Timeline for Why can't Mathematica evaluate this integral?
Current License: CC BY-SA 4.0
12 events
| when toggle format | what | by | license | comment | |
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| Oct 16, 2020 at 22:10 | vote | accept | Richard Burke-Ward | ||
| Oct 16, 2020 at 17:51 | answer | added | Akku14 | timeline score: 2 | |
| Oct 16, 2020 at 17:32 | answer | added | Anton Antonov | timeline score: 2 | |
| Oct 16, 2020 at 17:25 | comment | added | Richard Burke-Ward |
Hi @azerbajdzan. The issue must be with the definition, then; because if I predefine f and write f[x_, m_] := HeavisidePi[x - m];Integrate[f[x, 0], {x, -(1/2), 1/2}], I get Undefined...
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| Oct 16, 2020 at 17:04 | comment | added | Michael E2 |
My question was about f[x, 0], which does not evaluate correctly, imo. It was not about f[0,0]. In Integrate[], you're evaluating f[x,0], so understanding what it does is the key.
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| Oct 16, 2020 at 16:53 | comment | added | azerbajdzan |
When did you get Undefined? Integrate[HeavisidePi[x - m], {x, m - 1/2, m + 1/2}] /. m -> 0 nicely evaluates into 1.
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| Oct 16, 2020 at 16:44 | history | edited | Richard Burke-Ward | CC BY-SA 4.0 |
Further explanation
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| Oct 16, 2020 at 16:44 | comment | added | Richard Burke-Ward |
Hi all. In answer to your questions: (1) f[0,0] evaluates correctly to 1; (2) @Daniel, I'm not sure I understand the reservation: at least until the limit ->Infinity is reached, f is an entire (and therefore integrable) function - and in the limit the definite integral described clearly produces 1 (having said this, I did try Piecewise but that didn't work either, any least with the half-values at the discontinuities); (3) As mentioned in the OP, HeavisidePi also produces an Undefined result under integration.
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| Oct 16, 2020 at 16:36 | comment | added | azerbajdzan |
Why not using HeavisidePi? It keeps HeavisidePi[1/2] and HeavisidePi[-1/2] unevaluated. If some of your outputs contain HeavisidePi[1/2] you can simply replace it with desired value - like HeavisidePi[1/2]->1/2 or HeavisidePi[-1/2]->1/2.
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| Oct 16, 2020 at 16:28 | comment | added | Daniel Lichtblau |
That limit evaluates to ConditionalExpression[0, Log[2 x] > 0]. It's not clear to me what Integrate will be able to do with it. I will suggest that if you want a Heaviside function but with a value at the origin, create it with Piecewise.
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| Oct 16, 2020 at 16:13 | comment | added | Michael E2 |
Does f[x, 0] evaluate correctly?
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| Oct 16, 2020 at 16:03 | history | asked | Richard Burke-Ward | CC BY-SA 4.0 |