Timeline for Integral of complex exponential with assumptions
Current License: CC BY-SA 4.0
12 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Nov 24, 2023 at 2:28 | comment | added | Daniel Lichtblau | Here is one older version among several others. | |
| Nov 23, 2023 at 10:46 | history | edited | Domen | CC BY-SA 4.0 |
Improve readability
|
| Nov 23, 2023 at 10:44 | comment | added | Roman |
Another "by hand" solution: integrate over the reals and then specialize on the integers. Integrate[Exp[2π*I*q*t], {t,0,1}] == Sinc[2π*q] + I*Sinc[π*q]*Sin[π*q] // FullSimplify gives True. The imaginary part is zero $\forall q\in\mathbb{Z}$. The real part is 1 for $q=0$ and zero for all other integers.
|
|
| Nov 23, 2023 at 8:14 | comment | added | David G. Stork | Thanks... helpful... but again essentially a solution "by hand." | |
| Nov 23, 2023 at 4:20 | comment | added | Bob Hanlon |
int[k_, m_] = Assuming[{k, m} \[Element] NonNegativeIntegers, Piecewise[{{Assuming[k == m , Integrate[E^(2*Pi*I*(k - m)*t), {t, 0, 1}]], {k, m} \[Element] NonNegativeIntegers && k == m }, {Assuming[k != m , Integrate[E^(2*Pi*I*(k - m)*t), {t, 0, 1}]], {k, m} \[Element] NonNegativeIntegers && k != m}}]]
|
|
| Nov 22, 2023 at 23:23 | comment | added | eyorble |
Using FullSimplify on that result with the same assumptions reduces it to 0 though.
|
|
| Nov 22, 2023 at 23:22 | comment | added | eyorble |
On 13.3.0.0 with Integrate[Exp[2 \[Pi] I (k - m) t], {t, 0, 1}, Assumptions -> {k \[Element] NonNegativeIntegers, m \[Element] NonNegativeIntegers}] I get a non-zero expression (-((I (-1 + E^(2 I (k - m) \[Pi])))/(2 (k - m) \[Pi]))) which exhibits the appropriate behavior (assuming Limit is used for the $k=m$ case).
|
|
| Nov 22, 2023 at 22:57 | history | edited | David G. Stork | CC BY-SA 4.0 |
added 5 characters in body
|
| Nov 22, 2023 at 19:58 | comment | added | Roman | Should be a Kronecker delta, not a Dirac delta. | |
| Nov 22, 2023 at 19:50 | comment | added | David G. Stork | @Nasser: Thanks. The result $0$ isn't exactly "generic," but instead the typical case. I realize I could split the cases into $k=m$ and $k \neq m$, but I would have thought that Mathematica could do this explicitly. | |
| Nov 22, 2023 at 19:48 | comment | added | Nasser |
Mathematica gives result which is generic. You could always add extra assumption. Here is screen shot !Mathematica graphics which gives zero only when k!=m and gives 1 when k=m but I do not know if there is way to make Integrate do that on its own.
|
|
| Nov 22, 2023 at 19:24 | history | asked | David G. Stork | CC BY-SA 4.0 |
