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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.

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  • $\begingroup$ Distinguish between $p < 0$ and $p \geqslant 0$. (And one of the two converges.) $\endgroup$ – Daniel Fischer Dec 28 '13 at 14:24
  • $\begingroup$ $$\int_0^1 \frac{x^p}{1+x^p}\,dx$$ is entirely harmless. The integrand is continuous on $[0,1]$ (if $f(0)$ is appropriately chosen, $0,\frac12,1$, depending on whether $p > 0,\; p = 0,\; p < 0$). The integral over $[1,\infty)$ is what determines the overall behaviour. $\endgroup$ – Daniel Fischer Dec 28 '13 at 14:47
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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 $[0,1]$.

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Hint: $$\frac{x^p}{x^p+1}=\frac{x^p +1 -1 }{x^p+1}=\frac{x^p+1}{x^p+1}- \frac{1}{x^p+1}$$

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  • $\begingroup$ Yes I got that too. But I can't find suitable comparisons (a nonnegative expression that doesn't exceed $\frac 1{x^p+1}$ or $\frac 1{x^{-p}+1}$ on each domain of integration such that the expressions are not integrable over their respective domains) $\endgroup$ – Ryan Dec 28 '13 at 14:40
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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 $1$.

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  • $\begingroup$ Thanks, but I wish not to assume acquaintance with the beta function... $\endgroup$ – Ryan Dec 28 '13 at 14:46

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