[ANT] The Gamma Function

짤방

STEIN&SHAKARCHI의 Complex Analysis, Chapter 6의 연습문제 중
 
$$|\Gamma(1/2+it)|=\sqrt{\frac{2\pi}{e^{\pi t}+e^{-\pi t}}}$$
 
을 증명하는 문제가 있다. 이 문제를 해결하기 위해
 
$$\Gamma(\overline{z})=\overline{\Gamma(z)}$$
 
를 증명한다. 그리고 그 전에 감마함수의 해석적 연속에 대해 알아본다.
 

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For real $s>0$, the gamma function is defined by

$$\Gamma(s)=\int_0^\infty e^{-t}t^{s-1}dt$$

The gamma function, $\Gamma(s)$ is well-defined for all $s>0$ since for sufficiently chosen $\epsilon>0$, we can decompose the integral by

$$\Gamma(s)=\int_0^\infty e^{-t}t^{s-1}dt=\int_0^\epsilon e^{-t}t^{s-1}dt+\int_\epsilon^\infty e^{-t}t^{s-1}dt$$

For the first integral, from $0$ to $\epsilon$, the integrand is dominated by $t^{s-1}$ and since $t^{s-1}$ is integrable hence the integral converges. For the second integral, from $\epsilon$ to $\infty$, the integrand is dominated by the exponential decay $e^{-t}$ and hence converges. From this observation, we may extend the domain of the gamma function to the half plane $\text{Re}(s)>0$.

Theorem 1.  The gamma function $\Gamma(s)$ extends to an analytic function in the half plane $\text{Re}(s)>0$, and is still given there by the integral formula
 
$$\Gamma(s)=\int_0^\infty e^{-t}t^{s-1}dt$$

Before we prove Theorem 1. we need two theorems introduced in the Chapter 2 of [STEIN].

Theorem 5.2. If $\{f_n\}$ is a sequence of holomorphic function that converges locally uniformly to a function $f$, then $f$ is a holomorphic function.

proof. Suppose $f_n$ is a holomorphic function in $\Omega$. For any disc $D\subset\Omega$ whose closure is contained in $\Omega$ we have triangle $T\subset D$ and by Goursat, we have
 
$$\int_T f_n(z)dz=0$$

for all $n$. Since $\{f_n\}$ converges uniformly to $f$ in $D$, we have

$$\lim_{n\to\infty}\left(\int_T f_n(z)dz\right)=\int_T\left(\lim_{n\to\infty} f_n(z)dz\right)=\int_T f(z)dz=0$$

Now by Morera's theorem, $f$ is holomorphic in any disc $D\subset\Omega$. Then from Weierstrass, $f$ is holomorphic in all of $\Omega$. 증명 끝.

Theorem 5.4. Let $F(z, s)$ be defined for $(z, s)\in \Omega\times[a, b]$. Suppose $F(z, s)$ is holomorphic in $z$ for every fixed $s$ and $F$ is continuous on $\Omega\times[a, b]$. Then,

$$f(z)=\int_a^b F(z, s) ds$$
    
is holomorphic for every $z\in \Omega$.

proof. Likewise, take any disc $D\subset\Omega$ whose closure is contained in $\Omega$. Then for any triangle $T\subset D$, it suffices to prove

$$\int_T f(z)dz=\int_T\int_a^b F(z, s)dsdz=0$$

Since $F(z,s)$ is continuous on $\Omega\times[a,b]$ we may apply Fubini's theorem to get

$$\int_T\int_a^b F(z, s)dsdz=\int_a^b\int_T F(z, s)dzds$$

But since $F(z, s)$ is holomorphic for $z\in\Omega$ for every fixed $s$,

$$\int_T F(z, s)dz=0$$

by Goursat and thus we have

$$\int_T\int_a^b F(z, s)dsdz=\int_a^b\int_T F(z, s)dzds=0$$

Then by Morera, $f(z)$ is holomorphic in any disc $D\subset\Omega$ and hence is holomorphic in all of $\Omega$. 증명 끝.

Now we can prove Theorem 1.

proof. By Weierstrass theorem, we know it suffices to prove that the integral defines a holomorphic function in every strip

$$S_{\delta, M}=\{s\in\mathbb{C} : \delta<\text{Re}(s)<M\}$$

where $0<\delta<M$. Let $s=\sigma+i\tau$ then

$$|e^{-t} t^{s-1}|=e^{-t}|e^{(s-1)\log t}|=e^{-t}|e^{(\sigma-1)\log t}|\cdot|e^{i\tau \log t}|=e^{-t}t^{\sigma-1}$$

so we have

$$|\Gamma(s)|=\Bigg|\int_0^\infty e^{-t}t^{s-1} dt\Bigg|\le\int_0^\infty |e^{-t}t^{s-1}|dt=\int_0^\infty e^{-t}t^{\sigma-1} dt$$

This implies that the integral formula for $\Gamma(s)$ is well defined for $\text{Re}(s)>0$. Now we need to prove the holomorphicity of $\Gamma(s)$ in the given half plane. Define

$$F_\epsilon (s)=\int_\epsilon^{1/\epsilon} e^{-t}t^{s-1} dt$$

Then $F_\epsilon(s)$ is holomorphic in the strip $S_{\delta, M}$ by Theorem 5.4. Then by Theorem 5.2. it suffices to show that $F_\epsilon$ converges uniformly to $\Gamma$ on the strip $S_{\delta, M}$ then the theorem is proved. Observe that

$$|\Gamma(s)-F_\epsilon(s)|=\Bigg|\int_0^\epsilon e^{-t}t^{s-1}dt+\int_{1/\epsilon}^\infty e^{-t}t^{s-1}dt\Bigg| \le\int_0^\epsilon e^{-t}t^{\sigma-1}dt+\int_{1/\epsilon}^\infty e^{-t}t^{\sigma-1}dt$$

Note that $g_\epsilon$ converges uniformly to $g$ on $\Omega$ as $\epsilon\to 0$ is that we have $\epsilon_0>0$ where if $\epsilon<\epsilon_0$ then $|g(s)-g_\epsilon(s)|<\delta$ for all $s\in\Omega$ and $\delta>0$. This is equivalent to $|g(s)-g_\epsilon(s)|\to 0$ as $\epsilon\to 0$ for all $s\in\Omega$. From this point of view, it is clear that the first integral

$$\int_0^\epsilon e^{-t}t^{\sigma-1}dt$$

goes $0$ as $\epsilon\to 0$ for all $\delta<\sigma=\text{Re}(s)<M$ since for small $\epsilon>0$, the integrand is dominated by $t^{\sigma-1}$ and hence the integral can be majorized by $\epsilon^\sigma/\sigma$ which goes $0$ as $\epsilon\to 0$. For the second integral

$$\int_{1/\epsilon}^\infty e^{-t}t^{\sigma-1}dt$$

as $\epsilon\to 0$, we have

$$\int_{1/\epsilon}^\infty e^{-t}t^{\sigma-1}dt\le \int_{1/\epsilon}^\infty e^{-t}t^{M-1}dt\le \int_{1/\epsilon}^\infty e^{-t/2}dt$$

thus the integral goes $0$ as $\epsilon\to 0$. This proves that $F_\epsilon(s)$ converges uniformly to $\Gamma(s)$ on every strip $S_{\delta, M}$, as to be desired. 증명 끝.
 
Proposition 1. If $\text{Re}(s)>0$, $\Gamma(\overline{s})=\overline{\Gamma(s)}$.

proof. We know that for $\text{Re}(s)>0$, $\Gamma(s)$ is defined by

$$\Gamma(s)=\int_0^\infty e^{-t}t^{s-1}dt$$

But we have

$$\overline{\int_0^\infty e^{-t}t^{s-1}dt}=\int_0^\infty\overline{e^{-t}t^{s-1}}dt=\int_0^\infty e^{-t}\overline{t^{s-1}}dt$$

Here,

$$\overline{t^{s-1}}=\overline{e^{(s-1)\log t}}=\overline{e^{(\sigma-1)\log t}}\cdot\overline{e^{i\tau \log t}}=e^{(\sigma-1)\log t}\cdot e^{-i\tau \log t}=e^{(\overline{s}-1)\log t}=t^{\overline{s}-1}$$
    
so we have

$$\overline{\int_0^\infty e^{-t}t^{s-1}dt}=\int_0^\infty e^{-t}t^{\overline{s}-1}dt$$

and this implies $\Gamma(\overline{s})=\overline{\Gamma(s)}$, as to be desired. 증명 끝.
 

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다음으로
 
$$\Gamma(s+1)=s\Gamma(s)$$
 
를 보이면 $\Gamma(s)$의 복소평면 전체로의 해석적 연속을 얻을 수 있고,
 
$$\Gamma(s)\Gamma(1-s)=\frac{\pi}{\sin\pi s}$$
 
임을 증명할 수 있다. 이를 이용하면 $1/\Gamma$가 복소평면 전체에서 해석적이고 $s=0, -1, -2, \cdots$에서 영점을 가짐을 알 수 있다. 이후 $1/\Gamma$의 growth order가 2보다 작음을 보일 수 있고 그러면
https://whitemask.tistory.com/239

[복소해석학] 5.5. Hadamard's factorization theorem

Hadamard's factorization theorem에 의해 $1/\Gamma$, 따라서 $\Gamma$의 복소평면 전체로의 해석적 연속을
 
$$\frac{1}{\Gamma}=e^{\gamma s} s\prod_{n=1}^\infty \left(1+\frac{s}{n}\right)e^{-\frac{s}{n}}$$
 
으로 얻을 수 있음을 보일 수 있다. 여기서 $\gamma$는 물론 Euler-Mascheroni 상수이다. 이제 해석적 연속의 유일성에 의해 음의 정수가 아닌 임의의 복소수 $s$에 대해 $\Gamma(\overline{s})=\overline{\Gamma(s)}$이 성립함을 알 수 있다.