Timeline for Integral $ g(a)= \int_{0}^{\frac{\pi}{2}} \frac{\arctan(a \tan x)}{\tan x}dx $
Current License: CC BY-SA 4.0
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| Dec 16, 2019 at 19:13 | comment | added | Iosif Pinelis | @user64494 : Good find! | |
| Dec 16, 2019 at 19:05 | comment | added | user64494 | (ii) This is Demidovich 3734 dropbox.com/s/5wdh380ok6eqazj/Demidovich%203734.docx?dl=0 | |
| Dec 16, 2019 at 18:53 | comment | added | Iosif Pinelis | @user64494 : (i) Yes, Mathematica is imperfect. Here you have to work a bit to simplify the expression. Alternatively, you can use Integrate[ArcTan[a Tan[x]]/Tan[x], {x, 0, Pi/2}, Assumptions -> a > 0] to get 1/2 [Pi] Log[1 + a]. I have now removed the reference to Mathematica anyway, since I gave an explicit derivation without using Mathematica. (ii) As to whether this question is OK for MO, I am not sure; I could imagine it arise in some research. | |
| Dec 16, 2019 at 18:38 | history | edited | Iosif Pinelis | CC BY-SA 4.0 |
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| Dec 16, 2019 at 18:19 | comment | added | user64494 | Sorry, but my Mathematica 12.0 by Integrate[ArcTan[a*Tan[x]]/Tan[x], {x, 0, Pi/2}] produces ConditionalExpression[1/4 [Pi] (2 ArcTanh[Abs[a]]+Log[1-a^2]) Sign[a],a \ [Element] Reals]. I check it for $a=\frac 1 2$ numerically. In my personal opinion a right place for the question and answer is MSE. | |
| Dec 16, 2019 at 18:08 | history | edited | Iosif Pinelis | CC BY-SA 4.0 |
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| Dec 16, 2019 at 17:57 | history | edited | Iosif Pinelis | CC BY-SA 4.0 |
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| Dec 16, 2019 at 17:49 | history | edited | Iosif Pinelis | CC BY-SA 4.0 |
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| Dec 16, 2019 at 17:42 | history | edited | Iosif Pinelis | CC BY-SA 4.0 |
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| Dec 16, 2019 at 17:34 | history | answered | Iosif Pinelis | CC BY-SA 4.0 |