calculus - 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.

Answer

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