Integral <msubsup> &#x222B;<!-- ∫ --> 0 1 </msubsup> ln

Petrovcic2x

Petrovcic2x

Answered question

2022-06-24

Integral 0 1 ln x x 2 + 1 ln ( 3 x 2 + 1 x 2 + 3 ) d x
I need to evaluate the following integral:
0 1 ln x x 2 + 1 ln ( 3 x 2 + 1 x 2 + 3 ) d x .
Could you suggest how to find a closed form for it? I am not sure if there is one, but the integrand seems simple enough, so I hope it might exist.

Answer & Explanation

Kaydence Washington

Kaydence Washington

Beginner2022-06-25Added 32 answers

My calculation shows that
(1) 0 1 log x x 2 + 1 log ( t 2 x 2 + 1 x 2 + t 2 ) d x = π χ 2 ( 1 t 1 + t ) ,
where χ 2 stands for the Legendre chi function. I will post a detailed solution later, but I should note that the idea is very simple: denote this integral as I ( t ) and differentiate it to obtain
I ( t ) = π log t 1 t 2 .
Restricting | t | < 1 temporarily, this gives
I ( t ) = I ( 0 ) + 0 t I ( s ) d s = π { π 2 8 + log t artanh t χ 2 ( t ) } .
Then (1) restricted to | t | < 1 follows from the following identity:
χ 2 ( 1 t 1 + t ) + χ 2 ( t ) = π 2 8 + log t artanh t .Then the equality for the general t follows by analytic continuation. Here is a Mathematica code for testing this:

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