1. **State the problem:** We want to evaluate the limit as $R \to \infty$ of the sum over $n \in \mathbb{N}_0$ of the contour integrals over $\partial C_R$ of the function $$\frac{(-1)^n z^n}{(z^4 + 0.1\sigma) \int_0^\infty t^n e^{-t} dt} \, dz,$$ where $C_R = \{ z \in \mathbb{C} \mid \Im(z) > 0 \text{ and } 0 < |z| < R \}$. 2. **Recall important formulas and facts:** - The integral $\int_0^\infty t^n e^{-t} dt$ is the Gamma function $\Gamma(n+1) = n!$ for $n \in \mathbb{N}_0$. - The contour $\partial C_R$ is the boundary of the upper half disk of radius $R$. - For large $R$, the contour integral over $\partial C_R$ can be split into the integral over the semicircle of radius $R$ and the real axis segment. 3. **Rewrite the integral using the Gamma function:** $$\int_0^\infty t^n e^{-t} dt = n!$$ So the integrand becomes $$\frac{(-1)^n z^n}{(z^4 + 0.1\sigma) n!}.$$ 4. **Sum over $n$ inside the integral:** Consider the sum $$\sum_{n=0}^\infty \frac{(-1)^n z^n}{n!} = e^{-z}.$$ Therefore, the sum and integral can be interchanged (assuming uniform convergence) to write $$\sum_{n=0}^\infty \int_{\partial C_R} \frac{(-1)^n z^n}{(z^4 + 0.1\sigma) n!} dz = \int_{\partial C_R} \frac{e^{-z}}{z^4 + 0.1\sigma} dz.$$ 5. **Evaluate the limit as $R \to \infty$ of the contour integral:** The contour $\partial C_R$ is the semicircle in the upper half-plane plus the real axis segment. By Jordan's lemma, the integral over the large semicircle tends to zero if the integrand decays sufficiently fast. Since $e^{-z}$ decays exponentially in the upper half-plane and $|z^4 + 0.1\sigma|$ grows like $|z|^4$, the integral over the semicircle vanishes as $R \to \infty$. 6. **Apply the residue theorem:** The integral over $\partial C_R$ as $R \to \infty$ equals $2\pi i$ times the sum of residues of $$\frac{e^{-z}}{z^4 + 0.1\sigma}$$ at poles inside the upper half-plane. 7. **Find the poles:** The poles are the roots of $$z^4 + 0.1\sigma = 0 \implies z^4 = -0.1\sigma.$$ Let $r = (0.1|\sigma|)^{1/4}$ and $\theta = \frac{\pi + 2k\pi}{4}$ for $k=0,1,2,3$ be the arguments of the roots. The two roots in the upper half-plane correspond to $k=0$ and $k=1$: $$z_1 = r e^{i\pi/4}, \quad z_2 = r e^{i3\pi/4}.$$ 8. **Calculate residues:** The residue at a simple pole $z_0$ of $f(z) = \frac{e^{-z}}{z^4 + 0.1\sigma}$ is $$\text{Res}(f,z_0) = \lim_{z \to z_0} (z - z_0) f(z) = \frac{e^{-z_0}}{4 z_0^3}$$ because $\frac{d}{dz}(z^4 + 0.1\sigma) = 4 z^3$. 9. **Sum residues and multiply by $2\pi i$:** $$\int_{\partial C_R} \frac{e^{-z}}{z^4 + 0.1\sigma} dz = 2\pi i \left( \frac{e^{-z_1}}{4 z_1^3} + \frac{e^{-z_2}}{4 z_2^3} \right).$$ 10. **Final answer:** $$\boxed{\lim_{R \to \infty} \sum_{n=0}^\infty \int_{\partial C_R} \frac{(-1)^n z^n}{(z^4 + 0.1\sigma) n!} dz = \frac{\pi i}{2} \left( \frac{e^{-z_1}}{z_1^3} + \frac{e^{-z_2}}{z_2^3} \right)}.$$