1. **Problem statement:** Evaluate the improper integral $$\int_0^\infty x \sin x \, dx$$. 2. **Formula and approach:** We use integration by parts for integrals of the form $$\int x \sin x \, dx$$ and then analyze the limit as the upper bound approaches infinity. 3. **Integration by parts formula:** $$\int u \, dv = uv - \int v \, du$$ Choose: $$u = x \implies du = dx$$ $$dv = \sin x \, dx \implies v = -\cos x$$ 4. **Apply integration by parts:** $$\int x \sin x \, dx = -x \cos x + \int \cos x \, dx = -x \cos x + \sin x + C$$ 5. **Evaluate the improper integral:** $$\int_0^\infty x \sin x \, dx = \lim_{t \to \infty} \left[-x \cos x + \sin x \right]_0^t = \lim_{t \to \infty} (-t \cos t + \sin t) - (0 + 0)$$ 6. **Analyze the limit:** The term $$-t \cos t$$ oscillates between $$-t$$ and $$t$$ as $$\cos t$$ oscillates between $$-1$$ and $$1$$, and $$t$$ grows without bound. Therefore, the limit does not exist in the usual sense because the expression oscillates with increasing amplitude. 7. **Conclusion:** The improper integral $$\int_0^\infty x \sin x \, dx$$ is **divergent** and does not converge to a finite value. **Final answer:** The integral diverges and does not have a finite value.