1. **State the problem:** Evaluate the improper integral $$\int_0^{+\infty} 2 \ln(x) \, dx$$ and determine if it converges or diverges. 2. **Recall the integral and convergence rules:** The integral involves the natural logarithm function $\ln(x)$ multiplied by 2. Since the limits are from 0 to infinity, we must check the behavior near 0 and as $x \to +\infty$. 3. **Rewrite the integral:** $$\int_0^{+\infty} 2 \ln(x) \, dx = 2 \int_0^{+\infty} \ln(x) \, dx$$ 4. **Check convergence near 0:** As $x \to 0^+$, $\ln(x) \to -\infty$, so the integral near 0 is improper. Consider the limit: $$\lim_{a \to 0^+} \int_a^1 \ln(x) \, dx$$ 5. **Evaluate $\int \ln(x) \, dx$:** Use integration by parts: Let $u = \ln(x)$, $dv = dx$, then $du = \frac{1}{x} dx$, $v = x$. $$\int \ln(x) \, dx = x \ln(x) - \int x \cdot \frac{1}{x} dx = x \ln(x) - \int 1 \, dx = x \ln(x) - x + C$$ 6. **Evaluate the definite integral from $a$ to 1:** $$\int_a^1 \ln(x) \, dx = [x \ln(x) - x]_a^1 = (1 \cdot \ln(1) - 1) - (a \ln(a) - a) = -1 - (a \ln(a) - a)$$ 7. **Evaluate the limit as $a \to 0^+$:** Since $\lim_{a \to 0^+} a \ln(a) = 0$ and $\lim_{a \to 0^+} a = 0$, $$\lim_{a \to 0^+} (a \ln(a) - a) = 0$$ Therefore, $$\lim_{a \to 0^+} \int_a^1 \ln(x) \, dx = -1$$ 8. **Check convergence near infinity:** Consider the limit: $$\lim_{b \to +\infty} \int_1^b \ln(x) \, dx = \lim_{b \to +\infty} [x \ln(x) - x]_1^b = \lim_{b \to +\infty} (b \ln(b) - b) - (1 \cdot 0 - 1) = \lim_{b \to +\infty} (b \ln(b) - b) + 1$$ Since $b \ln(b)$ grows faster than $b$, this limit diverges to infinity. 9. **Conclusion:** The integral diverges because it does not converge at the upper limit. **Final answer:** The integral $$\int_0^{+\infty} 2 \ln(x) \, dx$$ diverges. Hence, the correct choice is **E. Diverges**.