1. **State the problem:** Calculate the integral $$\int_1^{\infty} x e^{-x} \, dx$$. 2. **Recall the formula and method:** We will use integration by parts, where $$\int u \, dv = uv - \int v \, du$$. 3. **Choose parts:** Let $$u = x$$ and $$dv = e^{-x} dx$$. 4. **Compute derivatives and integrals:** Then $$du = dx$$ and $$v = \int e^{-x} dx = -e^{-x}$$. 5. **Apply integration by parts:** $$\int x e^{-x} dx = -x e^{-x} - \int -e^{-x} dx = -x e^{-x} + \int e^{-x} dx$$ 6. **Integrate remaining integral:** $$\int e^{-x} dx = -e^{-x}$$ 7. **Combine results:** $$\int x e^{-x} dx = -x e^{-x} - e^{-x} + C = -(x+1)e^{-x} + C$$ 8. **Evaluate definite integral from 1 to infinity:** $$\int_1^{\infty} x e^{-x} dx = \lim_{t \to \infty} [-(x+1)e^{-x}]_1^t = \lim_{t \to \infty} [-(t+1)e^{-t} + 2 e^{-1}]$$ 9. **Evaluate limit:** As $$t \to \infty$$, $$e^{-t}$$ approaches 0 faster than $$t+1$$ grows, so $$\lim_{t \to \infty} (t+1)e^{-t} = 0$$. 10. **Final value:** $$\int_1^{\infty} x e^{-x} dx = 0 + 2 e^{-1} = \frac{2}{e}$$. **Answer:** $$\boxed{\frac{2}{e}}$$