Timeline for Proof of $\Re\int_0^{\infty}\exp(-ik-k/\sqrt{1-4ik})dk=\Im\int_0^{1/4}\exp(-k+ik/\sqrt{1-4k})dk$
Current License: CC BY-SA 4.0
9 events
| when toggle format | what | by | license | comment | |
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| Apr 15, 2023 at 15:18 | answer | added | Constantin-Nicolae Beli | timeline score: 2 | |
| Apr 15, 2023 at 4:14 | comment | added | Guoqing | @Constantin-NicolaeBeli Yes | |
| Apr 14, 2023 at 16:21 | comment | added | Constantin-Nicolae Beli | I believe I know how to do it, but first I have to know what you mean by $\sqrt{1-4ik}$. Usually $\sqrt x$ is defined for $x\geq 0$ and it is equal to the number $y\geq 0$ such that $x=y^2$. If you want to extend the definition to complex numbers you have to specify which of the two branches of the radical you consider. If I am to guess, then perhaps you choose $\sqrt{1-4ik}$ to be the complex number $z$ with $1-4ik=z^2$ and $\Re z\geq 0$. Is that right? | |
| Apr 14, 2023 at 11:50 | comment | added | Carlo Beenakker | clear enough, thanks. | |
| Apr 14, 2023 at 11:47 | comment | added | Guoqing | I think it's the high oscillatory term for large $k$ in the lhs, so it's necessary to increase the workprecision to decrease numerical errors. | |
| Apr 14, 2023 at 11:42 | comment | added | Guoqing | @CarloBeenakker I evaluate the lhs using MMA and get $-0.00395$ which matches your answer, but without setting the workprecision. Once I increase the WorkingPrecision I get $0.052$ that matches the rhs. | |
| Apr 14, 2023 at 11:35 | history | edited | Guoqing | CC BY-SA 4.0 |
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| Apr 14, 2023 at 11:33 | comment | added | Guoqing | @CarloBeenakker Idk how you perform these integrals. I seed these integrals to MMA and get 0.051997 for lhs and 0.051999 for rhs, with workingPrecision 200. | |
| Apr 14, 2023 at 9:39 | history | asked | Guoqing | CC BY-SA 4.0 |