1. **Problem Statement:** Determine whether the improper integral $$\int_{e^{3}}^{\infty} \frac{1}{x(\ln x)^4} \, dx$$ converges or diverges. If it converges, find its exact value or approximate to six decimal places. 2. **Formula and Approach:** To evaluate this integral, use the substitution method. Let $$t = \ln x$$, then $$dt = \frac{1}{x} dx$$. This transforms the integral into a simpler form. 3. **Change of Limits:** When $$x = e^{3}$$, $$t = 3$$. When $$x \to \infty$$, $$t \to \infty$$. 4. **Rewrite the Integral:** Substitute into the integral: $$\int_{e^{3}}^{\infty} \frac{1}{x(\ln x)^4} \, dx = \int_{3}^{\infty} \frac{1}{t^4} \, dt$$ 5. **Evaluate the Integral:** $$\int_{3}^{\infty} t^{-4} \, dt = \lim_{b \to \infty} \int_{3}^{b} t^{-4} \, dt$$ 6. **Integral of $$t^{-4}$$:** $$\int t^{-4} \, dt = \int t^{-4} \, dt = \frac{t^{-3}}{-3} + C = -\frac{1}{3t^{3}} + C$$ 7. **Apply Limits:** $$\lim_{b \to \infty} \left[-\frac{1}{3b^{3}} + \frac{1}{3 \cdot 3^{3}}\right] = 0 + \frac{1}{3 \cdot 27} = \frac{1}{81}$$ 8. **Conclusion:** The integral converges and its exact value is $$\frac{1}{81}$$. Hence, the improper integral $$\int_{e^{3}}^{\infty} \frac{1}{x(\ln x)^4} \, dx$$ converges to $$\frac{1}{81}$$.