TITLE: Equivalent of $\int_0^{\infty} \frac{\mathrm dx}{(1+x^3)^n},n\rightarrow\infty$
QUESTION [7 upvotes]: According to my calculations
$$ \int_0^\infty \frac{\mathrm dx}{(1+x^3)^n}=\frac{(3n-4)\times(3n-7)\times\cdots\times5\times2}{3^{n+1/2}(n-1)!}2\pi$$
How can an equivalent of $$ \int_0^\infty \frac{\mathrm dx}{(1+x^3)^n}$$ be derived from this formula?
(Given that my objective is to study the nature of the series $ \sum \int_0^\infty \frac{\mathrm dx}{(1+x^3)^n} $)
So my question is simply: is there a simple equivalent for $(3n-4)\times(3n-7)\times\cdots\times5\times2$ ?

REPLY [3 votes]: The integral is the beta function in disguise. 
Let $x=\left(\frac{z}{1-z}\right)^{1/3}$.
Then 
$$\begin{eqnarray*}
I_n &=& \int_0^\infty \frac{dx}{(1+x^3)^n} \\
    &=& \frac{1}{3} \int_0^1 dz\, z^{-2/3}(1-z)^{n-4/3} \\
    &=& \frac{1}{3} B(n-1/3,1/3) \\
    &=& \frac{1}{3} \Gamma(1/3) \frac{\Gamma(n-1/3)}{\Gamma(n)} \\
    &=& \Gamma(4/3) \frac{\Gamma(n-1/3)}{\Gamma(n)}. 
\end{eqnarray*}$$
If the upper limit is finite, a closed expression can be found for your sum, 
$$\sum_{n=1}^N I_n = \frac{3}{2} \Gamma(4/3) 
\frac{\Gamma(N+2/3)}{\Gamma(N)}. $$
In the limit $N\to\infty$ the sum is divergent.
It goes like 
$\frac{3}{2} \Gamma(4/3) N^{2/3}$, 
as hinted at by @RagibZaman.