The integral I_(4) =int_(0)^(1)ln(1-x)ln^(2)(ln((1)/(x)))(dx)/(x) can be expressed as I_(4)=zeta^{''}(2)-(gamma^(2)pi^(2))/(6)-(gamma pi^(2))/(3)ln ((2pi)/ (A^(12)))+(pi^(4))/(36) where A is the Glaisher-Kinkelin constant.

Owen Mathis

Owen Mathis

Answered question

2022-11-17

Log Log Integrals II
The integral
I 4 = 0 1 ln ( 1 x )   ln 2 ( ln ( 1 x ) )   d x x
can be expressed as
I 4 = ζ ( 2 ) γ 2 π 2 6 γ π 2 3 ln ( 2 π A 12 ) + π 4 36
where A is the Glaisher-Kinkelin constant.
The integrals
I 5 = 0 1 ln ( 1 x ) ln ( 1 x )   ln 2 ( ln ( 1 x ) )   d x x
and
I 6 = 0 1 ln ( 1 x ) ln 3 / 2 ( 1 x )   ln 2 ( ln ( 1 x ) )   d x x
can be expressed in terms of s 2 ζ ( s ) | s = 3 / 2 and s 2 ζ ( s ) | s = 1 / 2 , respectively. Can these integrals be evaluated in closed form expression without the use of derivatives of the Zeta function?
If the integrals can be evaluated in such a way what is the resulting value?

Answer & Explanation

reinleikcyo

reinleikcyo

Beginner2022-11-18Added 11 answers

For the first integral,
I 5 = 0 1 ln ( 1 x ) ln ( 1 x )   ln 2 ( ln ( 1 x ) )   d x x = π 2 ( π 2 + 2 γ + 4 log 2 ) 2 ζ ( 3 2 ) + 2 π ( γ + 2 log 2 ) ζ ( 3 2 ) π ζ ( 3 2 )
there is only one ζ term so it doesn't seem likely to be able to remove it.
The second integral does not converge. But you might be interested in
ζ ( 1 2 ) = 1 4 ( π + 2 γ + 2 log 8 π ) ζ ( 1 2 ) ζ ( 1 2 ) = ( 2 G + 5 π 2 + 4 γ π + 4 γ 2 16 + log 8 π 4 ( log 8 π + π + 2 γ ) ) ζ ( 1 2 )

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