Evaluation of $\int_{1}^{\infty}\frac{1}{x(\ln{x})^s} \, dx$ I am revising my maths and a bit rusty on what approach I need to use to solve this...
I'm pretty sure it should converge and I can't figure out where I have gone wrong!
I have tried 2 different approaches which both lead to the same answer.
Attempt 1: Substitution (First)
$$I = \int_{1}^{\infty}\frac{1}{x(\ln{x})^s}dx = \lim_{t\to\infty}\int^{t}_{1}\frac{(\ln{x})^{-s}}{x}dx$$
let $u = \ln{x}$ and let $du = \frac{1}{x}dx$,
$$I \to \lim_{t\to\infty}\int^{t}_{1}u^{-s}du = \lim_{t\to\infty}\left[\frac{u^{1-s}}{1-s}\right]^{t}_{1}$$
$$ = \lim_{t\to\infty}\left[\frac{(\ln{x})^{1-s}}{1-s}\right]^{t}_{1}$$
$$ = \lim_{t \to \infty}\left[\frac{(\ln{t})^{1-s}}{1-s} - \frac{(\ln{1})^{1-s}}{1-s}\right]$$
$$ = \lim_{t \to \infty}\left[\frac{(\ln{t})^{1-s}}{1-s} - 0\right]$$

*

*$s<1$ : $I \to \infty$

*$s=1$ : $I$ is undefined

*$s>1$ : $I \to \infty$
Attempt 2: Integration by Parts (First)
$$\int_{1}^{\infty}\frac{1}{x(\ln{x})^s}dx = \lim_{t\to\infty}\int^{t}_{1}\frac{1}{x(\ln{x})^s}dx$$
$$ I = \lim_{t\to\infty}\int^{t}_{1}{x^{-1}(\ln{x})^{-s}}dx$$
Integration by parts:
$$\int{u}{dv} = uv - \int{v}{du}$$
$$u = (\ln{x})^{-s} \qquad dv = x^{-1}dx$$
$$v = \int{x^{-1}dx = }\ln{x} \qquad du = \frac{-s{(\ln{x})^{-s-1}}}{x}dx$$
giving:
$$\int{udv} = (\ln{x})^{-s}\ln{x} - \int{\ln{x}\frac{-s{(\ln{x})^{-s-1}}}{x}}$$
again, integration by parts:
$$\int{\ln{x}\frac{-s{(\ln{x})^{-s-1}}}{x}}dx = \int{udv} = uv - \int{vdu}$$
$$dv = \frac{-s{(\ln{x})^{-s-1}}}{x}dx \qquad u = \ln{x}$$
$$v = \int{\frac{-s{(\ln{x})^{-s-1}}}{x}}dx = \ln{x}^{-s} \qquad du = \frac{1}{x}dx$$
giving:
$$\int{udv} = \ln{x}\ln{x}^{-s} - \int{\ln{x}^{-s}\frac{1}{x}dx}$$
$$ = \ln{x}^{1-s} - \int{\frac{\ln{x}^{-s}}{x}dx} $$
by substitution, let $u = \ln{x}$ and $du = \frac{1}{x}$,
$$ \int{\frac{\ln{x}^{-s}}{x}dx} \to \int{u^{-s}du} = \frac{u^{-s+1}}{1-s}$$
$$ = (\ln{x})^{1-s}(1-s)^{-1}$$
therefore, I get:
$$I = \lim_{t \to \infty}\left[(\ln{x})^{1-s} - (\ln{x})^{1-s} + (\ln{x})^{1-s}(1-s)^{-1}\right]^{t}_{1}$$
$$ = \lim_{t \to \infty}\left[\frac{(\ln{x})^{1-s}}{1-s}\right]^{t}_{1}$$
... follow through as above
Attempt 3: Fixing Integral Limits & Final Limit
$$I = \int_{x=1}^{x=\infty}\frac{1}{x(\ln{x})^s}dx = \lim_{t\to\infty}\int^{x=t}_{x=1}\frac{(\ln{x})^{-s}}{x}dx$$
let $u = \ln{x}$ and let $du = \frac{1}{x}dx$,
$$ I \to \lim_{t\to\infty}\int^{u=\ln{t}}_{u=\ln{1}}u^{-s}du = \lim_{t\to\infty}\left[\frac{u^{1-s}}{1-s}\right]^{u=\ln{t}}_{u=0} $$
convert back from $u \to x : x = e^{u}$...
$$ = \lim_{t\to\infty}\left[\frac{(\ln{x})^{1-s}}{1-s}\right]^{x=t}_{x=1} = \lim_{t \to \infty}\left[\frac{(\ln{t})^{1-s}}{1-s} - \frac{(\ln{1})^{1-s}}{1-s}\right]$$
$$ = \lim_{t \to \infty}\left[\frac{(\ln{t})^{1-s}}{1-s}\right]$$

*

*$s<1$ : $I \to \infty$

*$s=1$ : $I = \frac{1}{0}$

*$s>1$ : $(\ln{t})^{1-s} \to 0 \therefore I \to 0$
I have fixed my limit for $s>1$ but can't see the problem with $s<1$
 A: Your first attempt is the easiest approach. When using substitution, you can also substitute the limits (rather than change the variable back to $x$ at the end). So the limit $x=1$ corresponds to the limit $u=\ln(1)=0$ and the limit $x=\infty$ corresponds to $u=\ln(\infty)=\infty$. The problem you have above is that:
$$
I=\lim_{t \to\infty}\int_{x=1}^{x=t}u^{-s}du=\lim_{t\to\infty}\int_{u=0}^{u=\infty}u^{-s}du\ne\lim_{t\to\infty}\left[ \frac{u^{1-s}}{1-s}\right]_{0}^{t} \text{ when } s=1
$$
So you need to separate the evaluation of the integral into cases:
$$
I=\lim_{t\to\infty}\int_{0}^{\infty}u^{-s}du=\left\{ \begin{array} \\\lim_{t\to\infty} \left[\frac{u^{1-s}}{1-s}\right]_{0}^{t}\text{ , }s\ne1 \\ \lim_{t\to\infty} \left[\ln(u)\right]_{0}^{t}\text{ , }s=1 \end{array}  \right.
$$
Now you can see that for the $s=1$ case you also have a problem at the lower limit as $\ln(0)$ is undefined (as mentioned in the comments). There will also be a problem at the lower limit when $s>2$. In all cases, however, the integral will diverge. 
