Let $p\geq -1$. How do I show that $\int_0^\infty \frac{x^p}{1+x^p}dx$ diverges?
I thought to break up the integral into $\int_0^1 \frac{x^p}{1+x^p}dx+\int_1^\infty \frac{x^p}{1+x^p}dx$, but I can't find suitable comparisons (a nonnegative expression that doesn't exceed the integrand on each domain of integration such that the expressions are not integrable over their respective domains) to prove that neither term converges.
The elementary way:
For $p \geqslant 0$, we have
$$\frac{x^p}{1+x^p} \geqslant \frac{x^p}{x^p+x^p} = \frac{1}{2}$$
on $[1,\infty)$, and for $p <0$, we have
$$\frac{x^p}{1+x^p} \geqslant \frac{x^p}{1+1} = \frac{x^p}{2}$$
on $[1,\infty)$. Since $\int_1^\infty x^s\,dx$ converges if and only if $s < -1$, the result follows.
For $p < -1$, we also have
$$\frac{x^p}{1+x^p} < x^p,$$
which shows the convergence in that case.
The integral over $[0,1]$ is harmless in all cases, since $0 \leqslant f(x) \leqslant 1$ there, and for appropriately chosen $f(0)$ the integrand is continuous on $[0,1]$.
Hint: $$\frac{x^p}{x^p+1}=\frac{x^p +1 -1 }{x^p+1}=\frac{x^p+1}{x^p+1}- \frac{1}{x^p+1}$$
Hint: using the change of variables $t=1/(1+x^p)$ casts the integral in terms of th beta function. See a related problems.
Added: note that, for large $x$ the integrand behaves as $1$.
