$$\Gamma(z) := \int\limits_0^\infty x^{z-1} e^{-x} dx$$
However for the zeta function, it’s not clear how to get a nice form from the primary definition involving an infinite sum.
$$\zeta(z) := \sum\limits_{k=1}^\infty \frac{1}{k^z}$$
We can attempt to find a nicer form for the zeta function by starting with the product $\zeta(z) \Gamma(z)$ and continuing with their definitions…
$
\begin{eqnarray}
\zeta(z) \Gamma(z) & = & \sum\limits_{k=1}^\infty \frac{1}{k^z} \int\limits_0^\infty x^{z-1} e^{-x} dx \\
& = & \sum\limits_{k=1}^\infty \int\limits_0^\infty \frac{x^{z-1} e^{-x}}{k^{z-1}} \frac{dx}{k} \\
& = & \sum\limits_{k=1}^\infty \int\limits_0^\infty u^{z-1} e^{-k u} du \\
& = & \sum\limits_{k=0}^\infty \int\limits_0^\infty u^{z-1} e^{-u} e^{-k u} du \\
& = & \int\limits_0^\infty \frac{u^{z-1} e^{-u}}{1 – e^{-u}} du \\
& = & \int\limits_0^\infty \frac{u^{z-1}}{e^u – 1} du
\end{eqnarray}
$
And now we have a definition for the zeta function that works for $Re(z) > 1$:
$$\zeta(z) = \frac{1}{\Gamma(z)} \int\limits_0^\infty \frac{u^{z-1} du}{e^u – 1}$$
Of course, I just really like the $\zeta(z) \Gamma(z)$ integral:
$$\zeta(z) \Gamma(z) = \large\int\limits_{\small0}^{\small\infty} \! \frac{u^{z-1}}{e^u-1} \, \mathrm{d}u, \qquad Re(z) > 1$$