When is $\int_e^4 \frac{dt}{\ln^\alpha[\ln(t)]}$ finite?

Question

For which $\alpha \geq 0$ does this integral converge ?

$\int_e^4 \frac{dt}{\ln^\alpha[\ln(t)]}$

Answer 1

The integrand is continuous and positive in $(e,4]$, so check what happens near $e^+$: $$\frac{1}{\ln^\alpha[\ln(t)]}=\frac{1}{\ln^\alpha[1+\ln(t/e)]}\sim \frac{1}{[\ln(1+(t/e-1))]^\alpha}\sim \frac{1}{[t/e-1]^\alpha}=\frac{e^{\alpha}}{[t-e]^\alpha} $$ which means that integral is convergent iff it is convergent the integral of $\frac{e^{\alpha}}{[t-e]^\alpha}$ that is for $\alpha<1$. Notice that we used two times the fact that $\ln(1+t)\sim t$ for $t\to 0$.

Answer 2

Let $t=e^x$. Then, we see that

$$\int_{e}^4 \frac{1}{\left(\log(\log(t))\right)^\alpha}\,dt=\int_1^{\log(4)} \frac{e^x}{\log^\alpha(x)}\,dx$$

Now, in THIS ANSWER, I showed using only the limit definition of the exponential function and Bernoulli's Inequality that the logarithm function satisfies the inequalities

$$\frac{x-1}{x}\le \log(x)\le x-1$$

for $x>0$. Then, we see that

$$\int_1^{\log(4)} \frac{x^\alpha e^x}{(x-1)^\alpha}\,dx\ge \int_1^{\log(4)} \frac{e^x}{\log^\alpha(x)}\,dx\ge \int_1^{\log(4)} \frac{e^x}{(x-1)^\alpha}\,dx$$

Hence, the integral of interest converges for $\alpha <1$ and diverges for $\alpha \ge 1$ by comparison.