According to my calculations ∫ 0 ∞ d x ( 1 + x 3 )...

bandikizaui

bandikizaui

Answered

2022-07-09

According to my calculations
0 d x ( 1 + x 3 ) n = ( 3 n 4 ) × ( 3 n 7 ) × × 5 × 2 3 n + 1 / 2 ( n 1 ) ! 2 π

Answer & Explanation

Allison Pena

Allison Pena

Expert

2022-07-10Added 14 answers

k = 1 n 1 3 k 1 = 3 n 1 Γ ( n 1 / 3 ) Γ ( 2 / 3 )
so now you can use Stirling's series to find an asymptotic for your integral. I get that
0 1 ( 1 + x 3 ) n d x = Γ ( 4 / 3 ) n 3 ( 1 + 2 9 n + O ( n 2 ) ) .
So you can combine this result with k = 1 n k p = n p + 1 p + 1 + n p 2 + O ( n p 1 ) to find an asymptotic for your sum.
Audrina Jackson

Audrina Jackson

Expert

2022-07-11Added 4 answers

The integral is the beta function in disguise. Let x = ( z 1 z ) 1 / 3 . Then
I n = 0 d x ( 1 + x 3 ) n = 1 3 0 1 d z z 2 / 3 ( 1 z ) n 4 / 3 = 1 3 B ( n 1 / 3 , 1 / 3 ) = 1 3 Γ ( 1 / 3 ) Γ ( n 1 / 3 ) Γ ( n ) = Γ ( 4 / 3 ) Γ ( n 1 / 3 ) Γ ( n ) .
If the upper limit is finite, a closed expression can be found for your sum,
n = 1 N I n = 3 2 Γ ( 4 / 3 ) Γ ( N + 2 / 3 ) Γ ( N ) .

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