Equivalent of $\int_0^{\infty} \frac{\mathrm dx}{(1+x^3)^n},n\rightarrow\infty$ 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 $(3n-4)\times(3n-7)\times\cdots\times5\times2$ ?
 A: $$ \prod_{k=1}^{n-1} 3k-1 = \frac{3^{n-1} \Gamma (n-1/3) }{\Gamma(2/3)} $$ so now you can use Stirling's series to find an asymptotic for your integral. I get that $$\int^{\infty}_0 \frac{1}{(1+x^3)^n} dx = \frac{\Gamma(4/3) }{\sqrt[3]{n}} \left( 1 + \frac{2}{9n} + \mathcal{O}(n^{-2}) \right).$$ So you can combine this result with $\displaystyle \sum_{k=1}^n k^p = \frac{n^{p+1}}{p+1} + \frac{n^p}{2} + \mathcal{O}(n^{p-1})$ to find an asymptotic for your sum.
A: 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. 
