I am preparing for the complex analysis qualifying exam, and I recently came across this problem:
Show that $$F(z)=\int_1^\infty\frac{t^z}{\sqrt{1+t^3}}\,dt$$ defines an analytic function on $\{z\in\mathbb{C}|\text{Re }z<\frac{1}{2}\}$ that has a meromorphic extension to $\{z\in\mathbb{C}|\text{Re }z<\frac{3}{2}\}$.
The first portion is easy, the integral converges when $\text{Re }z<\frac{1}{2}$, and $F(z)$ is analytic on $\{z\in\mathbb{C}|\text{Re }z<\frac{1}{2}\}$ because we can interchange differentiation and integration. But I don't know where to start on the second part. The theorems I know on continuation are the monodromy theorem and Schwartz reflection principle. But this problem asks for meromorphic continuation. I need some help please!
Best Answer
Intuitively, the integrand is $\asymp t^{z-3/2}$ as $t\to\infty$, which (integrated) causes a pole at $z=1/2$. Formally, $$\frac{t^z}{\sqrt{1+t^3}}=t^{z-3/2}\big(1-f(t)\big),\quad f(t)=1-\frac{t^{3/2}}{\sqrt{1+t^3}}=\mathcal{O}(t^{-3})\quad(t\to\infty)$$ so that, for $\Re z<1/2$, we have $$F(z)+\frac{1}{z-1/2}=-\int_1^\infty t^{z-3/2}f(t)\,dt,$$ and the RHS is analytic not only on $\Re z<3/2$, but even on $\Re(z-3/2-3)<-1$, i.e. $\Re z<7/2$.
Thus, $F(z)+(z-1/2)^{-1}$ extends analytically (that is, $F(z)$ extends meromorphically) onto there.