Evaluation of $\lim_{n \to \infty} \int_{0}^\infty \frac{1}{1+x^n} dx$

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

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

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

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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

$$\int_0^\infty \frac1{1+x^n}\,dx=1+O\left(\frac1n\right)$$

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Taking the limit as $n\to \infty$, yields the coveted limit

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

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

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...