Convergence of $\int_3^{\infty} \frac{1}{(\ln(x))^2(x-\ln(x))}$ Does this integral converge?
$$\int_3^{\infty} \frac{1}{(\ln(x))^2(x-\ln(x))}$$
I've been trying to solve this for the past 2 hours...literally. I know the answer is fairly simple, but I just can't think of it
 A: We have $$\dfrac{x}e > \ln(x) \,\,\, \forall x > 0$$
Hence,
$$x - \ln(x) > x - \dfrac{x}e = x \left(1 - \dfrac1e\right)$$
Hence, we get that
$$I = \int_3^{\infty} \dfrac{dx}{\ln^2(x)(x-\ln(x))} < \left(\dfrac{e}{e-1} \right) \times \int_3^{\infty} \dfrac{dx}{x \ln^2(x)} = \dfrac{e}{(e-1)\ln(3)}$$
EDIT
A way to approximate the integral is as follows.
$$\dfrac1{x-\ln(x)} = \dfrac1x \sum_{k=0}^{\infty} \left(\dfrac{\ln(x)}{x} \right)^k$$
Hence,
$$I = \int_3^{\infty} \dfrac{dx}{\ln^2(x)(x-\ln(x))} = \sum_{k=0}^{\infty} \int_3^{\infty} \dfrac{\ln^{k-2}(x)}{x^{k+1}} dx$$
Let us evaluate each term now.
\begin{align}
f_k & = \int_3^{\infty} \dfrac{\ln^{k-2}(x)}{x^{k+1}} dx\\
& = \int_{\ln(3)}^{\infty} \dfrac{t^{k-2}}{e^{kt}} dt\\
& = \int_{\ln(3)/k}^{\infty} \dfrac{y^{k-2}}{k^{k-2}e^y} \dfrac{dy}k\\
& = \dfrac1{k^{k-1}} \int_{\ln(3)/k}^{\infty} y^{k-2} e^{-y} dy\\
& = \dfrac{\Gamma(k-1,\ln(3)/k)}{k^{k-1}}
\end{align}
where $\Gamma(m,z)$ is the incomplete $\Gamma$ function and there are many ways to compute incomplete $\Gamma$ function to arbitrary accuracy. For example, as shown here. Hence, we get that
$$I = \sum_{k=0}^{\infty} \dfrac{\Gamma(k-1,\ln(3)/k)}{k^{k-1}}$$which is an exponentially converging series and truncating this will provide us arbitrarily accurate answer.
A: Hint: Try comparing it to $$\int_{3}^\infty \frac{1}{x(\ln(x))^2} dx.$$
Secondary Hint: $$\frac{1}{x-\ln (x)}\leq \frac{2}{x}.$$
A: Yes, this converges.  The integrand is bounded above by $2/(x (\log x)^2)$.  Then, substitute $y=\log x$ to get the convergent integral $\int_{\log 3}^\infty 2 y^{-2} dy$.
