An integral involving the inverse of f ( x ) = log &#x2061;<!-- ⁡ --> x

uri2e4g 2022-07-09 Answered
An integral involving the inverse of f ( x ) = log x log cos x + x tan x
Let the function f : ( 0 , π 2 ) R be defined as
f ( x ) = log x log cos x + x tan x .
Let its inverse be denoted as f ( 1 ) : R ( 0 , π 2 ) , it satisfies
z R ,     f ( f ( 1 ) ( z ) ) = z .
Consider the integral
I = 0 1 z e z f ( 1 ) ( z ) d z .
Is it possible to evaluate the integral I in a closed form?
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Answers (1)

Antonio Dickerson
Answered 2022-07-10 Author has 5 answers
I know it has been some time now but we can transform this question into something more manageable. We'll just use the substitution z = f ( x ) with d z = f ( x ) d x. The bounds of integration change to the points x where f ( x ) is respectively 0 or 1. Call these points x 0 and x 1 .
Numerically they are
x 0 = 0.576412723
x 1 = 0.807626251
And so we have
I = x 0 x 1 f ( x ) e f ( x ) f ( 1 ) ( f ( x ) ) f ( x ) d x = x 0 x 1 x f ( x ) f ( x ) e f ( x ) d x
Now we perform integration by parts with
u = x f ( x ) , d u = ( f ( x ) + x f ( x ) ) d x
d v = f ( x ) e f ( x ) d x , v = e f ( x )
and the integral becomes
I = x 1 f ( x 1 ) e f ( x 1 ) x 0 f ( x 0 ) e f ( x 0 ) x 0 x 1 f ( x ) e f ( x ) d x x 0 x 1 x f ( x ) e f ( x ) d x
Since f ( x 0 ) = 0 and f ( x 1 ) = 1 we have
I = x 1 x 0 x 1 f ( x ) e f ( x ) d x x 0 x 1 x f ( x ) e f ( x ) d x
Now we perform integration by parts on the latter integral with
u = x , d u = d x
d v = f ( x ) e f ( x ) d x , v = e f ( x )
and we get
I = x 1 x 1 e f ( x 1 ) + x 0 e f ( x 0 ) x 0 x 1 f ( x ) e f ( x ) d x + x 0 x 1 e f ( x ) d x
It turns out we can perform the integration in the first integral since
e f ( x ) = x cos x e x tan x
and
e f ( x ) d x = e x tan x cos x
After cleaning up a bit
I = x 0 + e x 1 tan x 1 cos x 1 e x 0 tan x 0 cos x 0 x 0 x 1 f ( x ) e f ( x ) d x
and this is where I am getting stuck. Mathematica is having a problem with it too. On [ x 0 , x 1 ] we have
f ( x ) e f ( x ) = x e x tan ( x ) cos x ( ln ( x cos x ) + x tan x )
We also have another form
f ( x ) e f ( x ) = ( x 2 sin x cos 2 x + x cos x ln ( x cos x ) ) e x tan x
which suggest that we might have better luck looking at the following two integrals
J = x 0 x 1 x 2 sin x cos 2 x e x tan x d x
and
K = x 0 x 1 ln ( x cos x ) e x tan x d x
The other problem perhaps, is that x 0 and x 1 are given numerically. I have no idea if one could find a closed for expression for them or even perhaps something like a series representation.
EDIT1: I'm not sure if this could be called a reduction but using the equation defining f ( x ) we can say that
cos x 1 e x 1 tan x 1 = x 1 e 1 + 2 x 1 tan x 1
cos x 1 0 e x 0 tan x 0 = x 0 e 2 x 0 tan x 0

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