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

Beginner2022-10-08Added 11 answers

Answer:

$$\begin{array}{rl}{\int}_{1}^{a}\frac{n}{1+{x}^{n}}\phantom{\rule{thinmathspace}{0ex}}dx& ={\int}_{1}^{a}\frac{n}{{x}^{n}}\left(\frac{1}{1+(1/x{)}^{n}}\right)dx\\ & ={\int}_{1}^{a}\frac{n}{{x}^{n}}(1-\frac{1}{{x}^{n}}+\frac{1}{{x}^{2n}}-\cdots )dx\\ & =n{\int}_{1}^{a}\frac{1}{{x}^{n}}-\frac{1}{{x}^{2n}}+\frac{1}{{x}^{3n}}-\cdots \phantom{\rule{thinmathspace}{0ex}}dx\\ & =n[\frac{1}{n-1}-\frac{1}{2n-1}+\cdots ]-\underset{\to \phantom{\rule{thinmathspace}{0ex}}0}{\underset{\u23df}{n[\frac{{a}^{1-n}}{n-1}-\frac{{a}^{1-2n}}{2n-1}+\cdots ]}}\end{array}$$

$$\text{Since}\phantom{\rule{thickmathspace}{0ex}}\underset{n\to \mathrm{\infty}}{lim}\frac{n}{kn-1}=\frac{1}{k}:$$

$$\underset{n\to \mathrm{\infty}}{lim}{\int}_{1}^{a}\frac{n}{1+{x}^{n}}dx=1-\frac{1}{2}+\frac{1}{3}-\cdots =\mathrm{ln}2$$

$$\begin{array}{rl}{\int}_{1}^{a}\frac{n}{1+{x}^{n}}\phantom{\rule{thinmathspace}{0ex}}dx& ={\int}_{1}^{a}\frac{n}{{x}^{n}}\left(\frac{1}{1+(1/x{)}^{n}}\right)dx\\ & ={\int}_{1}^{a}\frac{n}{{x}^{n}}(1-\frac{1}{{x}^{n}}+\frac{1}{{x}^{2n}}-\cdots )dx\\ & =n{\int}_{1}^{a}\frac{1}{{x}^{n}}-\frac{1}{{x}^{2n}}+\frac{1}{{x}^{3n}}-\cdots \phantom{\rule{thinmathspace}{0ex}}dx\\ & =n[\frac{1}{n-1}-\frac{1}{2n-1}+\cdots ]-\underset{\to \phantom{\rule{thinmathspace}{0ex}}0}{\underset{\u23df}{n[\frac{{a}^{1-n}}{n-1}-\frac{{a}^{1-2n}}{2n-1}+\cdots ]}}\end{array}$$

$$\text{Since}\phantom{\rule{thickmathspace}{0ex}}\underset{n\to \mathrm{\infty}}{lim}\frac{n}{kn-1}=\frac{1}{k}:$$

$$\underset{n\to \mathrm{\infty}}{lim}{\int}_{1}^{a}\frac{n}{1+{x}^{n}}dx=1-\frac{1}{2}+\frac{1}{3}-\cdots =\mathrm{ln}2$$

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