bandikizaui

2022-07-09

According to my calculations
${\int }_{0}^{\mathrm{\infty }}\frac{\mathrm{d}x}{\left(1+{x}^{3}{\right)}^{n}}=\frac{\left(3n-4\right)×\left(3n-7\right)×\cdots ×5×2}{{3}^{n+1/2}\left(n-1\right)!}2\pi$

Allison Pena

Expert

$\prod _{k=1}^{n-1}3k-1=\frac{{3}^{n-1}\mathrm{\Gamma }\left(n-1/3\right)}{\mathrm{\Gamma }\left(2/3\right)}$
so now you can use Stirling's series to find an asymptotic for your integral. I get that
${\int }_{0}^{\mathrm{\infty }}\frac{1}{\left(1+{x}^{3}{\right)}^{n}}dx=\frac{\mathrm{\Gamma }\left(4/3\right)}{\sqrt[3]{n}}\left(1+\frac{2}{9n}+\mathcal{O}\left({n}^{-2}\right)\right).$
So you can combine this result with $\sum _{k=1}^{n}{k}^{p}=\frac{{n}^{p+1}}{p+1}+\frac{{n}^{p}}{2}+\mathcal{O}\left({n}^{p-1}\right)$ to find an asymptotic for your sum.

Audrina Jackson

Expert

The integral is the beta function in disguise. Let $x={\left(\frac{z}{1-z}\right)}^{1/3}$. Then
$\begin{array}{rcl}{I}_{n}& =& {\int }_{0}^{\mathrm{\infty }}\frac{dx}{\left(1+{x}^{3}{\right)}^{n}}\\ & =& \frac{1}{3}{\int }_{0}^{1}dz\phantom{\rule{thinmathspace}{0ex}}{z}^{-2/3}\left(1-z{\right)}^{n-4/3}\\ & =& \frac{1}{3}B\left(n-1/3,1/3\right)\\ & =& \frac{1}{3}\mathrm{\Gamma }\left(1/3\right)\frac{\mathrm{\Gamma }\left(n-1/3\right)}{\mathrm{\Gamma }\left(n\right)}\\ & =& \mathrm{\Gamma }\left(4/3\right)\frac{\mathrm{\Gamma }\left(n-1/3\right)}{\mathrm{\Gamma }\left(n\right)}.\end{array}$
If the upper limit is finite, a closed expression can be found for your sum,
$\sum _{n=1}^{N}{I}_{n}=\frac{3}{2}\mathrm{\Gamma }\left(4/3\right)\frac{\mathrm{\Gamma }\left(N+2/3\right)}{\mathrm{\Gamma }\left(N\right)}.$

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