1. **State the problem:** Evaluate the improper integral $$I = \int_{e}^{\infty} \frac{dx}{x(\ln x)^3}$$. 2. **Recall the formula and substitution:** For integrals involving $\ln x$, use substitution $t = \ln x$ which implies $dt = \frac{1}{x} dx$ or $dx = x dt$. 3. **Apply substitution:** $$I = \int_{x=e}^{\infty} \frac{dx}{x(\ln x)^3} = \int_{t=1}^{\infty} \frac{x dt}{x t^3} = \int_{1}^{\infty} \frac{dt}{t^3}$$ 4. **Simplify the integral:** $$\int_{1}^{\infty} t^{-3} dt$$ 5. **Integrate:** Use the power rule for integrals: $$\int t^n dt = \frac{t^{n+1}}{n+1} + C, \quad n \neq -1$$ Here, $n = -3$, so $$\int t^{-3} dt = \frac{t^{-2}}{-2} + C = -\frac{1}{2 t^2} + C$$ 6. **Evaluate the definite integral:** $$I = \lim_{b \to \infty} \left[-\frac{1}{2 t^2} \right]_{1}^{b} = \lim_{b \to \infty} \left(-\frac{1}{2 b^2} + \frac{1}{2 \cdot 1^2} \right) = 0 + \frac{1}{2} = \frac{1}{2}$$ 7. **Final answer:** $$\boxed{\frac{1}{2}}$$ This means the integral converges and its value is one half.