I noticed from here that $$\lim_{n\to\infty} \int_0^\infty \frac{1}{1+x^n}dx = 1$$
Moreover if in split the integral limits as $0\to 1$ and $1\to\infty$ the limit is so obvious that first term is $1$ and last term is $0$. I don't have much experience with Real Analysis. Can anyone show me why I can/cannot use this agrument?
$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ With $\ds{t \equiv {1 \over 1 + x^{n}}\quad\imp\quad x = \pars{{1 \over t} - 1}^{1/n}}$:
\begin{align} &\color{#00f}{\large\lim_{n \to \infty}\int_{0}^{\infty}{1 \over 1 + x^{n}}\,\dd x} =\lim_{n \to \infty}\int_{1}^{0}t\,{1 \over n}\pars{{1 \over t} - 1}^{1/n - 1} \pars{-\,{1 \over t^{2}}}\,\dd t \\[3mm]&= \lim_{n \to \infty}\bracks{% {1 \over n}\int_{0}^{1}t^{-1/n}\pars{1 - t}^{1/n - 1}\,\dd t} =\lim_{n \to \infty}\bracks{% {1 \over n}\,{\Gamma\pars{1 - 1/n}\Gamma\pars{1/n} \over \Gamma\pars{1}}} \\[3mm]&=\lim_{n \to \infty}\bracks{\pi/n \over\sin\pars{\pi/n}} =\color{#00f}{\LARGE 1} \end{align}
where we used well known properties of the Beta $\ds{{\rm B}\pars{x,y}}$ and Gamma $\ds{\Gamma\pars{z}}$ functions.
There is a theorem known as the Lebesgue Dominated Convergence Theorem that states: If there is exists a function $g(x)$ such that $|f_n(x)| \leq g(x)$ and $\int_0^\infty g(x) dx < \infty$, then $\lim_{n \to \infty} \int_0^\infty f_n(x) dx = \int_0^\infty \lim_{n \to \infty} f_n(x) dx$ provided the limit exists. Can you find such a $g(x)$ so that you can apply this theorem? (Ignore $n =1$ and use $g(x)=1/(1+x^2)$ for $x \geq 1$ and $g(x) = 1 otherwise).
Alternatively, we can do the following. Split the integral into three integrals, $I(n) = I_1(n) + I_2(n) + I_3(n)$, given by $$\int_0^\infty \frac{1}{1+x^n} dx = \int_0^{1-\epsilon} \frac{1}{1+x^n} dx + \int_{1-\epsilon}^{1+\epsilon}\frac{1}{1+x^n} dx + \int_{1+\epsilon}^\infty \frac{1}{1+x^n} dx.$$
For $I_1(n)$ we have $1 \geq 1/(1+x^n) \geq 1/(1+(1-\epsilon)^n)$, which gives uniform convergence and hence $I_1(n) = 1-\epsilon$ in the limit.
Similarly, for $I_3(n)$, $1/(1+x^n) \leq 1/(1+(1+\epsilon)^n)$ gives uniform convergence to $0$ and $I_3(n) = 0$ in the limit.
Finally, we can show $|I_2(n)| \leq \int_{1-\epsilon}^{1+\epsilon} 1 dx = 2\epsilon$. Putting this all together, we have $$\limsup_{n \to \infty} (I(n)-1) \leq \epsilon + 2\epsilon + 0.$$ Let $\epsilon \to 0$ and we see that the limit exists and is zero, hence the original limit is 1.
Why do we need to split it up into three intervals? The reason is so that we can get uniform convergence on two of them. Splitting into $(0,1)$ and $[1, \infty)$ does not give uniform convergence because $f_n(1)=1/2$ for all $n$. But if we leave a little wiggle room of order $\epsilon$, then we have uniform convergence. It then turns out that taking $\epsilon \to 0$ smooths everything out. Pay attention since this is a common trick.
