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

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: $$\begin{equation*} \Gamma(z) := \int_{0}^\infty e^{-t}t^{z-1}\,\,dt, \text{ for } \Re(z) \geq 0 \text{ and } z\neq 0?' \end{equation*}$$

• Perhaps the $t^{-1}$ term in $\int\limits_{0}^\infty e^{-t}t^{3i}t^{-1}\,\,dt$ causes problems Mar 27, 2019 at 9:53
• @Henry Okay cool. Mar 27, 2019 at 10:00

$$\int_1^{\infty} e^{-t} t^{z-1}\, dt$$ 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 $$\Gamma (i)$$.
• Okay. So is the functional equation $$\zeta(s) = 2^s\pi^{s-1}\sin \Bigl(\frac{s\pi}{2}\Bigr)\Gamma(1-s)\zeta(1-s), \text{ for } \Re(s) \leq 1 \text{ and } s \neq 0,1$$ valid at $s=1+3i$? Mar 27, 2019 at 10:05
• When I use Wolfram alpha to evaluate the integral, wolframalpha.com/input/… , it evaluates $\Gamma(i)$, i think. Is there an alternative definition to evaluate $\Gamma(z)$? Mar 27, 2019 at 10:12
• Before deriving the functional equation for $\zeta$ function one shows that there is a unique analytic function on $\mathbb C \setminus \{0,-1,-2,\cdots\}$ which coincides with the Gamma function defined by you for $\Re z >0$. The extended functions is also denoted by $\Gamma$. Mar 27, 2019 at 10:14