The Gamma Function

Since $K$ is compact in the open right half-plane, choose real numbers $0<a\le b$ with $a\le \mathrm{Re}z\le b$ for every $z\in K$. For $0<t\le 1$ one has $\mathrm{\mid }{t}^{z-1}{e}^{-t}\mathrm{\mid }=t^{\mathrm{Re}z-1}{e}^{-t}\le t^{a-1}$, and for $t\ge 1$ one has $\mathrm{\mid }{t}^{z-1}{e}^{-t}\mathrm{\mid }\le t^{b-1}{e}^{-t}$.

By [L1], both majorants from step 1.1 have convergent improper integrals on their respective ranges. The Weierstrass M-test on compact subsets therefore gives local-uniform convergence of the truncated integrals to the Gamma integral on $K$.

For $n\ge 1$, define ${\mathrm{\Gamma }}_n(z):={\int }_{1\mathrm{/}n}^n{t}^{z-1}{e}^{-t}dt\mathrm{.}$ The integrand is jointly continuous in $(t,z)$ on $[1\mathrm{/}n,n]\times \{z:\mathrm{Re}z>0\}$ and holomorphic in $z$ for each fixed $t>0$, so [L2] makes ${\mathrm{\Gamma }}_n$ holomorphic on the right half-plane.

By [L1], on every compact subset of the right half-plane the functions ${\mathrm{\Gamma }}_n$ converge uniformly to $\mathrm{\Gamma }$. Therefore [L3] makes $\mathrm{\Gamma }$ holomorphic on $\mathrm{Re}z>0$.

For $t>0$ and real $x$, the complex-analytic convention gives $t^{x-1}=\exp ((x-1)\log t)$, which is the ordinary real power.

Since $x>0$, [L2] says the integral converges, and [L1] names its value as the real Gamma function. By [L3], the complex Gamma function assigns exactly the same integral to $x$. Therefore the two constructions agree on $(0,\mathrm{\infty })$.

For $0<\epsilon <R$, apply [L1] on $[\epsilon ,R]$ with $u(t)=t^z$ and $v^{\mathrm{\prime }}(t)=e^{-t}$. This gives ${\int }_{\epsilon }^R{t}^z{e}^{-t}dt=[-t^z{e}^{-t}{]}_{\epsilon }^R+z{\int }_{\epsilon }^R{t}^{z-1}{e}^{-t}dt\mathrm{.}$

Since $\mathrm{Re}z>0$, one has $\mathrm{\mid }{\epsilon }^z{e}^{-\epsilon }\mathrm{\mid }={\epsilon }^{\mathrm{Re}z}{e}^{-\epsilon }\to 0$ as $\epsilon \downarrow 0$, and $\mathrm{\mid }{R}^z{e}^{-R}\mathrm{\mid }=R^{\mathrm{Re}z}{e}^{-R}\to 0$ as $R\uparrow \mathrm{\infty }$. Passing to the improper limits in step 1.1 and using [L2] and [L3] gives $\mathrm{\Gamma }(z+1)=z\mathrm{\Gamma }(z)$.

Directly from the defining integral, $\mathrm{\Gamma }(1)={\int }_0^{\mathrm{\infty }}{e}^{-t}dt=1$.

Repeatedly applying [L1] gives $\mathrm{\Gamma }(n+1)=n\mathrm{\Gamma }(n)=\cdots =n!\mathrm{\Gamma }(1)=n!$.

For each integer $m\ge 0$, define $G_m(z):=\frac{\mathrm{\Gamma }(z+m+1)}{z(z+1)\cdots (z+m)}$ on the half-plane $\mathrm{Re}z>-m-1$ with the nonpositive integers $0,-1,\dots ,-m$ removed. By repeated use of [L1], $G_m(z)=\mathrm{\Gamma }(z)$ whenever $\mathrm{Re}z>0$.

The functions $G_m$ and $G_{m+1}$ agree on their common domain because both equal $\mathrm{\Gamma }(z)$ on the nonempty open half-plane $\mathrm{Re}z>0$. Hence the $G_m$ glue to a meromorphic continuation of Gamma to $\mathbb{C}$, and step 1.1 is exactly the displayed continuation formula on the domain of $G_m$. The denominator in step 1.1 shows that the only possible poles are the nonpositive integers, and each is simple.

Near $z=-n$, take $m=n$. Then $\mathrm{\Gamma }(z)=\frac{\mathrm{\Gamma }(z+n+1)}{z(z+1)\cdots (z+n-1)(z+n)}\mathrm{.}$ Using [L2], the numerator tends to $\mathrm{\Gamma }(1)=1$ and the product excluding $z+n$ tends to $(-n)(-n+1)\cdots (-1)=(-1{)}^{n}n!$. Therefore $\mathrm{Res}(\mathrm{\Gamma },-n)=\frac{1}{(-1{)}^{n}n!}=\frac{(-1{)}^n}{n!}\mathrm{.}$

First assume $\mathrm{Re}z>0$. By [L2] and the functional equation inside [L1], $\frac{n!n^z}{z(z+1)\cdots (z+n)}=n^z{\int }_0^1{t}^{z-1}(1-t{)}^{n}dt={\int }_0^n{u}^{z-1}{(1-\frac{u}{n})}^{n}du\mathrm{.}$

For each fixed $u\ge 0$, $(1-u\mathrm{/}n{)}^n\to e^{-u}$, and on compact right-half-plane strips the integrands from step 1.1 are dominated by an integrable majorant. Therefore the integrals in step 1.1 converge locally uniformly to ${\int }_0^{\mathrm{\infty }}{u}^{z-1}{e}^{-u}du=\mathrm{\Gamma }(z)$. Hence the displayed limit formula holds on $\mathrm{Re}z>0$.

For $n\ge 1$, write $F_n(z):=\frac{n!n^z}{z(z+1)\cdots (z+n)}\mathrm{.}$ Let $K\subseteq \mathbb{C}\setminus \{0,-1,-2,\dots \}$ be compact. Choose $m\ge 0$ so that $K+m+1:=\{z+m+1:z\in K\}$ lies in the right half-plane. Repeatedly using $F_n(z)=\frac{z+n+1}{nz}{F}_n(z+1)$ gives $F_n(z)=\left({\prod }_{j=0}^m\frac{z+n+j+1}{n(z+j)}\right)F_n(z+m+1)\mathrm{.}$ On $K$, the prefactor converges uniformly to $1\mathrm{/}(z(z+1)\cdots (z+m))$, while step 2.1 applied on $K+m+1$ gives $F_n(z+m+1)\to \mathrm{\Gamma }(z+m+1)$ uniformly there. By [L1], $\mathrm{\Gamma }(z)=\frac{\mathrm{\Gamma }(z+m+1)}{z(z+1)\cdots (z+m)}$ on $K$, so $F_n\to \mathrm{\Gamma }$ uniformly on $K$. Since $K$ was arbitrary, the limit formula holds locally uniformly on the whole pole-free set.

Define $F(z):=z{e}^{\gamma z}{\prod }_{k\ge 1}(1+\frac{z}{k})e^{-z\mathrm{/}k}\mathrm{.}$ On a fixed compact set, after finitely many initial factors the logarithms of the remaining factors are $O(k^{-2})$ uniformly in $z$. Hence the product converges locally uniformly and defines an entire function. Its tail is zero-free, so its zeros are simple and occur exactly at $0,-1,-2,\dots$.

Let $x>0$. By [L3], $\mathrm{\Gamma }(x)$ is a positive real number. Taking reciprocals in [L1] is therefore valid at $x$, and rewriting the finite product gives $\frac{1}{\mathrm{\Gamma }(x)}={\lim }_{n\to \mathrm{\infty }}x{e}^{(H_n-\log n)x}{\prod }_{k=1}^n(1+\frac{x}{k})e^{-x\mathrm{/}k}=F(x),$ where [L2] supplies $H_n-\log n\to \gamma$.

On the right half-plane, both $F$ and $\mathrm{\Gamma }$ are holomorphic by [L4], so $F\mathrm{\Gamma }-1$ is holomorphic there. Step 2.1 makes it vanish on the positive real axis, which has accumulation points in that half-plane. Thus [L5] gives $F(z)\mathrm{\Gamma }(z)=1(\mathrm{Re}z>0)\mathrm{.}$

By [L4], the only poles of $\mathrm{\Gamma }$ are simple poles at the nonpositive integers. Step 1.1 gives $F$ a simple zero at each of those points, so every possible singularity of $F\mathrm{\Gamma }$ there is removable. The resulting entire function equals $1$ on the right half-plane by step 3.1, hence equals $1$ on all of $\mathbb{C}$ by [L5]. Therefore $F$ is the entire reciprocal of Gamma, which proves the displayed product formula everywhere.

In [L1], the exponential factors never vanish, while the factor $z$ gives a simple zero at $0$ and the factor $1+z\mathrm{/}n$ gives a simple zero at $z=-n$ for each $n\ge 1$. There are no other zeros.

Therefore $1\mathrm{/}\mathrm{\Gamma }$ vanishes exactly at the nonpositive integers, and Gamma has poles there and no zeros anywhere.

Apply [L1] at $1-z$. For the $N$th partial product, $\frac{1}{\mathrm{\Gamma }(1-z)}=(1-z)e^{\gamma (1-z)}{\lim }_{N\to \mathrm{\infty }}{\prod }_{n=1}^N(1+\frac{1-z}{n})e^{-(1-z)\mathrm{/}n}\mathrm{.}$ The identity $(1-z){\prod }_{n=1}^N(1+\frac{1-z}{n})=(N+1){\prod }_{n=1}^{N+1}(1-\frac{z}{n})$ therefore gives $\frac{1}{\mathrm{\Gamma }(1-z)}=e^{-\gamma z}{\lim }_{N\to \mathrm{\infty }}(N+1)e^{\gamma -H_N}{e}^{-z\mathrm{/}(N+1)}{\prod }_{n=1}^{N+1}(1-\frac{z}{n})e^{z\mathrm{/}n}\mathrm{.}$ By [L3], the scalar prefactor tends to $1$, so $\frac{1}{\mathrm{\Gamma }(1-z)}=e^{-\gamma z}{\prod }_{n\ge 1}(1-\frac{z}{n})e^{z\mathrm{/}n}\mathrm{.}$

Multiplying step 1.1 by the product for $1\mathrm{/}\mathrm{\Gamma }(z)$ from [L1], the exponential factors cancel and one gets $\frac{1}{\mathrm{\Gamma }(z)\mathrm{\Gamma }(1-z)}=z{\prod }_{n\ge 1}(1-\frac{z^2}{n^2})\mathrm{.}$ By [L2], the product on the right is $\sin (\pi z)\mathrm{/}\pi$. Therefore $\frac{1}{\mathrm{\Gamma }(z)\mathrm{\Gamma }(1-z)}=\frac{\sin (\pi z)}{\pi }$ on $\mathbb{C}\setminus \mathbb{Z}$, which is equivalent to the displayed formula.

Substituting $z=1\mathrm{/}2$ into [L3] gives $\mathrm{\Gamma }(1\mathrm{/}2{)}^2=\pi$, because $\sin (\pi \mathrm{/}2)=1$.

By [L1] and [L2], $\mathrm{\Gamma }(1\mathrm{/}2)$ is the positive real number $\sqrt{\pi }$. Step 1.1 leaves only the two square roots of $\pi$, so the positive one is the required value.

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.

Apply [L1] to each factor $\mathrm{\Gamma }(z+k\mathrm{/}m)$ and multiply. The denominator collapses by ${\prod }_{k=0}^{m-1}{\prod }_{j=0}^n(z+\frac{k}{m}+j)=m^{-m(n+1)}{\prod }_{r=0}^{m(n+1)-1}(mz+r)\mathrm{.}$ Therefore ${\prod }_{k=0}^{m-1}\mathrm{\Gamma }\text{\hspace{0.17em}⁣}(z+\frac{k}{m})=\mathrm{\Gamma }(mz){\lim }_{n\to \mathrm{\infty }}\frac{(n!{)}^m{m}^{m(n+1)}{n}^{mz+(m-1)\mathrm{/}2}}{(m(n+1)-1)!(m(n+1)-1{)}^{mz}}\mathrm{.}$

Apply [L2] to the factorial ratio in step 1.1. After the standard cancellation of the exponential and power terms, the limit becomes $(2\pi {)}^{(m-1)\mathrm{/}2}{m}^{1\mathrm{/}2-mz}$. Substituting this into step 1.1 yields the displayed multiplication formula.

Apply [L1] with $m=2$. This gives $\mathrm{\Gamma }(z)\mathrm{\Gamma }(z+1\mathrm{/}2)=(2\pi {)}^{1\mathrm{/}2}{2}^{1\mathrm{/}2-2z}\mathrm{\Gamma }(2z)$.

Step 1.1 is exactly the duplication formula after rewriting $(2\pi {)}^{1\mathrm{/}2}{2}^{1\mathrm{/}2-2z}$ as $2^{1-2z}\sqrt{\pi }$.

Taking logarithms in [L1] on the chosen sector gives [L1, given, algebra] $\log \mathrm{\Gamma }(z)=-\log z-\gamma z+{\sum }_{n\ge 1}(\frac{z}{n}-\mathrm{Log}(1+\frac{z}{n})),$ where $\mathrm{Log}$ denotes the principal logarithm.

For an integer $N\ge 1$, define [step 1.1, algebra] $I_N(z):={\int }_0^N\frac{\lfloor u\rfloor -u+1\mathrm{/}2}{u+z}du\mathrm{.}$ On each interval $[n,n+1]$ with $0\le n\le N-1$, one has ${\int }_n^{n+1}\frac{n-u+1\mathrm{/}2}{u+z}du=(n+\frac{1}{2}+z)\mathrm{Log}\frac{n+1+z}{n+z}-1.$ Summing these equalities and telescoping the logarithms yields $I_N(z)=(N+z-\frac{1}{2})\mathrm{Log}(N+z)-(z+\frac{1}{2})\mathrm{Log}z-N-\log (N-1)!-z{H}_{N-1}+{\sum }_{n=1}^{N-1}(\frac{z}{n}-\mathrm{Log}(1+\frac{z}{n}))\mathrm{.}$

For fixed $z$ in the sector, [L2] and [L3] give [L2, L3, step 1.1, step 2.1, algebra] $H_{N-1}=\log N+\gamma +o(1),\log (N-1)!=(N-\frac{1}{2})\log N-N+\frac{1}{2}\log (2\pi )+o(1),$ and also $\mathrm{Log}(N+z)=\log N+o(1)$ as $N\to \mathrm{\infty }$. Comparing the $N\to \mathrm{\infty }$ limit of step 2.1 with the partial sums in step 1.1 gives the Binet-type formula $\log \mathrm{\Gamma }(z)=(z-\frac{1}{2})\mathrm{Log}z-z+\frac{1}{2}\log (2\pi )+{\int }_0^{\mathrm{\infty }}\frac{\lfloor u\rfloor -u+1\mathrm{/}2}{u+z}du\mathrm{.}$

Let [step 3.1, algebra] $\mathrm{\Phi }(u):={\int }_0^u(\lfloor v\rfloor -v+1\mathrm{/}2)dv\mathrm{.}$ Then $\mathrm{\Phi }$ is $1$-periodic and therefore bounded. Integrating by parts in step 3.1 gives ${\int }_0^{\mathrm{\infty }}\frac{\lfloor u\rfloor -u+1\mathrm{/}2}{u+z}du={\int }_0^{\mathrm{\infty }}\frac{\mathrm{\Phi }(u)}{(u+z{)}^2}du\mathrm{.}$ On the closed sector $\mathrm{\mid }\arg z\mathrm{\mid }\le \pi -\delta$, one has $\mathrm{\mid }u+z\mathrm{\mid }\ge c_{\delta }(u+\mathrm{\mid }z\mathrm{\mid })$ for a constant $c_{\delta }>0$, so the integral above is $O_{\delta }(\mathrm{\mid }z{\mathrm{\mid }}^{-1})$ uniformly as $\mathrm{\mid }z\mathrm{\mid }\to \mathrm{\infty }$. Hence $\log \mathrm{\Gamma }(z)=(z-\frac{1}{2})\mathrm{Log}z-z+\frac{1}{2}\log (2\pi )+O_{\delta }(\mathrm{\mid }z{\mathrm{\mid }}^{-1}),$ and exponentiating yields $\mathrm{\Gamma }(z)=\sqrt{2\pi }{z}^{z-1\mathrm{/}2}{e}^{-z}(1+O_{\delta }(\mathrm{\mid }z{\mathrm{\mid }}^{-1}))\mathrm{.}$

First assume $0<\mathrm{Re}z<1$. On the small circle about $0$, the integrand has size $O(r^{-\mathrm{Re}z})$, so the circular contribution tends to $0$ as $r\downarrow 0$. On the two rays, $e^t$ decays exponentially as $t=-x$ with $x\to +\mathrm{\infty }$, so the contour integral converges absolutely.

Along the upper and lower sides of the cut, one has $(-x\pm i0{)}^{-z}=e^{\mp \pi iz}{x}^{-z}$. With the orientation from [L3], the lower ray contributes $e^{\pi iz}{\int }_0^{\mathrm{\infty }}{e}^{-x}{x}^{-z}dx$ and the upper ray contributes $-e^{-\pi iz}{\int }_0^{\mathrm{\infty }}{e}^{-x}{x}^{-z}dx$. Therefore step 1.1 yields ${\int }_H{e}^t{t}^{-z}dt=(e^{\pi iz}-e^{-\pi iz}){\int }_0^{\mathrm{\infty }}{e}^{-x}{x}^{-z}dx=2i\sin (\pi z)\mathrm{\Gamma }(1-z)\mathrm{.}$

Using [L1], the right-hand side of step 2.1 equals $2\pi i\mathrm{/}\mathrm{\Gamma }(z)$ on $0<\mathrm{Re}z<1$. Hence the displayed integral formula holds on that strip. Both sides are meromorphic in $z$, and [L2] extends the identity to all of $\mathbb{C}$.

Define $F(z):=\mathrm{\Gamma }(z)e^{\sin (2\pi z)}$. Since $\sin (2\pi (z+1))=\sin (2\pi z)$, the exponential factor is $1$-periodic. Hence [L1] gives $F(z+1)=zF(z)$.

For every integer $n$, $\sin (2\pi n)=0$, so $e^{\sin (2\pi n)}=1$. Thus [L2] gives $F(n+1)=\mathrm{\Gamma }(n+1)=n!$. But $e^{\sin (2\pi z)}\mathrm{\ne }1$ identically, so $F\mathrm{\ne }\mathrm{\Gamma }$. Therefore the stated data do not characterize Gamma.