Timeline for Is there a closed form for $\int_0^\infty\frac{\tanh^3(x)}{x^2}dx$?
Current License: CC BY-SA 4.0
12 events
| when toggle format | what | by | license | comment | |
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| Nov 18, 2021 at 18:23 | comment | added | LSpice | @IgorKhavkine's approach mentioned above. | |
| Nov 18, 2021 at 18:23 | history | edited | LSpice | CC BY-SA 4.0 |
Link to @WillSawin's comment while this is on the front page
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| Jun 8, 2017 at 16:47 | history | edited | T. Amdeberhan | CC BY-SA 3.0 |
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| Jun 8, 2017 at 5:48 | comment | added | Zurab Silagadze | Indeed very cute, as well as Igor Khavkine's approach above. | |
| Jun 7, 2017 at 23:32 | history | edited | T. Amdeberhan | CC BY-SA 3.0 |
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| Jun 7, 2017 at 22:48 | comment | added | Lewi_Sol | This is super, Ramanujan style! | |
| Jun 7, 2017 at 22:35 | history | edited | T. Amdeberhan | CC BY-SA 3.0 |
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| Jun 7, 2017 at 19:03 | comment | added | Will Sawin | $\sum_k (-1)^k k^c$ is $\left( \sum_k k^c \right) (-1 + 2^{1+c} )$ and the first term is of course $\zeta(-c)$. This is surely related to the appearance of the derivative of $\zeta$ in Igor's formula. | |
| Jun 7, 2017 at 7:17 | comment | added | T. Amdeberhan | One may start by finding a divergent series formula for $\sum_{k\geq2}(-1)^kk^c$ for a range of values of $c$ real. Take derivative to get $\sum_k(-1)^kk^c\log k$ and then put $c=1$. | |
| Jun 7, 2017 at 6:24 | history | edited | T. Amdeberhan | CC BY-SA 3.0 |
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| Jun 7, 2017 at 6:13 | history | edited | T. Amdeberhan | CC BY-SA 3.0 |
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| Jun 7, 2017 at 6:00 | history | answered | T. Amdeberhan | CC BY-SA 3.0 |