Timeline for Integral involving InverseErf: closed-form expression and asymptotics
Current License: CC BY-SA 3.0
14 events
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| Oct 17, 2017 at 17:01 | history | edited | Mariusz Iwaniuk | CC BY-SA 3.0 |
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| Feb 15, 2017 at 21:41 | history | edited | Mariusz Iwaniuk | CC BY-SA 3.0 |
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| Feb 15, 2017 at 17:03 | history | edited | Mariusz Iwaniuk | CC BY-SA 3.0 |
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| Feb 15, 2017 at 14:20 | history | edited | Mariusz Iwaniuk | CC BY-SA 3.0 |
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| Feb 15, 2017 at 14:12 | history | edited | Mariusz Iwaniuk | CC BY-SA 3.0 |
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| Feb 7, 2017 at 14:50 | history | edited | Mariusz Iwaniuk | CC BY-SA 3.0 |
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| Feb 7, 2017 at 14:49 | comment | added | Mariusz Iwaniuk | @Dr.WolfgangHintze. Yes you are right.Thanks :) | |
| Feb 7, 2017 at 14:36 | comment | added | Dr. Wolfgang Hintze | It seems that the formula in your update is incorrect. It looks as if it was meant as a consequence of the relations arctan(x) + arccot(x) = pi/2 and arccot(x) = arctan(1/x) but the argument in the arctan on the right hand side of your formula is wrong. | |
| Feb 6, 2017 at 22:29 | comment | added | Dr. Wolfgang Hintze | Call the integral B(s). Differentiating with respect to s I found this formula. Don't know if it is useful $$B(s) = \int_0^1 \frac{\arctan\left(\sqrt{\frac{s+1}{y^2+2}}\right)}{\left(y^2+1\right) \sqrt{y^2+2}} \, dy=-\int_0^s \frac{\cot^{-1}\left(\sqrt{z+3}\right)}{2 \sqrt{z+1} (z+2) \sqrt{z+3}} \, dz+\frac{ \pi}{4} \arctan\left(\sqrt{s+1}\right)-\frac{\pi ^2}{32}$$ At $s = 0$ we have $$B(s=0) = \int_0^1 \frac{\cot ^{-1}\left(\sqrt{y^2+2}\right)}{\left(y^2+1\right) \sqrt{y^2+2}} \, dy=\frac{\pi ^2}{32}$$ | |
| Feb 6, 2017 at 18:39 | history | edited | Mariusz Iwaniuk | CC BY-SA 3.0 |
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| Feb 6, 2017 at 11:13 | history | edited | Mariusz Iwaniuk | CC BY-SA 3.0 |
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| Feb 6, 2017 at 11:02 | history | edited | Mariusz Iwaniuk | CC BY-SA 3.0 |
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| Feb 6, 2017 at 10:53 | history | edited | Mariusz Iwaniuk | CC BY-SA 3.0 |
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| Feb 6, 2017 at 10:44 | history | answered | Mariusz Iwaniuk | CC BY-SA 3.0 |