path of the integral in the initial definition of gamma function

Question

Can the path of the integral in the initial definition of gamma function be altered to a straight line starting from $0$ to $\infty;e^{ia},a<\pi/2$)?

Answer 1

Yes. Fix $z$ such that $\operatorname{Re}z>0$. By Cauchy's theorem, the difference between the integral over line segment $[r,R]$ and the integral over line segment $[re^{ia},Re^{ia}]$ comes from the contribution of two circular arcs:

1. arc from $R$ to $Re^{ia}$
2. arc from $r$ to $re^{ia}$

Since $ t^{z-1} e^{-t}$ decays exponentially at infinity, the integral over the first arc tends to $0$ as $R\to 0$.

For $0\le \theta\le a$ we have
$$\left|(re^{i\theta})^{z-1}\right| = r^{\operatorname{Re}z-1} |e^{i\theta (z-1)}|$$ The factor $|e^{i\theta (z-1)}|$ is harmless: it is bounded by some $M$ that does not depend on $r$. Integration over the second arc contributes at most $$ar M r^{\operatorname{Re}z-1} = aMr^{\operatorname{Re}z}$$ which tends to $0$ as $ r\to 0$.