Timeline for Differentiating an integral that grows like log asymptotically
Current License: CC BY-SA 4.0
11 events
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| Jan 14, 2019 at 14:02 | comment | added | Iosif Pinelis | @Raziel : Oh, no, what you are suggesting is certainly not the kind of approximation that I had in mind. I have now given some details on that. I still think the approximation is easy, even if tedious; also, in such situations any degree of smoothness is quite inessential. | |
| Jan 14, 2019 at 13:58 | history | edited | Iosif Pinelis | CC BY-SA 4.0 |
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| Jan 14, 2019 at 13:06 | comment | added | Raziel | Nice example. The "approximation" part is not that trivial though. The easiest approximation given by setting $e^{j^2} = y$ yields the function $f(y) = \frac{2\sqrt{\log(y)}}{\exp((\sqrt{\log(y)}+1)^2)}$, whose integral tends to a constant. | |
| Jan 14, 2019 at 12:52 | comment | added | random_person | I am feeling so embarrassed that I have missed this observation...thank you so much for your patience and again your counter-example. | |
| Jan 14, 2019 at 12:48 | comment | added | Iosif Pinelis | @random_person : I have now added a sentence showing that, in the same example, the lower bound $\frac1{Ct}$ on $f(t)$ is impossible either. | |
| Jan 14, 2019 at 12:46 | history | edited | Iosif Pinelis | CC BY-SA 4.0 |
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| Jan 14, 2019 at 12:42 | comment | added | random_person | Oh I have asked a dumb question. I actually want to ask if a lower bound $f(t) \ge \frac{1}{Ct}$ is possible. | |
| Jan 14, 2019 at 12:41 | comment | added | Iosif Pinelis | @random_person : This very example shows that the upper bound $\frac Ct$ on $f(t)$ is impossible in general, as we have $f(t_{j+1}-)/\frac1{t_{j+1}}\to\infty$. | |
| Jan 14, 2019 at 12:38 | comment | added | random_person | Thanks for the nice counter-example. Would it still be possible to establish the upper bound $f(t) \le \frac{C}{t}$ though? | |
| Jan 14, 2019 at 12:34 | history | edited | Iosif Pinelis | CC BY-SA 4.0 |
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| Jan 14, 2019 at 12:26 | history | answered | Iosif Pinelis | CC BY-SA 4.0 |