Find the value of p for which the integral converges and evaluate integral for $\int ^\infty_e \frac{1}{x(\ln x)^p} dx$

Question

Find the value of p for which the integral converges and evaluate integral for $\int ^\infty_e \frac{1}{x(\ln x)^p} dx$

MY ATTEMPT: Given Integral: $I= \int ^\infty_e \frac{1}{x(\ln x)^p} dx$

put $\ln x =t\rightarrow \frac{1}{x}dx=dt$ and $x\rightarrow \infty,t\rightarrow \infty, x=e, t\rightarrow 1$

$\therefore I= \int ^\infty_e \frac{1}{x(\ln x)^p} dx=\int^\infty_1 \frac{1}{t^p} dt$

Hence the integral is convergent for p>1.. how to find the value of integral for those values of p

Answer 1

Putting $u= \ln x$ we get $du = \frac{1}{x} \, dx$. We have

$$\int ^\infty_e \frac{1}{x(\ln x)^p} dx = \int_1^\infty\frac{1}{u^p}\, du = \lim_{a \to \infty} \left[\frac{1}{1-p}u^{1-p} \right]_1^a, \, \text{for } p \neq 1.$$

Now consider the cases $1 - p > 0$ and $1 - p < 0$. Lastly, consider the case where $p=1$ which will involve a new antiderivative.