Are there other cases similar to Herglotz's integral <msubsup> &#x222B;<!-- ∫ --> 0

Antoine Hill

Antoine Hill

Answered question

2022-05-19

Are there other cases similar to Herglotz's integral 0 1 ln ( 1 + t 4 + 15 ) 1 + t   d t?
This post of Boris Bukh mentions amazing Gustav Herglotz's integral
0 1 ln ( 1 + t 4 + A 15 ) 1 + t   d t = π 2 12 ( 15 2 ) + ln 2 ln ( 3 + 5 ) + ln 1 + 5 2 ln ( 2 + 3 ) .
I wonder if there are other irrational real algebraic exponents α such that the integral
0 1 ln ( 1 + t α ) 1 + t d t
has a closed-form representation? Is there a general formula giving results for such cases?
Are there such algebraic α of degree > 2?

Answer & Explanation

komizmtk

komizmtk

Beginner2022-05-20Added 8 answers

Here is a list of some of these integrals:
0 1 log ( 1 + t 2 + 3 ) 1 + t d t = π 2 12 ( 1 3 ) + log 2 log ( 1 + 3 ) 0 1 log ( 1 + t 3 + 8 ) 1 + t d t = π 2 24 ( 3 32 ) + 1 2 log 2 log ( 2 ( 3 + 8 ) 3 / 2 ) 0 1 log ( 1 + t 4 + 15 ) 1 + t d t = π 2 12 ( 2 15 ) + log ( 1 + 5 2 ) log ( 2 + 3 ) + + log 2 log ( 3 + 5 ) 0 1 log ( 1 + t 5 + 24 ) 1 + t d t = π 2 24 ( 5 96 ) + 1 2 log ( 1 + 2 ) log ( 2 + 3 ) + + 1 2 log 2 log ( 2 ( 5 + 24 ) 3 / 2 ) 0 1 log ( 1 + t 6 + 35 ) 1 + t d t = π 2 12 ( 3 35 ) + log ( 1 + 5 2 ) log ( 8 + 3 7 ) + + log 2 log ( 5 + 7 ) 0 1 log ( 1 + t 8 + 63 ) 1 + t d t = π 2 12 ( 4 63 ) + log ( 5 + 21 2 ) log ( 2 + 3 ) + + log 2 log ( 3 + 7 ) 0 1 log ( 1 + t 11 + 120 ) 1 + t d t = π 2 24 ( 11 480 ) + 1 2 log ( 1 + 2 ) log ( 4 + 15 ) + + 1 2 log ( 2 + 3 ) log ( 3 + 10 ) + + 1 2 log ( 1 + 5 2 ) log ( 5 + 24 ) + + 1 2 log 2 log ( 2 ( 11 + 120 ) 3 / 2 ) 0 1 log ( 1 + t 12 + 143 ) 1 + t d t = π 2 12 ( 6 143 ) + log ( 3 + 13 2 ) log ( 10 + 3 11 ) + + log 2 log ( 11 + 13 ) 0 1 log ( 1 + t 13 + 168 ) 1 + t d t = π 2 24 ( 13 672 ) + 1 2 log ( 1 + 2 ) log ( 5 + 21 2 ) + + 1 4 log ( 2 + 3 ) log ( 15 + 224 ) + + 1 4 log ( 5 + 24 ) log ( 8 + 63 ) + + 1 2 log 2 log ( 2 ( 13 + 168 ) 3 / 2 ) 0 1 log ( 1 + t 14 + 195 ) 1 + t d t = π 2 12 ( 7 195 ) + log ( 1 + 5 2 ) log ( 25 + 4 39 ) + + log ( 3 + 13 2 ) log ( 4 + 15 ) + + log 2 log ( 15 + 13 )
Rachel Villa

Rachel Villa

Beginner2022-05-21Added 4 answers

One that follows from the one above is
0 1 log ( 1 + t 4 15 ) 1 + t d t = log ( 2 ) 2 0 1 log ( 1 + t 4 + 15 ) 1 + t d t
I will look for others.

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