Timeline for What is the integral of $r \frac{2^{r-1} \log (2) e^{-\frac{\sqrt{2^r-1}}{b}} \left(2^r-1\right)^{\frac{d}{2}-1}}{b^d \Gamma (d)}$?
Current License: CC BY-SA 4.0
7 events
| when toggle format | what | by | license | comment | |
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| Jul 20, 2020 at 1:08 | vote | accept | Felipe Augusto de Figueiredo | ||
| Jul 20, 2020 at 1:08 | answer | added | Felipe Augusto de Figueiredo | timeline score: 0 | |
| Jul 10, 2020 at 12:05 | comment | added | Felipe Augusto de Figueiredo | @LSpice,the expression is correct. Thanks! | |
| Jul 10, 2020 at 3:16 | comment | added | dvitek | Make the obvious substitution $u = \sqrt{2^r-1}$, and one is left with evaluating (up to multiplicative constants and my arithmetic errors) $$\int_0^{\infty} u^{d-1} \exp(-u/b) \log(1+u^2) \;du.$$ (@LSpice, the presence of $2^{r-1} \log(2)$ in the $du$ term suggests that the expression may well be "correct" as is, for whatever the appropriate definition of "correct" is in this problem.) It may not be possible to find a closed form for this integral in terms of $d$, but at the very least one should be able to find a useful series expansion... | |
| Jul 10, 2020 at 2:09 | comment | added | LSpice | Of course, $\frac{\log(2)}{b^d\Gamma(d)}$ is just clutter. You have $2^r - 1$ in two places and $2^{r - 1}$ in one; is that intentional? | |
| Jul 10, 2020 at 2:09 | history | edited | LSpice | CC BY-SA 4.0 |
Display formula
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| Jul 10, 2020 at 1:38 | history | asked | Felipe Augusto de Figueiredo | CC BY-SA 4.0 |