Timeline for Symbolic integral not computed
Current License: CC BY-SA 4.0
24 events
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| S Jul 22, 2018 at 22:00 | history | bounty ended | CommunityBot | ||
| S Jul 22, 2018 at 22:00 | history | notice removed | CommunityBot | ||
| Jul 15, 2018 at 22:43 | vote | accept | πρόσεχε | ||
| Jul 15, 2018 at 12:45 | answer | added | t-smart | timeline score: 1 | |
| Jul 15, 2018 at 8:46 | history | edited | m0nhawk | CC BY-SA 4.0 |
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| Jul 15, 2018 at 2:06 | comment | added | theorist | related?: mathematica.stackexchange.com/questions/15920/… [Though there the nested roots are purely numeric -- they don't include a variable.] | |
| Jul 14, 2018 at 23:04 | answer | added | Greg Hurst | timeline score: 7 | |
| Jul 14, 2018 at 22:03 | answer | added | AccidentalFourierTransform | timeline score: 5 | |
| S Jul 14, 2018 at 20:00 | history | bounty started | πρόσεχε | ||
| S Jul 14, 2018 at 20:00 | history | notice added | πρόσεχε | Authoritative reference needed | |
| Jul 13, 2018 at 21:45 | comment | added | Mariusz Iwaniuk | How you expand the double radical? Could you explain and update the question? | |
| Jul 13, 2018 at 18:06 | history | edited | πρόσεχε | CC BY-SA 4.0 |
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| Jul 13, 2018 at 10:08 | history | edited | πρόσεχε | CC BY-SA 4.0 |
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| Jul 13, 2018 at 8:44 | history | edited | πρόσεχε | CC BY-SA 4.0 |
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| Jul 13, 2018 at 8:38 | history | edited | πρόσεχε | CC BY-SA 4.0 |
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| Jul 13, 2018 at 6:14 | comment | added | Ulrich Neumann |
Note: Mathematica can evaluate the asymptotic part of the integrand Integrate[ Sin[x]/(2 Sqrt[x] ), {x, 0, Infinity} ](*Sqrt[\[Pi]/2]/2*)
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| Jul 13, 2018 at 0:42 | history | tweeted | twitter.com/StackMma/status/1017570012922556417 | ||
| Jul 12, 2018 at 20:35 | comment | added | AccidentalFourierTransform |
Note sure if useful, but your integral can also be written as a sum: -Cos[1] + Sqrt[Pi/2] FresnelC[Sqrt[2/Pi]] + Sum[((-1)^(1 + n) (-1 + n) Sqrt[2/Pi] Gamma[-(7/2) + 2 n] HypergeometricPFQ[{-(1/2) + n}, {3/2, 1/2 + n}, -(1/4)])/Gamma[2 n] - ((1/16 + I/16) I^n (-I + (-1)^n) (Gamma[5/4 - n/2] Gamma[-(1/2) + n] HypergeometricPFQRegularized[{5/4 - n/2}, {3/2, 9/4 - n/2}, -(1/4)] + 4 Sqrt[Pi] Sec[n Pi] Sin[1/4 (Pi + 2 n Pi)]))/Gamma[1 + n], {n, 1, Infinity}].
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| Jul 12, 2018 at 19:32 | comment | added | user64494 | The questions arise: isn't it art for art's sake? where is this integral applied? | |
| Jul 12, 2018 at 19:28 | comment | added | JungHwan Min | @MariuszIwaniuk You can verify the solution like that, but you cannot compute it (plus, being close does not necessarily mean they're equal). The assumption here is that you don't know the solution $\frac{1}{2}\,\sqrt{\frac{\pi}{2}}\,\frac{e-1}{e}$ a priori. | |
| Jul 12, 2018 at 18:56 | comment | added | Mariusz Iwaniuk |
Ok I correct the comment. You can check with NIntegrate code: Sqrt[Pi/2]/2*((E - 1)/E) - NIntegrate[(Sqrt[x] - Sqrt[Sqrt[1 + x^2] - 1])*Sin[x], {x, 0, Infinity}, WorkingPrecision -> 200] almost zero.
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| Jul 12, 2018 at 18:33 | history | edited | David G. Stork | CC BY-SA 4.0 |
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| Jul 12, 2018 at 17:42 | comment | added | JungHwan Min |
For the approximation, you could use NIntegrate.
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| Jul 12, 2018 at 17:32 | history | asked | πρόσεχε | CC BY-SA 4.0 |