Why is the Gamma function defined by the definite integral for $\Re(z) >0$?

Question

Gamma Function gives the integral definition of the gamma function as

\begin{equation*} \tag{1} \Gamma(z) := \int_{0}^\infty e^{-t}t^{z-1}\,\,dt, \text{ for } \Re(z) > 0. \end{equation*}

Why is $(1)$ defined for $\Re(z) > 0$? Could I not have $\Gamma(3i) = \int_{0}^\infty e^{-t}t^{-1+3i}\,\,dt$ for example? Can I write 'The integral definition of the gamma function is: \begin{equation*} \Gamma(z) := \int_{0}^\infty e^{-t}t^{z-1}\,\,dt, \text{ for } \Re(z) \geq 0 \text{ and } z\neq 0?' \end{equation*}

Answer 1

$\int_1^{\infty} e^{-t} t^{z-1}\, dt$ exists for all complex numbers $z$ but $\int_0^{1} e^{-t} |t^{z-1}|\, dt=\int_0^{1} e^{-t} t^{x-1}\, dt$ where $x$ is the real part of $z$ and this integral is finite iff $x>0$.

In particular $\int_0^{\infty} e^{-t} t^{i-1}\, dt$ does not exist so you cannot use this integral to define $\Gamma (i)$.