Integral <msubsup> &#x222B;<!-- ∫ --> 0 <mi mathvariant="normal">&#x221E;<!-- ∞ -

rigliztetbf

rigliztetbf

Answered question

2022-06-22

Integral 0 1 x 3 ( 1 + log 1 + e x 1 1 + e x ) d x
Is it possible to evaluate this integral in a closed form?

Answer & Explanation

Govorei9b

Govorei9b

Beginner2022-06-23Added 21 answers

Our initial observation is the identity of:
 0  x s  1 z e x  1  d x  =  0  z  1 x s  1 e  x 1  z  1 e  x  d x   =  n = 1  z  n  0  x s  1 e  n x  d x   = Γ ( s )  n = 1  z  n n s = Γ ( s ) L i s ( z  1 ) 
which initially holds for | z | > 1, and then extends holomorphically for z  1  ( 1 ,  ] since holomorphic functions are defined on this region on both sides. Nowadays, by integrating in pieces,
 0  1 x 3 ( 1 + log  1 + e x  1 1 + e x )  d x  = 3 2  0  x 2 / 3 ( 1 (  1 ) e x  1  1 (  e  1 ) e x  1 )  d x   = 3 2 Γ ( 5 3 ) { L i 5 / 3 (  1 )  L i 5 / 3 (  e ) }   =  3 2 Γ ( 5 3 ) { ( 1  2  2 / 3 ) ζ ( 5 / 3 ) + L i 5 / 3 (  e ) } . 
That is really unbelievable L i 5 / 3 (  e ) can be simplified further.

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