Evaluating improper integral <msubsup> &#x222B;<!-- ∫ --> 0 1 </msubsup>

Patatiniuh 2022-07-15 Answered
Evaluating improper integral 0 1 log ( x ) x + α d x for small positive α
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Answers (2)

alomjabpdl0
Answered 2022-07-16 Author has 12 answers
0 1 ln ( x ) a + x d x = 0 1 / a ln ( a y ) 1 + y d y
= ln ( 1 + y ) ln ( a y ) | 0 1 / a 0 0 1 / a ln ( 1 + y ) y d x = Li 2 ( y ) | 0 1 / a = Li 2 ( 1 / a )
and the result follows on using the dilogarithm inversion formula:
Li 2 ( 1 / z ) = π 2 6 1 2 ln 2 ( z ) Li 2 ( z ) .
To prove the last identity, differentiate Li 2 ( 1 / z ) then integrate back.
In geeneral we have
0 1 ln p ( x ) a + x d x = ( 1 ) p + 1 p ! Li p + 1 ( 1 / a )
which follows from the integral represenation of the polylogarithm function Li p + 1 ( z ) = ( 1 ) p p ! 0 1 z ln p ( t ) 1 z t d t
upon replacing z by 1 / a
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dikcijom2k
Answered 2022-07-17 Author has 6 answers
Changing the integration variable to t := x / α, we get
I = 0 1 / α log α + log t 1 + t d t .
The integration of the log α term is easy:
0 1 / α log α 1 + t d t = log α log 1 + α α .
For the log t term, changing the integration variable to u := 1 + t
0 1 / α log t 1 + t d t = 1 1 + 1 / α log ( u 1 ) u d u = [ Φ ( 1 + 1 α ) Φ ( 1 ) ]
where Φ is defined as in this Wikipedia article:
Φ ( x ) := 0 x | 1 u | u d u = { Li 2 ( x ) ( x 1 ) π 2 3 Li 2 ( 1 x ) 1 2 ( log ( x ) ) 2 ( x > 1 )
Combining the two and using Li 2 ( 1 ) = π 2 / 6 yield
I = 1 2 ( log α ) 2 π 2 6 + Li 2 ( α 1 + α ) + 1 2 ( log ( 1 + α ) ) 2 .
Finally, this identity:
Li 2 ( z ) + Li 2 ( z z 1 ) = 1 2 ( log ( 1 z ) ) 2
simplifies the last two terms into Li 2 ( α ), giving the desired result.
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