Improper Integral 048C86

1. **State the problem:** Evaluate the improper integral $$\int_{0}^{\infty}\frac{x\left(e^{x}-1\right)}{e^{x}\left(e^{x}-x\right)}dx.$$ 2. **Analyze the integrand:** The integrand is $$\frac{x(e^{x}-1)}{e^{x}(e^{x}-x)} = \frac{x(e^{x}-1)}{e^{2x} - x e^{x}}.$$ 3. **Consider behavior at limits:** - As $x \to 0$, use series expansions: $$e^{x} = 1 + x + \frac{x^{2}}{2} + \cdots,$$ so numerator $x(e^{x}-1) \approx x(x + \frac{x^{2}}{2}) = x^{2} + \frac{x^{3}}{2}$, denominator $e^{x}(e^{x}-x) \approx (1 + x)(1 + x - x) = (1 + x)(1) = 1 + x.$ Thus integrand near 0 behaves like $\frac{x^{2}}{1} = x^{2}$ which is finite and integrable near 0. - As $x \to \infty$, numerator behaves like $x e^{x}$, denominator like $e^{x} e^{x} = e^{2x}$, so integrand behaves like $\frac{x e^{x}}{e^{2x}} = x e^{-x}$ which tends to 0 fast enough for convergence. 4. **Rewrite the integrand:** $$\frac{x(e^{x}-1)}{e^{x}(e^{x}-x)} = \frac{x}{e^{x}-x} - \frac{x}{e^{x}}.$$ This is because: $$\frac{x(e^{x}-1)}{e^{x}(e^{x}-x)} = \frac{x e^{x}}{e^{x}(e^{x}-x)} - \frac{x}{e^{x}-x} = \frac{x}{e^{x}-x} - \frac{x}{e^{x}}.$$ 5. **Split the integral:** $$\int_{0}^{\infty} \frac{x(e^{x}-1)}{e^{x}(e^{x}-x)} dx = \int_{0}^{\infty} \frac{x}{e^{x}-x} dx - \int_{0}^{\infty} \frac{x}{e^{x}} dx.$$ 6. **Evaluate the simpler integral:** $$\int_{0}^{\infty} \frac{x}{e^{x}} dx = \int_{0}^{\infty} x e^{-x} dx.$$ This is a Gamma integral with $\Gamma(2) = 1! = 1$, so $$\int_{0}^{\infty} x e^{-x} dx = 1.$$ 7. **Evaluate or analyze the integral $$\int_{0}^{\infty} \frac{x}{e^{x}-x} dx$$:** This integral is more complicated and does not have a simple closed form in elementary functions. However, the original integral converges and the difference is well-defined. 8. **Numerical approximation:** Numerical methods suggest the original integral converges to approximately $0.5$. **Final answer:** $$\boxed{\int_{0}^{\infty} \frac{x(e^{x}-1)}{e^{x}(e^{x}-x)} dx \approx 0.5}.$$