Evaluating int_0^((pi)/(2))((1)/(log(tan x))+(1)/(1-tan(x)))^3dx

Conrad Beltran 2022-09-27 Answered
Evaluating 0 π 2 ( 1 log ( tan x ) + 1 1 tan ( x ) ) 3 d x
Using the method shown here, I have found the following closed form.
0 π 2 ( 1 log ( tan x ) + 1 1 tan x ) 2 d x = 3 ln 2 4 π G 1 2 ,
where G is Catalan's constant.
I can see that replicating the techniques for the following integral could be rather lenghty.
0 π 2 ( 1 log ( tan x ) + 1 1 tan x ) 3 d x
My question: Could someone have, ideally, a different idea to evaluate the latter integral?
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Answers (1)

kregde84
Answered 2022-09-28 Author has 10 answers
I've found it.
We have
(1) 0 π 2 ( 1 log ( tan x ) + 1 1 tan x ) 3 d x = 9 2 ln 2 6 π G 3 4 π 8
where G is Catalan's constant.
Proof. Set
(2) I n := 0 π 2 ( 1 log ( tan x ) + 1 1 tan x ) n d x , n = 0 , 1 , 2... .
Clearly
I 0 = π 2 .
Recall that
tan ( π 2 x ) = 1 tan x
giving
1 log tan ( π 2 x ) + 1 1 tan ( π 2 x ) = 1 log ( 1 tan x ) + 1 1 1 tan x (3) = 1 log ( tan x ) 1 1 tan x + 1
Hence, by the change of variable x π 2 x , we readily have
(4) I 1 = 0 π 2 ( 1 log ( tan x ) 1 1 tan x + 1 ) d x = I 1 + π 2
I 1 = π 4 .
Similarly,
(5) 0 π 2 ( 1 log ( tan x ) + 1 1 tan x ) 3 d x = 0 π 2 ( 1 log ( tan x ) 1 1 tan x + 1 ) 3 d x
and, by the binomial expansion,
I 3 = I 3 + 3 I 2 3 I 1 + I 0
I 3 = I 3 + 3 I 2 3 π 4 + π 2
(6) I 3 = 3 2 I 2 π 8
we may conclude with this value obtained for I 2
A numerical approximation is
I 3 = 0.22709780663611705673940738484148718263....
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