Timeline for Integration of products of spherical Hankel functions yield different results
Current License: CC BY-SA 4.0
13 events
| when toggle format | what | by | license | comment | |
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| Jun 4, 2022 at 20:11 | history | edited | J. M.'s missing motivation♦ |
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| Jan 4, 2022 at 2:09 | history | became hot network question | |||
| Jan 3, 2022 at 20:48 | comment | added | ExtremOPS | @Domen: thank you very much; also to the others.Your help was very fast and solved/answered my question. | |
| Jan 3, 2022 at 20:47 | vote | accept | ExtremOPS | ||
| Jan 3, 2022 at 18:44 | answer | added | Michael E2 | timeline score: 5 | |
| Jan 3, 2022 at 18:29 | comment | added | Domen |
Therefore, increase the precision of the boundary in the integration: Integrate[J[z, (1 + I)*10^(-8), 0, 1, 0], {z, 1`50, 1.1`50}]. This yields -0.0333333329833333333333333 - 3.499999970577778*10^-10 I.
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| Jan 3, 2022 at 18:22 | comment | added | Domen |
It probably has to do with the inadequate precision due to extremely large/small terms in the integral. Take a look at the analytic integral (int = Integrate[J[z, (1 + I)*10^(-8), 0, 1, 0], z]) and its plot Plot[Evaluate@Re@int, {z, .9, 1.2}]. The plot looks much nices if you increase the precision Plot[Evaluate@Re@int, {z, .9, 1.2}, WorkingPrecision -> 30].
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| Jan 3, 2022 at 18:20 | comment | added | ExtremOPS |
Thank you @bbgodfrey. I changed it. Also, I cannot use the $$ formatting to make the math nicer to view.
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| Jan 3, 2022 at 18:16 | history | edited | ExtremOPS | CC BY-SA 4.0 |
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| Jan 3, 2022 at 18:16 | comment | added | bbgodfrey | Integrate[J[z, (1+I) * 10^(-8), 0, 1 0], {z, 1, 1.1}]; should be Integrate[J[z, (1+I) * 10^(-8), 0, 1, 0], {z, 1, 1.1}]; | |
| Jan 3, 2022 at 18:10 | history | edited | ExtremOPS | CC BY-SA 4.0 |
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| S Jan 3, 2022 at 18:03 | review | First questions | |||
| Jan 3, 2022 at 18:39 | |||||
| S Jan 3, 2022 at 18:03 | history | asked | ExtremOPS | CC BY-SA 4.0 |