I wish to verify that
$$ g(w_1, w_2) = e^{x(z_1+1+z_2)}\frac{\Gamma (z_2) \Gamma (z_1)}{\Gamma (z_1+z_2 + 2)} \in L^1(\mathbb{R^2}) $$
Where $z_1 = - a - iw_1$ and $a <0$, $z_2 = -b - iw_2$ and $b < 0$, and $x$ a known constant. But the upper bounds I make are to extreme i.e
$$ \int_{\mathbb{R}} \int_{\mathbb{R}} \left| e^{x(z_1+1+z_2)} \frac{\Gamma(z_2) \Gamma(z_1)}{\Gamma(z_1 + z_2 + 2)} \right| dw_1 dw_2 $$
and since $\Re(z_1) > 0, \Re(z_2) >0$ the Gamma integrals exists and converges to the quotient can be bounded by some constant $M$ $$ \int_{\mathbb{R}} \int_{\mathbb{R}} \left| e^{x(z_1+1+z_2)} \frac{\Gamma(z_2) \Gamma(z_1)}{\Gamma(z_1 + z_2 + 2)} \right| dw_1 dw_2 < \\ \int_{\mathbb{R}} \int_{\mathbb{R}} \left| e^{x(z_1+1+z_2)} M \right| dw_1 dw_2 $$ But This is bounded to high.
Edit I got the following idea $$ \|g(w_1, w_2)\|_1 = \int_{\mathbb{R}} \int_{\mathbb{R}} \left| e^{x(z_1+1+z_2)}\frac{\Gamma (z_2) \Gamma (z_1)}{\Gamma (z_1+z_2 + 2)} \right| dw_1 dw_2 \leq \\e^{-ax -bx +x} \int_{\mathbb{R}} \int_{\mathbb{R}} \frac{\Gamma (z_2) \Gamma (z_1)}{\Gamma (z_1+z_2 + 2)} = \\ e^{-ax -bx +x} \int_{\mathbb{R}} \int_{\mathbb{R}}\left| \frac{(z_1-1)!(z_2-1)!}{(z_1 + z_2+1)(z_1+z_2)!)} \right| dw_1 dw_2 \sim \\ \frac{C}{(z_1 + z_2+1)} \text{ At the boundary} $$ Where $\sim$ means, behaves like. Since the real part of both $z_1, z_2$ is greater than zero, the integral feels very convergent.
A professor at my university came up with the following \begin{equation*} \begin{gathered} f_x(w) = \left| \Gamma (x + iw) \right| = \sqrt{2 \pi} e^{-\frac{\pi}{2} |w|} |w|^{x - \frac 12} \left( 1 + O \left( \frac{1}{|w|} \right) \right) \end{gathered} \end{equation*}
under the condition that $x$ is fixed and that $w \in \mathbb{R}$. Since $f_x$ is continuous and doesn't have any zeros it follows that there exists constants $0 \leq c_x \leq C_x$ such that \begin{equation*} c_x e^{-\frac{\pi}{2}|w|} \sqrt{1 + w^2}^{x- \frac 12} \leq f_x(w) \leq C_x e^{-\frac{\pi}{2}|w|} \sqrt{1 + w^2}^{x- \frac 12} \end{equation*}
By using these approximations on the quotient of Gamma functions in $g$ we obtain that
\begin{equation*} \begin{gathered} |g| \leq C \frac{\sqrt{1+w_1^2}^{-a - \frac 12} \sqrt{1+w_2^2}^{-b - \frac 12}}{\sqrt{1 + (w_1+w_2)^2}^{-a - b + \frac 32}} e^{- \frac {\pi}{2}(|w_1| + |w_2| - |w_1+w_2|)} \end{gathered} \end{equation*} Let $0 < \delta <1$ and \begin{equation*} M_{\delta} = \{ (w_1, w_2) | \delta \cdot ( \sqrt{1+ w_1^2} + \sqrt{1+w_2}) > \sqrt{1 + w_1} + \sqrt{1+w_2} - |w_1 + w_2| \geq 0. \end{equation*}
Then for $(w_1, w_2) \in \mathbb{R}^2 \backslash M_{\delta}$,
\begin{equation*} \begin{gathered} |g| \leq C \frac{\sqrt{1+w_1^2}^{-a - \frac 12} \sqrt{1+w_2^2}^{-b - \frac 12}}{\sqrt{1 + (w_1+w_2)^2}^{-a - b + \frac 32}} e^{- \frac {\pi}{2}(|w_1| + |w_2| - |w_1+w_2|)} \leq \\ C_1 \frac{\sqrt{1+w_1^2}^{-a - \frac 12} \sqrt{1+w_2^2}^{-b - \frac 12}}{\sqrt{1 + (w_1+w_2)^2}^{-a - b + \frac 32}} e^{- \delta \frac {\pi}{2}(|w_1| + |w_2|)} \leq \\ C_1 \sqrt{1 + w_1^2}^{-a - \frac 12}e^{-\delta \frac{\pi}{2}|w_1|} \cdot \sqrt{1 + w_2^2}^{-b - \frac 12} e^{-\delta \frac {\pi}{2} |w_2|} \in L^1(\mathbb{R}^2) \end{gathered} \end{equation*}
On the set $M_{\delta}$ we instead have the following upper bound
\begin{equation*} \begin{gathered} |g| \leq C \frac{\sqrt{1+w_1^2}^{-a - \frac 12} \sqrt{1+w_2^2}^{-b - \frac 12}}{\sqrt{1 + (w_1+w_2)^2}^{-a - b + \frac 32}} e^{- \frac {\pi}{2}(|w_1| + |w_2| - |w_1+w_2|)} \leq \\ C_2 \frac{\sqrt{1 + w_1^2}^{-a - \frac 12} \sqrt{1 + w_2^2}^{-b - \frac 12}}{\sqrt{1 + (w_1 + w_2)^2}^{-a - b + \frac 32}} \end{gathered} \end{equation*} Since $\sqrt{1 + t^2}$ is an even function of $t$, and with the property that $ \sqrt{1 + \delta ^2t^2} \geq \delta \sqrt{1 + t^2}$ it holds on $M_{\delta}$ that $\sqrt{1 + (w_1 + w_2)^2} \geq \sqrt{1 + \delta ^2 w_j^2} \geq \delta \sqrt{1+ w_j^2}$ with $j=1 \vee 2$. This implies that
\begin{equation*} \begin{gathered} |g | C_2 \frac{\sqrt{1 + w_1^2}^{-a - \frac 12} \sqrt{1 + w_2^2}^{-b - \frac 12}}{\sqrt{1 + (w_1 + w_2)^2}^{-a - b + \frac 32}} \leq \\ \frac{C_2}{\delta^{-a - b + \frac 32}} \frac{\sqrt{1 + w_1^2}^{-a - \frac 12} \sqrt{1 + w_2^2}^{-b - \frac 12}}{\sqrt{1+w_1^2}^{-a - \frac 34} \sqrt{1 + w_2^2}^{-b + \frac 34}} = C_{\delta} \sqrt{1 + w_1^2}^{-\frac 54} \sqrt{1+w_2^2}^{-\frac 54} \in L^1(\mathbb{R}^2). \end{gathered} \end{equation*}
The final conclusion is then that
\begin{equation*} \begin{gathered} \|g \|_1 = \int_{\mathbb{R}} \int_{\mathbb{R}} |g| dw_1 dw_2 = \\ \underbrace{\int \int_{M_{\delta}} |g| dw_1 dw_2 + \int \int_{\mathbb{R}^2 \backslash M_{\delta}} |g| dw_1 dw_2}_{\in L^1(\mathbb{R}^2)} < \infty \Rightarrow \\ g \in L^1(\mathbb{R}^2) \end{gathered} \end{equation*}