Deborah Proctor

2022-10-18

If:
$\lambda ={\int }_{0}^{1}\frac{dx}{1+{x}^{3}};$
Then evaluate:
$p=\underset{n\to \mathrm{\infty }}{lim}{\left(\frac{\prod _{r=1}^{n}\left({n}^{3}+{r}^{3}\right)}{{n}^{3n}}\right)}^{1/n};$

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Expert

First, to evaluate $\lambda$, use partial fractions:
$\frac{1}{1+{x}^{3}}=\frac{1}{3}\left(\frac{1}{1+x}-\frac{x-2}{{x}^{2}-x+1}\right)$
The second piece in the parentheses is more amenable to integration when expressed as follows:
$\frac{x-2}{{x}^{2}-x+1}=\frac{x-1/2}{\left(x-1/2{\right)}^{2}+3/4}-\frac{3/2}{\left(x-1/2{\right)}^{2}+3/4}$
Now, let
$A={\int }_{0}^{1}\frac{dx}{1+x}=\mathrm{log}2$
$B={\int }_{0}^{1}dx\phantom{\rule{mediummathspace}{0ex}}\frac{x-1/2}{\left(x-1/2{\right)}^{2}+3/4}={\left[\mathrm{log}\left[{\left(x-\frac{1}{2}\right)}^{2}+\frac{3}{4}\right]\right]}_{0}^{1}=0$
$C={\int }_{0}^{1}dx\phantom{\rule{mediummathspace}{0ex}}\frac{3/2}{\left(x-1/2{\right)}^{2}+3/4}=\sqrt{3}{\left[\mathrm{arctan}\left[\frac{2}{\sqrt{3}}\left(x-\frac{1}{2}\right)\right]\right]}_{0}^{1}=\frac{\pi }{\sqrt{3}}$
Therefore
$\lambda =\frac{1}{3}\left(A-B+C\right)=\frac{1}{3}\mathrm{log}2+\frac{\pi }{3\sqrt{3}}$
Now for the limit, which is evaluated by taking logs of both sides:
$\mathrm{log}p=\underset{n\to \mathrm{\infty }}{lim}\frac{1}{n}\sum _{r=0}^{n}\mathrm{log}\left(1+\frac{{r}^{3}}{{n}^{3}}\right)$
This is a Riemann sum, in the sense that
$\underset{n\to \mathrm{\infty }}{lim}\frac{1}{n}\sum _{r=0}^{n}f\left(\frac{r}{n}\right)={\int }_{0}^{1}dx\phantom{\rule{mediummathspace}{0ex}}f\left(x\right)$
Therefore
$\begin{array}{rl}\mathrm{log}p& ={\int }_{0}^{1}dx\phantom{\rule{mediummathspace}{0ex}}\mathrm{log}\left(1+{x}^{3}\right)\\ & =\left[x\mathrm{log}\left(1+{x}^{3}\right){\right]}_{0}^{1}-{\int }_{0}^{1}dx\phantom{\rule{mediummathspace}{0ex}}\frac{3{x}^{3}}{1+{x}^{3}}\\ & =\mathrm{log}2-3{\int }_{0}^{1}dx\phantom{\rule{mediummathspace}{0ex}}\left(1-\frac{1}{1+{x}^{3}}\right)\\ & =\mathrm{log}2-3+3\lambda \\ & =2\mathrm{log}2-3+\frac{\pi }{\sqrt{3}}\\ \therefore p& =4{e}^{-\left(3-\frac{\pi }{\sqrt{3}}\right)}\end{array}$

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