The Beta-Gamma identity

Fix a real number $q_0>0$. For $n\ge 2$, define $B_n(p):={\int }_{1\mathrm{/}n}^{1-1\mathrm{/}n}{t}^{p-1}(1-t{)}^{q_0-1}dt\mathrm{.}$ By [L4], each $B_n$ is holomorphic on $\mathrm{Re}p>0$. On a compact set $K\subseteq \{p:\mathrm{Re}p>0\}$ choose $a>0$ with $\mathrm{Re}p\ge a$ on $K$; then the omitted tails are dominated by $t^{a-1}(1-t{)}^{q_0-1}$ near $0$ and by $(1-t{)}^{q_0-1}$ near $1$, so $B_n\to B(\cdot ,q_0)$ locally uniformly on $\mathrm{Re}p>0$. Hence [L6] makes $p\mapsto B(p,q_0)$ holomorphic there.

The function $H_{q_0}(p):=\mathrm{\Gamma }(p+q_0)B(p,q_0)-\mathrm{\Gamma }(p)\mathrm{\Gamma }(q_0)$ is holomorphic on $\mathrm{Re}p>0$ by step 1.1 and [L5]. If $p>0$ is real, then [L1] and [L2] identify the complex and real formulas, so $H_{q_0}(p)=0$. The positive real axis has an accumulation point in the right half-plane, so [L3] gives $H_{q_0}\equiv 0$. Thus $\mathrm{\Gamma }(p+q_0)B(p,q_0)=\mathrm{\Gamma }(p)\mathrm{\Gamma }(q_0)(\mathrm{Re}p>0,q_0>0\text{ real})\mathrm{.}$

Now fix $p$ with $\mathrm{Re}p>0$. Repeating step 1.1 with the roles of $p$ and $q$ reversed shows that $q\mapsto B(p,q)$ is holomorphic on $\mathrm{Re}q>0$. Therefore $K_p(q):=\mathrm{\Gamma }(p+q)B(p,q)-\mathrm{\Gamma }(p)\mathrm{\Gamma }(q)$ is holomorphic on $\mathrm{Re}q>0$ by [L5]. Step 2.1 shows $K_p(q)=0$ for every positive real $q$, so [L3] gives $K_p\equiv 0$. This is exactly the displayed identity.