Study the following integral: $\int_0^\infty \frac{\mathrm{d} x}{x \cdot \ln x \cdot \ln^{(2)} x \cdot \ln^{(3)} x ... (\ln^{(k)} x)^s }$ How do I calculate for which values of $s$ the following integral converges?

$$\int\limits_{0}^{\infty} \frac{\mathrm{d} x}{x \cdot \ln x \cdot \ln^{(2)}  x \cdot \ln^{(3)}  x \cdots (\ln^{(k)} x)^s }$$

$$\ln^{(k)}=\underbrace{\ln(\ln(\ln \cdots (x)))}_{\text{k times}}$$
 A: Hint:
To simply the notation define $\ln_k(x) \equiv \ln\ln\cdots\ln(x)$ where there are $k$ logs on the right hand side. Note that $\ln_k(x)$ is only well defined for $x > 0,1,e,e^e,e^{e^e},\ldots$ for $k=1,2,3,4\ldots$. With this notation your integral can be written
$$\int^\infty\frac{dx}{x\ln_1(x)\ln_2 (x)\cdots \ln_{k-1}(x)(\ln_k(x))^s}$$
From the definition above it follows that $\ln_{k}(x) = \ln(\ln_{k-1}(x))$ so
$$\frac{d\ln_{k}(x)}{dx} = \frac{1}{\ln_{k-1}(x)}\frac{d\ln_{k-1}(x)}{dx}$$
Applying this formula $k$ times (using $\ln_1(x) = \ln(x)$) to get
$$\frac{d\ln_k(x)}{dx} = \frac{1}{x}\frac{1}{\ln_1 (x)}\frac{1}{\ln_2(x)} \cdots \frac{1}{\ln_{k-1}(x)}$$
Now apply the substitution $z=\ln_k(x)$ in your integral , using the result above, to get a much simpler integral for which you can apply your favoritte test to determine the convergence properties.
As mentioned in the comments you should take the lower limit of integration to be larger than $e_k = e^{e^{e^{\ldots e}}}$ in order for all the terms in the integrand to be well defined in the whole region of integration. 
