Evaluation of $\lim_{n \to \infty} \int_{0}^\infty \frac{1}{1+x^n} dx$ $$L=\lim_{n \to \infty} \int_{0}^\infty \frac{1}{1+x^n} dx$$
$$\phi(x)=\lim_{n \to \infty}\frac{1}{1+x^n}$$
So I was just playing around with $\int_{0}^\infty \frac{1}{1+x^2} dx $ and it equals to $\frac{\pi}{2}$. So I thought if this integral converges then higher powers of x it must also. And for that matter what is the limit as power becomes infinitely large? 
1) I tried graphing $\phi(x)$ and I have a at x=0 the function equals 1 but it also equals 1 for all $x \in [0,1)$
2) And at 1 it equals 1/2 and it equals 0 for, $x \in (1,\infty)$
Based on 1) and 2) I have a hunch that $L=1$.
Any rigorous proofs?
Thanks in advance. And tell me about this question's or this type's formal name because I didn't know what to search on google.
 A: You can also show with a beautiful contour integral that for $n \geq 2$

$$
\int_{0}^{+\infty}\frac{\text{d}x}{1+x^n}=\frac{\pi}{n\sin\left(\displaystyle \frac{\pi}{n}\right)}
$$

using that
$$
\sin\left(\frac{\pi}{n}\right) \underset{(+\infty)}{\sim}\frac{\pi}{n}
$$
You find that

$$
\int_{0}^{+\infty}\frac{\text{d}x}{1+x^n} \underset{n \rightarrow +\infty}{\rightarrow}1
$$

A: 
I thought it would be instructive to present a way forward that relies only on elementary calculus tools.  To that end, we now proceed.


Enforcing the substitution $x\mapsto x^{1/n}$, we see that for $n>1$
$$\begin{align}
\int_0^\infty \frac1{1+x^n}\,dx&=\frac1n \int_0^\infty \frac{x^{1/n}}{x(1+x)}\,dx\tag 1
\end{align}$$
Writing the integral on the right-hand side of $(1)$ as the sum 
$$\begin{align}
\int_0^\infty \frac{x^{1/n}}{x(1+x)}\,dx&=\int_0^1 \frac{x^{1/n}}{x(1+x)}\,dx+\int_1^\infty \frac{x^{1/n}}{x(1+x)}\,dx\tag2
\end{align}$$
and enforcing the substitution $x\mapsto 1/x$ in the second integral on the right-hand side of $(2)$ reveals
$$\begin{align}
\int_0^\infty \frac1{1+x^n}\,dx&=\frac1n \int_0^1 \frac{x^{1/n}+x^{1-1/n}}{x(1+x)}\,dx\tag3
\end{align}$$
Next, using partial fraction expansion, we find that 
$$\begin{align}
\int_0^\infty \frac1{1+x^n}\,dx&=\color{blue}{\frac1n\int_0^1 \frac{x^{1/n}+x^{1-1/n}}{x}\,dx}-\frac1n\int_0^1 \frac{x^{1/n}+x^{1-1/n}}{1+x}\,dx\\\\
&=\color{blue}{1+\frac1{n-1}}-\frac1n\int_0^1 \frac{x^{1/n}+x^{1-1/n}}{1+x}\,dx\tag4
\end{align}$$
Inasmuch as the integral on the right-hand side of $(4)$ is trivially seen to be bounded in absolute value by $2$, we find that 
$$\int_0^\infty \frac1{1+x^n}\,dx=1+O\left(\frac1n\right)$$

Taking the limit as $n\to \infty$, yields the coveted limit
$$\bbox[5px,border:2px solid #C0A000]{\lim_{n\to \infty}\int_0^\infty \frac1{1+x^n}\,dx=1}$$


TOOLS USED:  Elementary Integral Theorems, Substitution, Partial Fraction Expansion

A: Lebesgue dominated convergence theorem with dominating function $f(x)=1$ on $[0,1]$ and $\frac{1}{1+x^2}$ on $(1,+\infty)$ will give to you 
$$ f(x)=\int_0^{+\infty}\lim_n \frac{1}{1+x^n} d x= \int_0^{+\infty}\phi(x) d x=1.$$
For the computation of $\phi$, you have to show that $x^n\rightarrow 0$ if $0\leq x <1$, $1$ if $x=1$, $+\infty $ if $x>1$, which is straightforward from the definitions...
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
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\begin{align}
L & \equiv \lim_{n \to \infty}\int_{0}^{\infty}{1 \over 1 + x^{n}}\,\dd x =
\lim_{n \to \infty}\pars{{1 \over n}
\int_{0}^{\infty}{x^{1/n - 1} \over 1 + x}\,\dd x}
\\[5mm] &=
\lim_{n \to \infty}\bracks{{1 \over n}
\int_{0}^{\infty}x^{1/n - 1}\int_{0}^{\infty}\expo{-\pars{1 + x}t}
\,\dd t\,\dd x} =
\lim_{n \to \infty}\bracks{{1 \over n}
\int_{0}^{\infty}\expo{-t}\
\overbrace{\int_{0}^{\infty}x^{1/n - 1}\expo{-tx}
\,\dd x}^{\ds{\Gamma\pars{1/n} \over t^{1/n}}}\ \,\dd t}
\\[5mm] & =
\lim_{n \to \infty}\bracks{{\Gamma\pars{1/n} \over n}
\int_{0}^{\infty}t^{-1/n}\expo{-t}\,\dd t} =
\lim_{n \to \infty}\bracks{{\Gamma\pars{1/n} \over n}\,
\Gamma\pars{-\,{1 \over n} + 1}} =
\lim_{n \to \infty}\bracks{{1 \over n}\,
{\pi \over \sin\pars{\pi/n}}}
\\[5mm] & = \bbx{1}
\end{align}
A: 
In a comment under the answer posted by @Atmos, there was a supposition that one could use real analysis only to show that $\int_0^\infty \frac1{1+x^n}\,dx=\frac{\pi/n}{\sin(\pi/n)}$.  We now proceed to validate that hypothesis.


We begin by enforcing the substitution $x\mapsto x^{1/n}$ to find
$$\int_0^\infty \frac1{1+x^n}\,dx=\frac1n\int_0^\infty \frac{x^{1/n-1}}{1+x}\,dx\tag1$$
Next, enforcing the substitution $x\mapsto \frac{x}{1-x}$ in the integral on the right-hand side of $(1)$ reveals
$$\int_0^\infty \frac1{1+x^n}\,dx=\frac1n\int_0^1 x^{1/n-1}(1-x)^{-1/n}\,dx\tag2$$
We recognize that the right-hand side of $(2)$ is the Beta function, $B(x,y)$, evaluated at $(1/n,1-1/n)$.  Using the relationship, $B(x,y)=\frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}$ (SEE THE NOTES IN THIS ANSWER), in $(2)$, we find that 
$$\int_0^\infty \frac1{1+x^n}\,dx=\frac1n\Gamma(1/n)\Gamma(1-1/n)\tag3$$
Finally, using Euler's reflection principal, $\Gamma(z)\Gamma(1-z)=\frac{\pi}{\sin(\pi z)}$ (SEE THE NOTES IN THIS ANSWER), in $(3)$ yields the coveted result
$$\bbox[5px,border:2px solid #C0A000]{\int_0^\infty \frac1{1+x^n}\,dx=\frac{\pi/n}{\sin(\pi/n)}}$$

NOTE:
Another way forward is to use the approach used in the OP of THIS QUESTION. 
A: Here is a super-elementary proof.
I will show that
$1-\dfrac{1}{n+1}-\dfrac{\ln(n)}{n+1}
\lt\int_0^{\infty} \dfrac{dx}{1+x^n}
\lt 1+\dfrac1{n-2}
$.
First, $\int_0^1$.
$\begin{array}\\
\int_0^{n^{-1/n}} \dfrac{dx}{1+x^n}
&>\int_0^{n^{-1/n}} \dfrac{dx}{1+1/n}
\qquad x^n < 1/n\\
&=\dfrac{n(n^{-1/n})}{n+1}\\
&=\dfrac{n(e^{-\ln(n)/n})}{n+1}\\
&>\dfrac{n(1-\ln(n)/n)}{n+1}
\qquad\text{since } e^x > 1+x\\
&=\dfrac{n-\ln(n)}{n+1}\\
&=\dfrac{n}{n+1}-\dfrac{\ln(n)}{n+1}\\
&=1-\dfrac{1}{n+1}-\dfrac{\ln(n)}{n+1}\\
\text{so}\\
\int_0^1 \dfrac{dx}{1+x^n}
&>1-\dfrac{1}{n+1}-\dfrac{\ln(n)}{n+1}\\
\text{and}\\
\int_0^1 \dfrac{dx}{1+x^n}
&< 1\\
\end{array}
$
Since
$1+z \ge 2\sqrt{z}$
if $z > 0$,
$1+x^n \ge 2x^{n/2}
$
so
$\dfrac1{1+x^n}
\le \dfrac1{2x^{n/2}}
$
so
$\begin{array}\\
\int_1^{\infty} \dfrac{dx}{1+x^n}
&\le \int_1^{\infty} \dfrac{dx}{2x^{n/2}}\\
&= \frac12\int_1^{\infty} x^{-n/2}dx\\
&= \frac12\dfrac{x^{-n/2+1}}{-n/2+1}|_1^{\infty}\\
&=\dfrac{1}{n-2}\\
&\to 0\\
\end{array}
$
