Timeline for Interchanging sums and integrals in a specific instance
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Nov 2, 2017 at 3:30 | comment | added | user78249 | Thanks for your answer. I apologize for misusing the term contour. I wrote this question a little too fast. | |
| Nov 2, 2017 at 2:03 | vote | accept | CommunityBot | ||
| Nov 2, 2017 at 2:01 | vote | accept | CommunityBot | ||
| Nov 2, 2017 at 2:02 | |||||
| Oct 31, 2017 at 19:10 | comment | added | Alexandre Eremenko | I agree that the question is very confusing. First of all it is not explained what is $\int f$. Is this $\int fdz$ or $\int f |dz|$ :-) | |
| Oct 31, 2017 at 15:41 | comment | added | js21 | @Alexandre Emerenko: OP's terminology confused me as well. If I understand correctly by "contour" he or she means "path" and by "closed contour" he or she means what most of us would simply call "contour". Besides that, you are right, I do not know why I absolutely wanted a non-closed path of integration in my counterexample. | |
| Oct 31, 2017 at 13:22 | comment | added | Alexandre Eremenko | This holds trivially because $\int_Cf_n=0$ for every analytic function. | |
| Oct 31, 2017 at 13:18 | comment | added | js21 | @Alexandre Emerenko: The OP required $\sum_{n=0}^\infty |\int_C f_n| < \infty$ | |
| Oct 31, 2017 at 13:15 | comment | added | Alexandre Eremenko | Even simpler example is $f_n=z^n$. | |
| Oct 31, 2017 at 9:04 | history | answered | js21 | CC BY-SA 3.0 |