Comparison Test for Improper Integrals — Question 5

Question 5 – Solution

Step 1: Bound the logarithm. The logarithm grows more slowly than every positive power of $x$. In particular, $$\ln x\le\sqrt{x}$$ for all sufficiently large $x$.

Step 2: Divide by $x^2>0$. $$0\le\frac{\ln x}{x^2}\le\frac{\sqrt{x}}{x^2} =\frac1{x^{3/2}}.$$ Step 3: Test the upper comparison. $$\int_2^\infty x^{-3/2}\,dx$$ converges because it is a $p$-integral with $p=3/2>1$.

Step 4: Apply direct comparison. Therefore, the original integral converges. $$\boxed{\text{converges}}$$