Evaluate the Modulus of a Complex Contour Integral Limit with Power Series and Gamma Function

Question

Evaluate the modulus of the limit as $R \to \infty$ of the sum over $n \in \mathbb{N}_0$ of the contour integral around $\partial C_R$ of $\frac{(-1)^n z^n}{(z^4 + 0.1\sigma) \int_0^\infty t^n e^{-t} dt} dz$, where $C_R = \{ z \in \mathbb{C} \mid \Im(z) > 0 \ \land \ 0 < |z| < R \}$ is the upper half-disk of radius $R$ (excluding the origin and real axis boundary).

Accepted Answer

$0$

Answer

Step 1: Simplify the denominator integral. Recall that the Gamma function is defined as $\Gamma(n+1) = \int_0^\infty t^n e^{-t} dt$, and for non-negative integers $n \in \mathbb{N}_0$, $\Gamma(n+1) = n!$. So the integrand becomes $\frac{(-1)^n z^n}{(z^4 + 0.1\sigma) n!}$. Step 2: Recognize the power series sum. The sum $\sum_{n=0}^\infty \frac{(-1)^n z^n}{n!}$ is the Taylor series expansion of $e^{-z}$, which converges uniformly for all finite $z \in \mathbb{C}$. So the integrand simplifies to $\frac{e^{-z}}{z^4 + 0.1\sigma}$ inside the contour. Step 3: Analyze the contour integral over $\partial C_R$. The contour $\partial C_R$ consists of three parts: the upper semicircle $\Gamma_R$ of radius $R$ in the upper half-plane, the line segment from $-R$ to $\epsilon$ on the real axis, and the small semicircle $\gamma_\epsilon$ around the origin (radius $\epsilon \to 0$). Step 4: Estimate the integral over the large semicircle $\Gamma_R$. Use the ML-inequality: $\left| \int_{\Gamma_R} \frac{e^{-z}}{z^4 + 0.1\sigma} dz \right| \leq \frac{\max_{z \in \Gamma_R} |e^{-z}| \cdot \pi R}{\min_{z \in \Gamma_R} |z^4 + 0.1\sigma|}$. For $z = Re^{i\theta}$, $0 < \theta < \pi$, $|e^{-z}| = e^{-R\cos\theta}$, and $|z^4 + 0.1\sigma| \geq R^4 - 0.1|\sigma|$. As $R \to \infty$, $e^{-R\cos\theta}$ decays exponentially for $0 \leq \theta < \pi/2$, and is bounded for $\pi/2 \leq \theta \leq \pi$. The integral over $\Gamma_R$ tends to $0$ as $R \to \infty$ by the Jordan lemma for upper half-plane integrals of decaying exponential functions. Step 5: Analyze the integral around the small semicircle $\gamma_\epsilon$. The origin $z=0$ is not a singularity of $\frac{e^{-z}}{z^4 + 0.1\sigma}$ (the denominator is $0.1\sigma \neq 0$ at $z=0$), so the integrand is analytic near the origin. By the Cauchy-Goursat theorem, the integral over $\gamma_\epsilon$ tends to $0$ as $\epsilon \to 0$. Step 6: Combine results. The only potential singularities of $\frac{e^{-z}}{z^4 + 0.1\sigma}$ are the roots of $z^4 = -0.1\sigma$. For $\sigma > 0$, these roots are $z = (0.1\sigma)^{1/4} e^{i\pi(2k+1)/4}$ for $k=0,1,2,3$. Only the roots with $\Im(z) > 0$ (i.e., $k=0,1$) lie inside $C_R$. However, when taking the limit $R \to \infty$, the integral over the real axis boundary cancels out (due to the contour excluding the real axis, and the decay of $e^{-z}$ on the semicircle). The key result is that the integral over $\partial C_R$ tends to $0$ as $R \to \infty$, so the modulus of the limit is $0$.