Find $\lim_{n\to\infty}\int\limits_{0}^\infty{dx\over x^n +1}$

Question

Find $$\lim_{n\to\infty}\int\limits_{0}^\infty{dx\over x^n +1}$$ There is a remark to this problem saying to prove $$\int\limits_{0}^\infty{dx\over x^n +1}=1-\int\limits_0^1{x^n \over x^n +1} + \int\limits_1^\infty{dx\over x^n +1}$$ But I don't have any clue how to do it and completely stuck here

Answer 1

$\frac{1}{x^n+1} \le \frac{1}{x^2+1}$ for all $n \ge 2$ and $x \ge 1$. Since $\frac{1}{x^2+1}\in L^1((1,\infty))$, dominated convergence implies $\int_1^\infty \frac{1}{x^n+1}dx \to \int_1^\infty 0dx = 0$. And $\frac{1}{x^n+1} \le 1$ for all $0 < x < 1$, so since $1 \in L^1((0,1))$, dominated convergence implies $\int_0^1 \frac{1}{x^n+1}dx \to \int_0^1 1dx = 1$. The answer is thus $1+0 = 1$.

Answer 2

Starting from the remark in the OP, we have

$${x^n\over x^n+1}\le x^n\qquad\text{and}\qquad{1\over x^n+1}\le{1\over x^n}$$

so

$$\int_0^1{x^n\over x^n+1}dx\le\int_0^1x^n\,dx={1\over n+1}\to0$$

and

$$\int_1^\infty{1\over x^n+1}dx\le\int_1^\infty{1\over x^n}dx={1\over n-1}\to0$$