do that, hence the need for Fubini's Theorem.

$\endgroup$
16
  • 1
    $\begingroup$ Interchanging limit operations like that is a function of absolute convergence, this is Fubini's theorem. en.wikipedia.org/wiki/Fubini's_theorem $\endgroup$ Dec 7 '16 at 22:35
  • $\begingroup$ There are reasons to swap integration and summation if you already know a sum converges. It's much easier to tell that a sum does converge than to see what it converges to, Once you know a sum converges, integrating the terms may help you evaluate it. $\endgroup$ Dec 7 '16 at 23:33
  • $\begingroup$ $\bullet$ denotes pointwise multiplication of mappings? $\endgroup$
    – GFauxPas
    Dec 7 '16 at 23:48
  • $\begingroup$ @GFauxPas dot product between the coefficient and basis function. $\endgroup$
    – iwantmyphd
    Dec 8 '16 at 0:11
  • $\begingroup$ It hugely depends on the size of the b_n's,c_n's. Why do you think $dy/d\theta$ is so simple to compute? $\endgroup$
    – zhw.
    Jul 18 '17 at 22:47
2
$\begingroup$

To elaborate on my comment, the Fubini-Tonelli theorem states the conditions for interchange of the limit operations, in particular for integrals. Summation is integration on a discrete space, so this is exactly the conditions for use of the theorem. As always, there are ways to break this if you do not have absolute convergence.

Beyond just the general case, Robert Israel correctly notes that the usual situation in complex analysis is that the series of functions converges absolutely and uniformly on compact subsets of some domain, in particular if the contour is a rectifiable curve in $\Bbb C$, the conditions are satisfied.

$\endgroup$
5
  • 1
    $\begingroup$ In particular, the usual situation in complex analysis is that the series converges absolutely and unifomly on compact subsets of some domain, in particular on the contour $C$ which is a rectifiable curve, and then the conditions are satisfied. $\endgroup$ Dec 7 '16 at 23:01
  • $\begingroup$ @RobertIsrael That's absolutely worth adding to the general case, I agree. Thanks for noting it. $\endgroup$ Dec 7 '16 at 23:43
  • $\begingroup$ @AdamHughes Perhaps I'm not understanding how absolute convergence applies to a function with no input value. Let's say An*Sin[n*x], doesn't the condition of absolute convergence depend on the portion of the path around which you are integrating? For some parts of the contour, it would absolutely converge, for other it might only conditionally converge. So I'm not clear on how you can test for this without choosing a portion of the path, i.e. the bounds of the integral. $\endgroup$
    – iwantmyphd
    Dec 8 '16 at 0:17
  • $\begingroup$ @iwantmyphd you make it absolutely pointwise, that's how typical normal modes of usual convergence extend to functions. $\endgroup$ Dec 8 '16 at 0:22
  • $\begingroup$ @AdamHughes Perhaps I've asked the wrong question; I really don't follow what you mean. $\endgroup$
    – iwantmyphd
    Dec 8 '16 at 0:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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