Comparison Test for Improper Integrals — Question 8

Question 8 – Solution

Step 1: Consider $p>0$. Compare with $g(x)=1/x^p$: $$\begin{align*} L&=\lim_{x\to\infty} \frac{1/(x^p+1)}{1/x^p}\\ &=\lim_{x\to\infty}\frac{x^p}{x^p+1}\\ &=\lim_{x\to\infty}\frac1{1+1/x^p}=1. \end{align*}$$ Thus the original integral has the same behavior as the $p$-integral $\int_1^\infty x^{-p}\,dx$, which converges exactly when $p>1$.

Step 2: Consider $p=0$. The integrand is $1/2$, so the integral diverges.

Step 3: Consider $p<0$. Then $x^p\to0$, so $$\frac1{x^p+1}\to1\ne0.$$ For $x\ge1$ and $p<0$, $x^p\le1$, so the integrand is at least $1/2$. Comparison with $\int_1^\infty (1/2)\,dx$ proves divergence. $$\boxed{p>1}$$