Long run

The sequence of functions $f_n(x)=\frac{1}{1+x^n}$ converges pointwise to $f(x)=1$ for $x<1, f(1)=\frac{1}{2}$ and $f(x)=0$ for $x>1$ . By the dominated convergence theorem , the limit we seek is $\int_{0}^{\infty} f(x)dx=\boxed{1}$ .

This problem is an Inspiration of $\int_{0}^{\infty}\dfrac{1}{1+x^{2018}}\,dx$ Generalization For all $n>1$ we observe that $I(n)= \int_{0}^{\infty}\dfrac{\,dx}{1+x^n}$ is convergent. Now making u-substitution method of the integrand we get $I(n) = -\int_{1}^{0} u\dfrac{(1+x^n)^2}{nx^{n-1}}\,du$ as $u=\,(1+x^n)^{-1}$ and reversing the limits we have then $\begin{aligned} I(n) & = \dfrac{1}{n} \int_{0}^{1}\dfrac{u}{u^2}\cdot\dfrac{u^{-\frac{1}{n}+1}}{(1-u)^{-\frac{1}{n} +1}}\,du \\&= \dfrac{1}{n}\int_{0}^{1}\dfrac{u^{-\frac{1}{n}}}{(1-u)^{-\frac{1}{n}+1}}\,du\end{aligned}$ giving us $\dfrac{B}{n}\left(1-\dfrac{1}{n} , \dfrac{1}{n}\right)=\dfrac{1}{n}\dfrac{\Gamma\left(1-\frac{1}{n}\right)\Gamma\left(\frac{1}{n}\right)}{\Gamma(1)}$ and hence by Euler reflection formula we have $I(n) =\dfrac{1}{n}\dfrac{\pi}{\sin \frac{\pi}{n}}$ And hence $I(n) =\lim_{n\to \infty} \int_{0}^{\infty} \dfrac{1}{1+x^n}\,dx =1 \ \ \forall n>1$