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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 $(3n-4)\times(3n-7)\times\cdots\times5\times2$ ?

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@Norbert Sure. $n!$ is divergent, but is has an asymptotic formula. That is what Chon wants. –  Pedro Tamaroff Jun 3 '12 at 18:06

2 Answers 2

up vote 4 down vote accepted

$$ \prod_{k=1}^{n-1} 3k-1 = \frac{3^{n-1} \Gamma (n-1/3) }{\Gamma(2/3)} $$ 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.

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How did you get $\Gamma(4/3)$ in the numerator? I found the equivalent: $$ \int_0^{\infty} \frac{\mathrm dx}{(1+x^3)^n} \sim \frac{2\pi}{\Gamma(2/3)3^{3/2}n^{1/3}}$$ So $$\Gamma(4/3)\Gamma(2/3)=\frac{2\pi}{3^{3/2}} $$ ? –  Chon Jun 3 '12 at 20:23
    
@Chon That is correct. –  Ragib Zaman Jun 4 '12 at 0:56

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.

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