bandikizaui

Answered

2022-07-09

According to my calculations

${\int}_{0}^{\mathrm{\infty}}\frac{\mathrm{d}x}{(1+{x}^{3}{)}^{n}}=\frac{(3n-4)\times (3n-7)\times \cdots \times 5\times 2}{{3}^{n+1/2}(n-1)!}2\pi $

${\int}_{0}^{\mathrm{\infty}}\frac{\mathrm{d}x}{(1+{x}^{3}{)}^{n}}=\frac{(3n-4)\times (3n-7)\times \cdots \times 5\times 2}{{3}^{n+1/2}(n-1)!}2\pi $

Answer & Explanation

Allison Pena

Expert

2022-07-10Added 14 answers

$\prod _{k=1}^{n-1}3k-1=\frac{{3}^{n-1}\mathrm{\Gamma}(n-1/3)}{\mathrm{\Gamma}(2/3)}$

so now you can use Stirling's series to find an asymptotic for your integral. I get that

${\int}_{0}^{\mathrm{\infty}}\frac{1}{(1+{x}^{3}{)}^{n}}dx=\frac{\mathrm{\Gamma}(4/3)}{\sqrt[3]{n}}(1+\frac{2}{9n}+\mathcal{O}({n}^{-2})).$

So you can combine this result with $\sum _{k=1}^{n}{k}^{p}=\frac{{n}^{p+1}}{p+1}+\frac{{n}^{p}}{2}+\mathcal{O}({n}^{p-1})$ to find an asymptotic for your sum.

so now you can use Stirling's series to find an asymptotic for your integral. I get that

${\int}_{0}^{\mathrm{\infty}}\frac{1}{(1+{x}^{3}{)}^{n}}dx=\frac{\mathrm{\Gamma}(4/3)}{\sqrt[3]{n}}(1+\frac{2}{9n}+\mathcal{O}({n}^{-2})).$

So you can combine this result with $\sum _{k=1}^{n}{k}^{p}=\frac{{n}^{p+1}}{p+1}+\frac{{n}^{p}}{2}+\mathcal{O}({n}^{p-1})$ to find an asymptotic for your sum.

Audrina Jackson

Expert

2022-07-11Added 4 answers

The integral is the beta function in disguise. Let $x={\left(\frac{z}{1-z}\right)}^{1/3}$. Then

$\begin{array}{rcl}{I}_{n}& =& {\int}_{0}^{\mathrm{\infty}}\frac{dx}{(1+{x}^{3}{)}^{n}}\\ & =& \frac{1}{3}{\int}_{0}^{1}dz\phantom{\rule{thinmathspace}{0ex}}{z}^{-2/3}(1-z{)}^{n-4/3}\\ & =& \frac{1}{3}B(n-1/3,1/3)\\ & =& \frac{1}{3}\mathrm{\Gamma}(1/3)\frac{\mathrm{\Gamma}(n-1/3)}{\mathrm{\Gamma}(n)}\\ & =& \mathrm{\Gamma}(4/3)\frac{\mathrm{\Gamma}(n-1/3)}{\mathrm{\Gamma}(n)}.\end{array}$

If the upper limit is finite, a closed expression can be found for your sum,

$\sum _{n=1}^{N}{I}_{n}=\frac{3}{2}\mathrm{\Gamma}(4/3)\frac{\mathrm{\Gamma}(N+2/3)}{\mathrm{\Gamma}(N)}.$

$\begin{array}{rcl}{I}_{n}& =& {\int}_{0}^{\mathrm{\infty}}\frac{dx}{(1+{x}^{3}{)}^{n}}\\ & =& \frac{1}{3}{\int}_{0}^{1}dz\phantom{\rule{thinmathspace}{0ex}}{z}^{-2/3}(1-z{)}^{n-4/3}\\ & =& \frac{1}{3}B(n-1/3,1/3)\\ & =& \frac{1}{3}\mathrm{\Gamma}(1/3)\frac{\mathrm{\Gamma}(n-1/3)}{\mathrm{\Gamma}(n)}\\ & =& \mathrm{\Gamma}(4/3)\frac{\mathrm{\Gamma}(n-1/3)}{\mathrm{\Gamma}(n)}.\end{array}$

If the upper limit is finite, a closed expression can be found for your sum,

$\sum _{n=1}^{N}{I}_{n}=\frac{3}{2}\mathrm{\Gamma}(4/3)\frac{\mathrm{\Gamma}(N+2/3)}{\mathrm{\Gamma}(N)}.$

Most Popular Questions