Let's consider a function $$f(v) = \frac{v^{z-1}}{1+v}, \qquad 0<\text{Re}(z) < 1.$$
Let's also define a circle $C_R$ with a radius equal to $R$ and the centre at the beginning of the coordinate system.
We thing of $v^{z-1}$ as $\, \exp((z-1) \log v)$, where $\log$ denotes the complex logarithm such that $\log(-1) = \pi i$ and $\log$ is continuous on $\mathbb{C} \setminus [0, \infty)$.
I would like to compute the line integral over $C_R$ when $R \to \infty$. I suppose that it is equal to zero but I have no idea how can it be shown. I would appreciate any hints or tips.
This is a rough approximation, but can be made more complete. If $z = x + yi$ is fixed and $v = Re^{i\theta}$, then the norm of $v^{z-1}$ is equal to $R^{x-1}e^{-y\theta}$. Since $y$ is fixed and $\theta \in (0,2\pi)$, this means that the norm of the numerator grows (up to a constant) like $R^{x-1}$ as $R$ becomes large. For the numerator, clearly for $v$ with large norms $R$, $|v+1|$ is approximately $R$. Thus the norm of the function $f(v)$ grows asymptotically like $R^{x-2}$. Thus we have the bound $|f(v)| \leq c\cdot R^{x-2}$ for $|v| = R$ large enough for some constant $c$, and so $$\left|\int_{C_R}f(v)dv \right| \leq \int_{C_R}|f(v)|dv \leq c\cdot R^{x-2}\cdot 2\pi R = 2\pi c R^{x-1}.$$ Since $x < 1$, we see that the integral goes to 0.