Prove in_0^1(t^2-1)/((t^2+1)log t)dt=2log((2Gamma(5/4))/(Gamma(3/4)))

omgespit9q

omgespit9q

Answered question

2022-10-20

Prove 0 1 t 2 1 ( t 2 + 1 ) log t d t = 2 log ( 2 Γ ( 5 4 ) Γ ( 3 4 ) )

Answer & Explanation

Layne Murillo

Layne Murillo

Beginner2022-10-21Added 14 answers

Let's introduce the parameter α, and then differentiate with respect to α that yields
I ( α ) = 0 1 t α 1 ( t 2 + 1 ) ln t d t
I ( α ) = 0 1 t α ( t 2 + 1 ) d t = 1 4 ( ψ 0 ( 1 + α 4 ) + ψ 0 ( 3 + α 4 ) )
Then
I ( α ) = 1 4 ( ψ 0 ( 1 + α 4 ) + ψ 0 ( 3 + α 4 ) ) d α =
(1) I ( α ) = ( ln Γ ( 3 + α 4 ) ln Γ ( 1 + α 4 ) ) + C
If letting α = 2, then
I ( 2 ) = ln ( Γ ( 5 4 ) Γ ( 3 4 ) ) + C
On the other hand, by letting α = 0 in (1) we get
C = ln ( Γ ( 1 4 ) Γ ( 3 4 ) )
Thus
0 1 t 2 1 ( t 2 + 1 ) ln t d t = ln ( Γ ( 5 4 ) Γ ( 1 4 ) Γ 2 ( 3 4 ) )
Brianna Schmidt

Brianna Schmidt

Beginner2022-10-22Added 6 answers

Answer:
0 1 t + 1 t 2 + 1 t 1 log ( t ) d t = 0 1 t + 1 t 2 + 1 0 1 t x d x d t = 0 1 0 1 t x + 1 + t x t 2 + 1 d t d x = 0 1 ( 1 x + 1 + 1 x + 2 1 x + 3 1 x + 4 + ) d x = ( log ( 2 1 ) + log ( 3 2 ) ) ( log ( 4 3 ) + log ( 5 4 ) ) + = log ( 3 1 ) log ( 5 3 ) + log ( 7 5 ) log ( 9 7 ) + log ( 11 9 ) = log ( 3 1 3 5 7 5 7 9 11 9 ) = lim n log ( Γ ( 5 4 ) 2 Γ ( 3 4 ) 2 Γ ( 4 n + 3 4 ) 2 Γ ( 4 n + 5 4 ) 2 ( 4 n + 3 ) ) = 2 log ( 2 Γ ( 5 4 ) Γ ( 3 4 ) )
The last equality is due to Gautschi's inequality.

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