Aryan Lowery

2022-10-07

Finding $\underset{n\to \mathrm{\infty }}{lim}{\int }_{1}^{a}\frac{n}{1+{x}^{n}}\phantom{\rule{thinmathspace}{0ex}}dx$

barquegese2

$\begin{array}{rl}{\int }_{1}^{a}\frac{n}{1+{x}^{n}}\phantom{\rule{thinmathspace}{0ex}}dx& ={\int }_{1}^{a}\frac{n}{{x}^{n}}\left(\frac{1}{1+\left(1/x{\right)}^{n}}\right)dx\\ & ={\int }_{1}^{a}\frac{n}{{x}^{n}}\left(1-\frac{1}{{x}^{n}}+\frac{1}{{x}^{2n}}-\cdots \right)dx\\ & =n{\int }_{1}^{a}\frac{1}{{x}^{n}}-\frac{1}{{x}^{2n}}+\frac{1}{{x}^{3n}}-\cdots \phantom{\rule{thinmathspace}{0ex}}dx\\ & =n\left[\frac{1}{n-1}-\frac{1}{2n-1}+\cdots \right]-\underset{\to \phantom{\rule{thinmathspace}{0ex}}0}{\underset{⏟}{n\left[\frac{{a}^{1-n}}{n-1}-\frac{{a}^{1-2n}}{2n-1}+\cdots \right]}}\end{array}$
$\text{Since}\phantom{\rule{thickmathspace}{0ex}}\underset{n\to \mathrm{\infty }}{lim}\frac{n}{kn-1}=\frac{1}{k}:$
$\underset{n\to \mathrm{\infty }}{lim}{\int }_{1}^{a}\frac{n}{1+{x}^{n}}dx=1-\frac{1}{2}+\frac{1}{3}-\cdots =\mathrm{ln}2$

Do you have a similar question?