Cynthia Bell

2022-01-04

Solving limits:
$f\left(x\right)=\underset{n\to \mathrm{\infty }}{lim}\left(n{\int }_{0}^{\frac{\pi }{4}}{\left(\mathrm{tan}x\right)}^{n}dx\right)$

Hector Roberts

You may perform the change of variable $u=\mathrm{tan}x$ to get easily
${I}_{n}\phantom{\rule{0.222em}{0ex}}={\int }_{0}^{\frac{\pi }{4}}{\left(\mathrm{tan}x\right)}^{n}dx={\int }_{0}^{1}\frac{{u}^{n}}{1+{u}^{2}}du$
Then you may just integrate by parts,
${I}_{n}={\int }_{0}^{1}\frac{{u}^{n}}{1+{u}^{2}}du=\frac{{u}^{n+1}}{\left(n+1\right)}\frac{1}{1+{u}^{2}}{\mid }_{0}^{1}+\frac{2}{\left(n+1\right)}{\int }_{0}^{1}\frac{{u}^{n+2}}{{\left(1+{u}^{2}\right)}^{2}}du$
$=\frac{1}{2}\frac{1}{\left(n+1\right)}+\frac{2}{\left(n+1\right)}{\int }_{0}^{1}\frac{{u}^{n+2}}{{\left(1+{u}^{2}\right)}^{2}}du$ (1)
Observing that
$0\le {\int }_{0}^{1}\frac{{u}^{n+2}}{{\left(1+{u}^{2}\right)}^{2}}du\le {\int }_{0}^{1}{u}^{n}du=\frac{1}{n+1}$
gives
$0\le \frac{2}{\left(n+1\right)}{\int }_{0}^{1}\frac{{u}^{n}}{{\left(1+{u}^{2}\right)}^{2}}du\le \frac{2}{{\left(n+1\right)}^{2}}$ (2)
Then combining (1) and (2) leads to
$\underset{n\to +\mathrm{\infty }}{lim}n{\int }_{0}^{\frac{\pi }{4}}{\left(\mathrm{tan}x\right)}^{n}dx=\underset{n\to +\mathrm{\infty }}{lim}n{I}_{n}=\frac{12}{}$

Orlando Paz

Heres

karton

After letting $\mathrm{tan}\left(x\right)↦x$ and using the elementary limit $\underset{n\to \mathrm{\infty }}{lim}n{\int }_{0}^{1}{x}^{n}f\left(x\right)dx=f\left(1\right)$ where f(x) is continuous, we conclude that
$\underset{n\to \mathrm{\infty }}{lim}\left(n{\int }_{0}^{\pi /4}\left(\mathrm{tan}x{\right)}^{n}dx\right)=\frac{1}{2}$

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