Calculate $H= \int_0^\infty \frac{x^{n+1}}{1+x^n}dx, \, n>2$

Question

Calculate the following integral: $H=\displaystyle \int \limits_0^\infty \dfrac{x^{n+1}}{1+x^n}dx, \, n>2$

I wanna use Beta function $B(x,y)=\displaystyle \int \limits_0^\infty \frac{t^{x-1}}{(1+t)^{x+y}}dt, \,x,y>0$. As follows,

Set $x^n=t \Rightarrow x=t^{\tfrac{1}{n}}\Rightarrow dx=\dfrac{t^{\tfrac{1}{n}-1}}{n}dt$

\begin{align*} H&=\displaystyle \int \limits_0^\infty \frac{t^{\tfrac{1}{n}+1}}{1+t}.\frac{t^{\tfrac{1}{n}-1}}{n}dt\\ &=\frac{1}{n}\displaystyle \int \limits_0^\infty \frac{t^\tfrac{2}{n}}{1+t}dt \end{align*} So, $x=\frac{2}{n}+1 >0$ and $y=-\frac{2}{n}<0.$ So, cann't use Beta function.

Answer 1

We have that $\text{deg}\left(x^{n+1}\right)>\text{deg}\left(1+x^{n}\right)$, so $\lim_{x\to\infty}\frac{x^{n+1}}{1+x^n}=\infty$. This implies that there is a positive number $N$ such that $\frac{x^{n+1}}{1+x^n}>1$ for all $x>N$.

It's clear that the integral

$$\int_{N}^{\infty}1\text{ }dx$$

diverges, so $\int_{N}^{\infty}\frac{x^{n+1}}{1+x^n}dx$ diverges by comparison. It follows that $\int_{0}^{\infty}\frac{x^{n+1}}{1+x^n}dx$ is also divergent.

Answer 2

The given integral diverges. See $$H=\int_{0}^{\infty} \frac{x^{n+1} dx}{1+x^n} \sim \int_{0}^{\infty} x dx \rightarrow \infty$$