Timeline for Defining integrals by residue theorem
Current License: CC BY-SA 4.0
14 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| May 1, 2019 at 13:04 | vote | accept | Penchez | ||
| Apr 24, 2019 at 5:01 | history | bumped | CommunityBot | This question has answers that may be good or bad; the system has marked it active so that they can be reviewed. | |
| Mar 25, 2019 at 4:14 | answer | added | Count Iblis | timeline score: 1 | |
| Feb 22, 2019 at 16:46 | comment | added | David Hughes | You can use the function $y(x)=\tan \left( \frac{2x-a-b}{b-a} \frac{\pi}{2} \right)$ to map any interval $(a,b)$ to $(-\infty,\infty)$ in order to define definite integrals: $$\int_a^b f(x)dx = \frac{b-a}{\pi} \int_{-\infty}^\infty \frac{f(x(y))}{1+y^2}dy.$$ This gives a way to extend the definition for definit integrals and provided $f$ is meromorphic and not too wild on $(a,b)$ it should all work. | |
| Feb 22, 2019 at 14:42 | comment | added | Fedor Petrov | If you deal with formal Laurent series $f(z)$ (over arbitrary commutative ring), the residue $[z^{-1}]f$ of $f$ at point 0 is a good substitute of integral, which shares many its crucial properties. The main property is of course that $[z^{-1}]f'=0$. | |
| Feb 22, 2019 at 14:05 | review | Close votes | |||
| Feb 26, 2019 at 11:32 | |||||
| Feb 22, 2019 at 13:45 | comment | added | Alexandre Eremenko | Unlike the general definition of integral, this definition will work only for a very narrow class of functions. | |
| Feb 22, 2019 at 11:50 | comment | added | Penchez | @CarloBeenakker one alternative is suggested by user44191. Another could be just defining integrals over [-oo,+oo] for a class of functions large enough to contain functions with compact supports (which could be problematic if functions are required to be meromorphic). | |
| Feb 22, 2019 at 11:32 | comment | added | user44191 | @CarloBeenakker Presumably, by assuming the existence of a path between $0$ and $1$ (possibly including the "curve at infinity") such that $f$ decays fast enough (for some idea of "fast enough") near that curve? E.g. only allow $f$ such that $f((tanh(x) + 1)/2)$ decays "fast enough". | |
| Feb 22, 2019 at 11:05 | comment | added | Carlo Beenakker | but how would you express the definite integral $\int_0^1 dx$ by the residue theorem? | |
| Feb 22, 2019 at 11:04 | comment | added | Penchez | @CarloBeenakker that isn't a problem: I'm interested only in definite integrals. | |
| Feb 22, 2019 at 10:39 | comment | added | Carlo Beenakker | you would not be able to express an indefinite integral in this way, would you? | |
| Feb 22, 2019 at 10:25 | review | First posts | |||
| Feb 22, 2019 at 11:33 | |||||
| Feb 22, 2019 at 10:22 | history | asked | Penchez | CC BY-SA 4.0 |