The Real Gamma and Beta Functions

Suppose $s>0$. On $(0,1]$, $e^{-1}\le e^{-t}\le 1$, and [F5] with the fundamental theorem gives ${\int }_0^1{t}^{s-1}dt=1\mathrm{/}s<\mathrm{\infty }$. Thus [F2] gives convergence at zero.

If $s\le 0$, then $t^{s-1}\ge t^{-1}$ on $(0,1]$, while $e^{-t}\ge e^{-1}$. At $s=0$ this is exactly the logarithmic threshold, and [F3] and [F4] show ${\int }_0^{1}dt\mathrm{/}t$ diverges; the same lower comparison proves divergence for $s<0$.

For arbitrary real $s$, choose a natural $m\ge \max \{s-1,0\}$. For $t\ge 1$, $t^{s-1}{e}^{-t}\le t^m{e}^{-t}$, and [F1] with $a=1\mathrm{/}2$ makes $t^m{e}^{-t}\le e^{-t\mathrm{/}2}$ eventually. Since the latter has a convergent improper integral, [F2] gives convergence at infinity.

Steps 1.1 and 1.3 prove convergence for $s>0$, while step 1.2 proves divergence for every $s\le 0$. Hence the two improper ends converge simultaneously exactly on the positive real axis.

On $(0,1\mathrm{/}2]$, the continuous positive factor $(1-t{)}^{q-1}$ is bounded above and below by positive constants. Thus [F1] reduces convergence at zero to that of ${\int }_0^{1\mathrm{/}2}{t}^{p-1}dt$, whose primitive from [F4] converges exactly for $p>0$; at $p=0$, [F2] and [F3] give logarithmic divergence, and for $p<0$ the integrand dominates the same threshold.

The substitution $u=1-t$ changes the end $t\uparrow 1$ into $u\downarrow 0$. There $t^{p-1}$ is bounded above and below by positive constants, so [F4] and the same comparison give convergence exactly for $q>0$, with logarithmic divergence at $q=0$.

Both endpoint integrals converge exactly when $p>0$ and $q>0$. If either parameter is nonpositive, the corresponding endpoint diverges by step 1.1 or step 1.2.

Apply [F1] on $[\epsilon ,R]$ with $u(t)=t^s$ and $v^{\mathrm{\prime }}(t)=e^{-t}$: ${\int }_{\epsilon }^R{t}^s{e}^{-t}dt=[-t^s{e}^{-t}{]}_{\epsilon }^R+s{\int }_{\epsilon }^R{t}^{s-1}{e}^{-t}dt$.

Since $s>0$, ${\epsilon }^s{e}^{-\epsilon }\to 0$ as $\epsilon \downarrow 0$. At infinity choose a natural $m>s$; then $R^s{e}^{-R}\le R^m{e}^{-R}\to 0$ by exponential domination.

Letting the two truncations approach their improper ends in step 1.1 and using [F2] gives $\mathrm{\Gamma }(s+1)=s\mathrm{\Gamma }(s)$.

At $s=1$, $\mathrm{\Gamma }(1)={\int }_0^{\mathrm{\infty }}{e}^{-t}dt=[-e^{-t}{]}_0^{\mathrm{\infty }}=1$.

At $n=0$, [F1] gives $\mathrm{\Gamma }(1)=1=0!$.

Assume $\mathrm{\Gamma }(n+1)=n!$. Then [F1] gives $\mathrm{\Gamma }(n+2)=(n+1)\mathrm{\Gamma }(n+1)=(n+1)n!=(n+1)!$.

The induction principle therefore proves $\mathrm{\Gamma }(n+1)=n!$ for every $n\in \mathbb{N}$.

On $(0,1]$, put $u=-\log t$. Uniformly for $s\in [a,b]$, the absolute $k$th parameter derivative becomes $u^k{e}^{-su}\le u^k{e}^{-au}$ after substitution. By [F2] this is eventually bounded by $e^{-au\mathrm{/}2}$ and is integrable.

On $[1,\mathrm{\infty })$, $s\le b$ and $\log t\le t$ allow $\mathrm{\mid }(\log t{)}^k{t}^{s-1}{e}^{-t}\mathrm{\mid }$ to be bounded by $t^m{e}^{-t}$ for one natural $m$ depending only on $b,k$. By [F2] this has an integrable exponential majorant.

The case $k=0$ is the defining integral, convergent by [F4] and absolutely convergent because its integrand is nonnegative. If the displayed formula holds at order $k$, its integrand and parameter derivative are continuous; steps 1.1 and 1.2, applied at orders $k$ and $k+1$, give one absolutely integrable slice and an integrable uniform derivative bound. Thus [F1] differentiates once more. By [F3], the formula holds for every natural $k$.

Every $s>0$ lies in a compact interval $[a,b]\subset (0,\mathrm{\infty })$, so step 2.1 proves the formulas and smoothness throughout the positive axis.

Fix $c>0$ and write $c^s\mathrm{\Gamma }(s)={\int }_0^{\mathrm{\infty }}{t}^{-1}{e}^{-t}\exp (s\log (ct))dt$. By [F1], the integrand at $(1-\lambda )x+\lambda y$ is at most the corresponding convex combination, with strict inequality except at the single point $t=1\mathrm{/}c$; integration makes $s\mapsto c^s\mathrm{\Gamma }(s)$ strictly convex.

Choose $c=(\mathrm{\Gamma }(x)\mathrm{/}\mathrm{\Gamma }(y){)}^{1\mathrm{/}(y-x)}>0$. Then $c^x\mathrm{\Gamma }(x)=c^y\mathrm{\Gamma }(y)$.

Apply step 1.1 at $x,y$ with the $c$ from step 1.2. After cancelling $c^{(1-\lambda )x+\lambda y}$, one gets $\mathrm{\Gamma }((1-\lambda )x+\lambda y)<\mathrm{\Gamma }(x{)}^{1-\lambda }\mathrm{\Gamma }(y{)}^{\lambda }$, which is strict log-convexity by [F2].

By continuity at $1$ and [F1], $s\mathrm{\Gamma }(s)=\mathrm{\Gamma }(s+1)\to \mathrm{\Gamma }(1)=1$ as $s\downarrow 0$. Since $s\to 0^{+}$, this also gives $\mathrm{\Gamma }(s)\to +\mathrm{\infty }$.

One has $g(1)=g(2)=0$, while strict convexity [F2] gives $g(x)<0$ for $1<x<2$. Fixing such an $x$ and applying [F6] on $[1,x]$ and on $[x,2]$ gives $\xi \in (1,x)$ with $g^{\mathrm{\prime }}(\xi )<0$ and $\eta \in (x,2)$ with $g^{\mathrm{\prime }}(\eta )>0$.

By [F2] and [F5], $g^{\mathrm{\prime }}$ is nondecreasing. If $g^{\mathrm{\prime }}(u)=g^{\mathrm{\prime }}(v)$ for some $u<v$, then $g^{\mathrm{\prime }}$ is constant on $[u,v]$, so $g$ is affine there, contradicting the strict convexity of [F2]; hence $g^{\mathrm{\prime }}$ is strictly increasing and has at most one zero. By [F3] it is continuous, so the sign change in step 1.2 gives exactly one zero $s_0\in (1,2)$. Strict increase makes $g^{\mathrm{\prime }}$ negative before $s_0$ and positive after it, so [F6] makes $g$, and hence $\mathrm{\Gamma }$, decreasing on $(0,s_0]$ and increasing on $[s_0,\mathrm{\infty })$, and $s_0$ is the unique global minimum.

By [F4], $\mathrm{\Gamma }(n+1)=n!$. For $n\ge 3$, the ratios of $(n-1)!\mathrm{/}(n+1)$ grow by a factor exceeding $2$, so this sequence tends to infinity. By the eventual increase from step 2.1, if $n\le s<n+1$ then $\mathrm{\Gamma }(s)\ge \mathrm{\Gamma }(n)=(n-1)!$ and $\mathrm{\Gamma }(s)\mathrm{/}s\ge (n-1)!\mathrm{/}(n+1)$. Thus both quantities tend to infinity.

Log-convexity between $x$ and $x+1$ gives $\mathrm{\Gamma }(x+s)\le \mathrm{\Gamma }(x{)}^{1-s}\mathrm{\Gamma }(x+1{)}^s=x^s\mathrm{\Gamma }(x)$, strictly when $0<s<1$.

Since $x+1=s(x+s)+(1-s)(x+s+1)$, log-convexity gives $\mathrm{\Gamma }(x+1)\le \mathrm{\Gamma }(x+s{)}^s\mathrm{\Gamma }(x+s+1{)}^{1-s}=(x+s{)}^{1-s}\mathrm{\Gamma }(x+s)\le (x+1{)}^{1-s}\mathrm{\Gamma }(x+s)$.

Divide the inequalities in steps 1.1 and 1.2 by positive Gamma values and use [F2]. This gives the displayed bounds and the stated strictness. Direct substitution shows the lower equality at $s=0$ and both equalities at $s=1$.

By [F1] the function $\log f$ is convex, and the recurrence with $f(n)=(n-1)!$ makes its secant slopes $s(n-1,n)=\log (n-1)$ and $s(n,n+1)=\log n$. For $0<x<1$, [F3] at $n-1<n<n+x$ gives $s(n-1,n)\le s(n,n+x)$ and [F3] at $n<n+x<n+1$ gives $s(n,n+x)\le s(n,n+1)$; at $x=1$ the middle slope is $s(n,n+1)=\log n$ itself and $\log (n-1)\le \log n$. Either way $\log (n-1)\le (\log f(n+x)-\log f(n))\mathrm{/}x\le \log n$, and exponentiation yields $(n-1{)}^x(n-1)!\le f(n+x)\le n^x(n-1)!$.

Iterating the recurrence gives $f(n+x)=x(x+1)\cdots (x+n-1)f(x)$ and, by induction from [F2], $f(n)=(n-1)!$.

Divide the bounds of step 1.1 by the positive product in step 1.2. Use the upper bound at $n$ and the lower bound with $n$ replaced by $n+1$; both then have the common term $G_n(x)$, and they become $G_n(x)\le f(x)\le ((n+x)\mathrm{/}n)G_n(x)$.

Gamma is positive by its Euler integrand, normalized and recurrent by [F2], and log-convex by [F3]. Thus it satisfies the characterizing properties.

Fix $0<x\le 1$. Apply [F1] to $f$ and to Gamma. Both lie between $G_n(x)$ and $((n+x)\mathrm{/}n)G_n(x)$ for every $n\ge 2$, and the ratio of these bounds tends to $1$. The squeeze theorem therefore gives $f(x)=\mathrm{\Gamma }(x)$.

By [F4], every positive real $y$ is an integer shift of a unique $x\in (0,1]$. Iterating the common recurrence from [F2] and the hypothesis on $f$ extends the equality of step 2.1 from that strip to $y$.

Step 1.1 proves that Gamma has the properties, and steps 2.1 and 3.1 prove that every function with them equals Gamma. This is the claimed characterization.

In the convergent integral [F2], the decreasing substitution $u=1-t$ interchanges $p$ and $q$. By [F1], $B(p,q)=B(q,p)$.

On compact interior truncations use $t={\sin }^2\theta$, with $dt=2\sin \theta \cos \theta d\theta$ and $1-t={\cos }^2\theta$. The transformed integrand is $2{\sin }^{2p-1}\theta {\cos }^{2q-1}\theta$.

Let the truncations tend to $0$ and $\pi \mathrm{/}2$. Convergence from [F2] and [F1] yields the full trigonometric integral and completes the displayed equality.

On compact rectangles inside the first quadrant, [F3] factors the integral as a product of one-variable integrals. Passing through rectangular exhaustion by [F1] gives total improper integral $\mathrm{\Gamma }(p)\mathrm{\Gamma }(q)$.

On compact rectangles inside $(0,\mathrm{\infty })\times (0,1)$ use $x=rt$, $y=r(1-t)$. The map is injective, its absolute Jacobian is $r$, and [F2] transforms the integrand times Jacobian into $r^{p+q-1}{e}^{-r}{t}^{p-1}(1-t{)}^{q-1}$.

A fixed nested family of such compact rectangles maps to a cofinal exhaustion of the first quadrant. By [F1] the limit is independent of this exhaustion, and by [F3] the transformed integral factors as $\mathrm{\Gamma }(p+q)B(p,q)$.

Gamma is positive on the positive axis, so division of the equality in step 2.1 by $\mathrm{\Gamma }(p+q)$ gives the claimed identity.

By [F1] and [F2], $B(p+1,q)=\mathrm{\Gamma }(p+1)\mathrm{\Gamma }(q)\mathrm{/}\mathrm{\Gamma }(p+q+1)=pB(p,q)\mathrm{/}(p+q)$.

Similarly, $B(p,q+1)=qB(p,q)\mathrm{/}(p+q)$.

Adding steps 1.1 and 1.2 and using $p+q>0$ gives $B(p+1,q)+B(p,q+1)=B(p,q)$.

In $\mathrm{\Gamma }(1\mathrm{/}2)={\int }_0^{\mathrm{\infty }}{t}^{-1\mathrm{/}2}{e}^{-t}dt$, use $t=u^2$ on proper truncations. By [F1], the improper limit is $2{\int }_0^{\mathrm{\infty }}{e}^{-u^2}du$.

The integrand is even, so splitting [F2] at zero shows $2{\int }_0^{\mathrm{\infty }}{e}^{-u^2}du=\sqrt{\pi }$.

Combining the two identities gives $\mathrm{\Gamma }(1\mathrm{/}2)=\sqrt{\pi }$, with the positive square root selected because Gamma is positive.

Iterating [F4] gives $\mathrm{\Gamma }(n+1\mathrm{/}2)=\mathrm{\Gamma }(1\mathrm{/}2){\prod }_{k=0}^{n-1}(k+1\mathrm{/}2)=((2n)!\mathrm{/}(4^{n}n!))\mathrm{\Gamma }(1\mathrm{/}2)$, with the empty product valid at $n=0$.

Apply [F1] with $x=n$ and $s=1\mathrm{/}2$. After inversion, $\sqrt{n\mathrm{/}(n+1)}\le \sqrt{n}\mathrm{\Gamma }(n+1\mathrm{/}2)\mathrm{/}n!\le 1$, so this middle sequence tends to $1$.

By step 1.1 and [F3], the middle sequence is $\mathrm{\Gamma }(1\mathrm{/}2)\sqrt{n}\left(\genfrac{}{}{0px}{}{2n}{n}\right)\mathrm{/}{4}^n$. Fact [F2] makes its limit $\mathrm{\Gamma }(1\mathrm{/}2)\mathrm{/}\sqrt{\pi }$, while step 1.2 makes the same limit $1$.

Positivity of Gamma and uniqueness of limits therefore give $\mathrm{\Gamma }(1\mathrm{/}2)=\sqrt{\pi }$.

Put $e_r:=\log r-{\int }_{r-1\mathrm{/}2}^{r+1\mathrm{/}2}\log tdt$. After $t=r+u$, expand $\log (1+u\mathrm{/}r)$ by [F1]. Integration over the symmetric interval cancels the odd powers, and the remaining absolutely convergent even series gives $\mathrm{\mid }{e}_r\mathrm{\mid }\le K\mathrm{/}{r}^2$ for one constant $K$ and all $r\ge 1$.

By step 1.1 and [F2] with $p=2$, the series ${\sum }_{r\ge 1}{e}_r$ converges absolutely.

Summing the definition of $e_r$ from $1$ to $n$ telescopes the integrals to ${\int }_{1\mathrm{/}2}^{n+1\mathrm{/}2}\log tdt$. Fact [F3] gives the primitive $t\log t-t$, and comparison of $n+1\mathrm{/}2$ with $n$ shows that $\log (n!)-((n+1\mathrm{/}2)\log n-n)$ converges to a real constant $c$.

Exponentiating step 3.1 and putting $C=e^c>0$ gives $n!\mathrm{/}(\sqrt{n}(n\mathrm{/}e{)}^n)\to C$, which is the stated asymptotic.