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Petrovcic2x

Petrovcic2x

Answered question

2022-06-22

Need help with 0 e x ln ln ( e x + e 2 x 1 ) d x
I need help with this integral:
0 e x ln ln ( e x + e 2 x 1 ) d x 0.20597312051214...
Is it possible to evaluated it in a closed form?

Answer & Explanation

Punktatsp

Punktatsp

Beginner2022-06-23Added 22 answers

A useful identity here is a r c c o s h z = ln ( z + z 2 1 ). Therefore
0 e x ln ln ( e x + e 2 x 1 ) d x = 0 e x ln a r c c o s h e x d x = 1 ln a r c c o s h y y 2 d y
Change integration variable z = a r c c o s h y,
= 0 ln z sinh z cosh 2 z d z .
Integrate by parts and use the definition of the Euler's γ constant and Γ
= γ + ln [ Γ ( 1 4 ) Γ ( 5 4 ) Γ 2 ( 3 4 ) ] .
We can further simplify this using Γ ( 1 z ) Γ ( z ) = π / sin π z and Γ ( z ) Γ ( z + 1 2 ) = 2 1 2 z π Γ ( 2 z ) for z = 3 / 4 and sin ( 3 π / 4 ) = 1 / 2 . This gives
= γ 3 ln 2 2 ln π + 4 ln Γ ( 1 / 4 ) .

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