What is the real part of \int_(0)^(pi/2)(x^2)/(x^2+log^2(-2cos x))dx?

Danika Mckay

Danika Mckay

Answered question

2022-10-31

What is the real part of 0 π / 2 x 2 x 2 + log 2 ( 2 cos x ) d x
This is a new integral that I propose to evaluate in closed form:
R 0 π / 2 x 2 x 2 + log 2 ( 2 cos x ) d x
where denotes the real part and log ( z ) denotes the principal value of the logarithm defined for z 0 by
log ( z ) = ln | z | + i A r g z , π < A r g z π .

Answer & Explanation

Ostrakodec3

Ostrakodec3

Beginner2022-11-01Added 18 answers

0 π / 2 x 2 x 2 + log 2 ( 2 cos x ) d x = π 8 ( 1 ln 2 γ )
For a proof, I will continue from
Where
J = 0 1 x 2 + π 2 [ x 1 e x + ln ( e x 1 ) ] d x
Integration by part shows
J = π 0 arctan x e π x x ( e π x 1 ) 2 d x
Invoke the Binet second formula:
ln Γ ( z ) = ( z 1 2 ) ln z z + ln ( 2 π ) 2 + 2 z 0 arctan t e 2 π t z 1 d t
The value of J immdiately follows from differentiating with respect to z, and then set z = 1 / 2
J = γ 2
blackdivcp

blackdivcp

Beginner2022-11-02Added 4 answers

I don't think a closed form exists after computing that integral numerically in Mathematica, and looking up in the Inverse Symbolic Calculator. It is approximately equal to: 0.10617124113817 If you need more digits just ask.

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