Comparison test on $ \int_2 ^\infty \frac{\ln(t)}{t^{3/2}} \ dt$

Question

How would one use the comparison test to deduce whether the following function converges or diverges? I’ve attempted to use $ \ln(t) < t $ for $ t > 1$ but fail to conclude whether the desired function converges. I’ve also attempted to use $ \ln (t) > 1 $ for $ t > e $ but also encounter a similar issue once again.

Answer 1

The standard trick is use limit comparison test with $\frac{1}{t^{\alpha}} $ for $1<\alpha<\frac32$, as for example by the convergent

$$\int_2 ^\infty \frac{1}{t^{5/4}} \ dt$$

indeed as $x\to \infty$

$$\frac{\frac{\ln t}{t^{3/2}}}{\frac{1}{t^{5/4}} }=\frac{\ln t}{t^{1/4}}\to 0 $$

and thus the given integral converges.

Answer 2

If I've understood you correctly, you want to know whether $\int_2^\infty t^{-3/2}\ln t \,dt=\int_{\ln 2}^\infty ue^{-u/2}\,du$ converges. Well, it does because $\int_{\ln 2}^\infty ue^{-u/2}\,du\in \left[0,\,\int_0^\infty ue^{-u/2}\,du\right]=\left[0,\,\frac{1!}{(1/2)^2}\right]=[0,\,4]$.