1. The problem is to evaluate the sum of integrals: $$\sum_{k=1}^\infty \int_0^\infty x^3 e^{-kx} \, dx$$ 2. First, consider the integral for a fixed $k$: $$I(k) = \int_0^\infty x^3 e^{-kx} \, dx$$ 3. Use the Gamma function formula: $$\int_0^\infty x^m e^{-ax} \, dx = \frac{\Gamma(m+1)}{a^{m+1}}$$ where $m=3$ and $a=k$. 4. Since $\Gamma(4) = 3! = 6$, we have: $$I(k) = \frac{6}{k^4}$$ 5. Substitute back into the sum: $$\sum_{k=1}^\infty I(k) = \sum_{k=1}^\infty \frac{6}{k^4} = 6 \sum_{k=1}^\infty \frac{1}{k^4}$$ 6. The series $\sum_{k=1}^\infty \frac{1}{k^4}$ is the Riemann zeta function $\zeta(4)$, which equals $\frac{\pi^4}{90}$. 7. Therefore: $$\sum_{k=1}^\infty \int_0^\infty x^3 e^{-kx} \, dx = 6 \times \frac{\pi^4}{90} = \frac{\pi^4}{15}$$ **Final answer:** $$\boxed{\frac{\pi^4}{15}}$$