I am having trouble calculating this integral:
$$ g(a) = \int_{0}^{\frac{\pi}{2}} \frac{\arctan(a \tan x)}{\tan x}dx $$
I tried calculating $g'(a)$ but then I get stuck.
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$\begingroup$ Yes, I am sorry. $\endgroup$– nene123Commented Dec 16, 2019 at 17:07
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$\begingroup$ A right place fo such questions is MSE. $\endgroup$– user64494Commented Dec 16, 2019 at 18:23
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$\begingroup$ Both Approach0 and SearchOnMath lead to this post: Derivating with respect the parameter calcule $\int_0^{\frac{\pi}{2}}\frac{\arctan(a\tan x)}{\tan x}dx$ (It is closed, but still you can see a solution there.) $\endgroup$– Martin SleziakCommented Dec 16, 2019 at 18:45
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1 Answer
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The answer is $$g(a)=\frac{\pi}{2} \, \text{sgn}(a) \ln(1+|a|) $$ for real $a$.
This can be obtained as follows: $g(0)=0$ and $$g'(a)=\int_0^{\pi/2}\frac{dx}{1+a^2\tan^2 x}=\int_0^\infty\frac{du}{(1+a^2u^2)(1+u^2)}=\frac{\pi/2}{1+|a|}, $$ where we used the substitution $u=\tan x$; the latter integral can be taken by partial fraction decomposition: $$\frac1{(1+a^2u^2)(1+u^2)}=\frac{1}{\left(1-a^2\right) \left(1+u^2\right)} - \frac{a^2}{(1-a^2) \left(1+a^2 u^2\right)} $$ for real $a$ with $|a|\ne1$; the case $|a|=1$ can be obtained by continuity.
answered Dec 16, 2019 at 17:34
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$\begingroup$ 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. $\endgroup$ Commented Dec 16, 2019 at 18:19
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2$\begingroup$ @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. $\endgroup$ Commented Dec 16, 2019 at 18:53
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$\begingroup$ (ii) This is Demidovich 3734 dropbox.com/s/5wdh380ok6eqazj/Demidovich%203734.docx?dl=0 $\endgroup$ Commented Dec 16, 2019 at 19:05
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