Binet integral identity for the right-half-plane log-Gamma branch

Define $f(u)=\lfloor u\rfloor -u+1\mathrm{/}2$. For $u=n+t$ with integer $n\ge 0$ and $0\le t<1$, $f(u)=1\mathrm{/}2-t$. Each full unit interval has integral zero, and its partial integral is $t(1-t)\mathrm{/}2$. Therefore $\mathrm{\Phi }(n+t)=t(1-t)\mathrm{/}2$, $\mathrm{\Phi }(n)=0$, and $0\le \mathrm{\Phi }\le 1\mathrm{/}8$.

On $[n,n+1]$, ${\mathrm{\Phi }}^{\mathrm{\prime }}=f$ and [F7] gives ${\int }_n^{n+1}\frac{f(u)}{u+z}du={\left[\frac{\mathrm{\Phi }(u)}{u+z}\right]}_n^{n+1}+{\int }_n^{n+1}\frac{\mathrm{\Phi }(u)}{(u+z{)}^2}du={\int }_n^{n+1}\frac{\mathrm{\Phi }(u)}{(u+z{)}^2}du,$ since $\mathrm{\Phi }(n)=\mathrm{\Phi }(n+1)=0$. Summing over $n<N$ gives $I_N(z):={\int }_0^N\frac{f(u)}{u+z}du={\int }_0^N\frac{\mathrm{\Phi }(u)}{(u+z{)}^2}du\mathrm{.}$ Because $\mathrm{\mid }u+z\mathrm{\mid }\ge u+\mathrm{Re}z$ and $\mathrm{\Phi }\le 1\mathrm{/}8$, the right-hand side converges absolutely as $N\to \mathrm{\infty }$.

On each $[n,n+1]$, write $f(u)=n+\frac{1}{2}-u=(n+z+\frac{1}{2})-(u+z)$. By [F5] the function $(n+z+\frac{1}{2})\mathrm{Log}(u+z)-u$ has derivative $f(u)\mathrm{/}(u+z)$; its derivative is continuous and integrable on the compact interval by [F6]. Applying [F6] and summing the resulting endpoint differences yields $I_N(z)=(N+z-\frac{1}{2})\mathrm{Log}(N+z)-(z+\frac{1}{2})\mathrm{Log}z-N-\log ((N-1)!)-z{H}_{N-1}+{\sum }_{k=1}^{N-1}(\frac{z}{k}-\mathrm{Log}(1+z\mathrm{/}k))\mathrm{.}$ The coefficient of each internal $\mathrm{Log}(k+z)$ in the endpoint sum is $-1$. For $k\ge 1$, $\mathrm{Log}(k+z)=\log k+\mathrm{Log}(1+z\mathrm{/}k)$ because $k>0$ is real and multiplication by it preserves the principal argument; [F9] gives ${\sum }_{k=1}^{N-1}\log k=\log ((N-1)!)$. These identities give the displayed form.

Add $-\mathrm{Log}z-{\gamma }_{E}z$ to the finite sum in step 2.2. By [F1]--[F3], the resulting left side converges to ${\mathrm{\ell }}_{+}(z)$, while ${\mathrm{\ell }}_{+}(z)=I_N(z)-(N+z-\frac{1}{2})\mathrm{Log}(N+z)+(z-\frac{1}{2})\mathrm{Log}z+N+\log ((N-1)!)+z(H_{N-1}-{\gamma }_E)+o(1)\mathrm{.}$

Taking logarithms in [F4] is valid by continuity [F8]. The laws [F9] and $e=\exp (1)$ give $\log (n!)=(n+\frac{1}{2})\log n-n+\frac{1}{2}\log (2\pi )+o(1)\mathrm{.}$ At $n=N-1$, since $\log (1-1\mathrm{/}N)=-1\mathrm{/}N+O(N^{-2})$, this becomes $\log ((N-1)!)=(N-\frac{1}{2})\log N-N+\frac{1}{2}\log (2\pi )+o(1)\mathrm{.}$ The logarithm estimate follows by integrating $1\mathrm{/}(1-t)$ on $[0,1\mathrm{/}N]$; the difference from $1\mathrm{/}N$ is bounded by $1\mathrm{/}(N^2(1-1\mathrm{/}N))$. Also [F3] gives $H_{N-1}-{\gamma }_E=\log N+o(1)$. Finally, for fixed $z$ and $N>2\mathrm{\mid }z\mathrm{\mid }$, $\mathrm{Log}(N+z)=\log N+z\mathrm{/}N+O_z(N^{-2}),$ since $N>0$ and $1+z\mathrm{/}N,N+z$ lie in the right half-plane, so $\mathrm{Log}(N+z)=\log N+\mathrm{Log}(1+z\mathrm{/}N)$. By [F5] and [F6], $\mathrm{Log}(1+w)-w=-{\int }_0^{1}t{w}^2\mathrm{/}(1+tw)dt$ for $w=z\mathrm{/}N$; here $\mathrm{\mid }1+tw\mathrm{\mid }\ge 1$ when $\mathrm{Re}z>0$. Therefore $(N+z-\frac{1}{2})\mathrm{Log}(N+z)=(N+z-\frac{1}{2})\log N+z+o(1)$.

Substitute these asymptotics into step 3.1. The coefficients of $\log N$ cancel, and $N$ cancels the $-N$ in the factorial term, leaving ${\mathrm{\ell }}_{+}(z)={\lim }_{N\to \mathrm{\infty }}{I}_N(z)+(z-\frac{1}{2})\mathrm{Log}z-z+\frac{1}{2}\log (2\pi )\mathrm{.}$ By step 2.1, the limit is the absolutely convergent integral of $\mathrm{\Phi }(u)\mathrm{/}(u+z{)}^2$. This is the claimed identity.