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

Question

$$L=\lim_{n \to \infty} \int_{0}^\infty \frac{1}{1+x^n} dx$$ $$\phi(x)=\lim_{n \to \infty}\frac{1}{1+x^n}$$

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 1

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

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

Answer 2

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

Answer 3

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...

Answer 4

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \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}

Answer 5

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.

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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

$$\bbox[5px,border:2px solid #C0A000]{\int_0^\infty \frac1{1+x^n}\,dx=\frac{\pi/n}{\sin(\pi/n)}}$$

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NOTE:

Another way forward is to use the approach used in the OP of THIS QUESTION.

Answer 6

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} $