Timeline for Getting asymptotic behaviour of an integral?
Current License: CC BY-SA 3.0
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jan 8, 2017 at 16:03 | comment | added | PhoenixPerson | I have given the bounty because this looks to be alright, although for various reasons I haven't resolved all the details, what I am saying is, I may have more questions still ;) | |
| Jan 8, 2017 at 16:02 | history | bounty ended | PhoenixPerson | ||
| Jan 7, 2017 at 19:24 | comment | added | PhoenixPerson | Much appreciated, I will have a thorough look | |
| Jan 7, 2017 at 18:59 | comment | added | Jan-Christoph Schlage-Puchta | I expanded the computations, stretching the performance of MathJax to the limits of my computer. | |
| Jan 7, 2017 at 18:58 | history | edited | Jan-Christoph Schlage-Puchta | CC BY-SA 3.0 |
Expanded the computations.
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| Jan 7, 2017 at 17:16 | comment | added | PhoenixPerson | I don't wanna annoy you but everything that goes beyond where you say "Now ..." puzzles me. My knowledge of asymptotic analysis is very elementary and I would appreciate more indications | |
| Jan 7, 2017 at 17:07 | comment | added | PhoenixPerson | Good thanks. Now, how do you justify $\int_{\epsilon}^{\epsilon^{-1}}\frac{\sin t}{t^{1.9}[\log\rho^{-1}+\mathcal{O}(\log\epsilon^{-1})]}dt=\frac{1}{\log\rho^{-1}}\int_{\epsilon}^{\epsilon^{-1}}\frac{\sin t}{t^{1.9}}dt+\mathcal{O}\bigg(\frac{\log\epsilon^{-1}}{\log^2\rho^{-1}}\int_{\epsilon}^{\epsilon^{-1}}\frac{|\sin t|}{t^{1.9}}dt\bigg)$? | |
| Jan 7, 2017 at 16:49 | comment | added | Jan-Christoph Schlage-Puchta | We have $\epsilon\leq t\leq\epsilon^{-1}$, thus $\log(t/\rho)=\log\rho^{-1}+\log t=\log\rho^{-1}+\mathcal{O}(\log\epsilon^{-1})$. | |
| Jan 7, 2017 at 15:15 | comment | added | PhoenixPerson | How do you justify $\int_{\epsilon}^{\epsilon^{-1}}\frac{\sin t}{t^{1.9}\log\frac{t}{\rho}}dt=\int_{\epsilon}^{\epsilon^{-1}}\frac{\sin t}{t^{1.9}[\log\rho^{-1}+\mathcal{O}(\log\epsilon^{-1})]}dt$? | |
| Jan 7, 2017 at 10:16 | history | answered | Jan-Christoph Schlage-Puchta | CC BY-SA 3.0 |