伽玛函数的积分定义、在半整数的值$Γ(n/2)$

hbghlyj

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Extending $n!$

While factorials are usually used with integers, they can be generalized for complex values (except negative integers) and expressed by an integral: $$ n!=\Gamma (n+1)=\int _{ 0 }^{ \infty }{ \frac { { x }^{ n } }{ { e }^{ x } } } dx, $$ where $\Gamma(n)$ is defined as $(n-1)!$. This is also written as $\displaystyle z!=\Gamma (z+1)=\int _{ 0 }^{ \infty }{ { t }^{ z }{ e }^{ -t } } dt$.

Proof

A very nice derivation of this formula begins with the integral $$\int { \frac { 1 }{ { e }^{ kx } } } dx=-\frac { { e }^{ -kx } }{ k } +C,$$ and hence $$\int_{0}^{\infty} { \frac { 1 }{ { e }^{ kx } } } dx=\frac{1}{k}. $$ Then differentiate $n$ times with respect to $k$ to obtain $$ \int_{0}^{\infty} { \frac { {(-x)}^{n} }{ { e }^{ kx } } } dx=\frac{{(-1)}^{n}n!}{{k}^{n+1}}. $$ Observe that ${(-1)}^{n}$ can be factored out from both sides, and then taking $k=1,$ we arrive at $$ n!=\int _{ 0 }^{ \infty }{ \frac { { x }^{ n } }{ { e }^{ x } } } dx.\ _\square$$

From this integral definition, some nice properties of $\Gamma\left(\frac{n}{2}\right)$ can be derived.

First, we need to compute $\left(-\frac{1}{2}\right)!=\Gamma\left(\frac{1}{2}\right)$.

Proof

This is found through the integral$$\Gamma\left(\frac{1}{2}\right)=\intop_{t=0}^{+\infty}t^{\frac{1}{2}-1}e^{-t}dt=\intop_{t=0}^{+\infty}\frac{e^{-t}}{\sqrt{t}}dt,$$ and with $y=\sqrt{t}$, $dy=\frac{dt}{2\sqrt{t}}$, we get$$\Gamma\left(\frac{1}{2}\right)=2\intop_{y=0}^{+\infty}e^{-y^{2}}dy=\intop_{y=-\infty}^{+\infty}e^{-y^{2}}dy=\sqrt{\pi}.$$

Using this combined with $\Gamma(n+1)=n\Gamma(n),$ you can arrive at $$\left(n-\frac{1}{2}\right)!=\sqrt{\pi}\left(\frac{1}{2}\right)\left(\frac{3}{2}\right)...\left(n-\frac{1}{2}\right)=\sqrt{\pi}\prod _{ j=1 }^{ n }\left( j-\frac { 1 }{ 2 } \right). $$ All that is left is to simplify $\prod _{ j=1 }^{ n }\left( j-\frac { 1 }{ 2 } \right)$. This is easier once you express each term as a fraction: $$\prod _{ j=1 }^n \left( j-\frac { 1 }{ 2 } \right)= \prod _{ j=1 }^{ n }{ \frac { 2j-1 }{ 2 } } = \frac{(2n)!}{{2}^{n}\prod _{ j=1 }^{ n }{ 2j } } = \frac{(2n)!}{{4}^{n}n!}. $$ Hence, the final result for $\left(n-\dfrac{1}{2}\right)!$ is $\sqrt{\pi}\dfrac{(2n)!}{{4}^{n}n!} $.

hbghlyj

由$$(n-1)!<\sqrt{\pi}\dfrac{(2n)!}{{4}^{n}n!}<n!$$得到
$$\frac1n<\sqrt{\pi}\dfrac{(2n)!}{{4}^{n}(n!)^2}<1$$即
$$\frac{4^n}{n\sqrt{\pi}}<\binom{2n}n<\frac{4^n}{\sqrt{\pi}}$$而由Stirling公式得$\binom{2n}{n}\sim{4^n\over\sqrt{nπ}}$

Czhang271828

Indeed, one should be familiar with $\Gamma(s)\Gamma(1-s)=\dfrac{\pi}{\sin \pi s}$.

The proof is simple. One might notice that

\[
\begin{align}
\Gamma(s)\Gamma(1-s)&=\int_0^\infty e^{-t}t^{s-1}\mathrm dt\int_0^\infty e^{-u}u^{-s}\mathrm du\\
&=\int_0^\infty e^{-t}t^{s-1}\mathrm dt\int_0^\infty e^{-tv}(tv)^{-s}\mathrm d(tv)\\
&=\int_0^\infty \int_0^\infty e^{-t(1+v)}v^{-s}\mathrm dt\mathrm dv\\
&=\int_0^\infty \dfrac{v^{-s}}{1+v}\mathrm dv.
\end{align}
\]

In order to determine the integral

\[
{\int_0^\infty\dfrac{x^{p-1}}{(1+x)^m}\mathrm dx}=\dfrac{\pi}{\sin p\pi}\cdot \prod_{j=1}^{m-1}(1-p/j),
\]
with $m\in\mathbb Z_{\geq 1}$ and $p\in (0,m)\setminus  \mathbb Z$. We shall consider the multi-valued function $f(z)=\dfrac{e^{(p-1)\log z}}{(1+z)^m}$, which can be locally well-defined in the following region.

One observes that
1. $\int_{\gamma_R}f\to 0$ as $R\to\infty$,
2. $\int_{\gamma_\rho}f\to 0$ as $\rho\to 0$,
3. the residue on $z=-1$ is $\frac{1}{(m-1)!}\prod_{j=1}^{m-1} (p-j)$,
4. $-e^{2 p\pi i}\int_{\gamma_+} f=\int_{\gamma_-}f$ as $\|\gamma_-- \gamma_+\|\to 0$.

The residue formulae shows that
\[
\int_0^\infty\dfrac{x^{p-1}}{(1+x)^m}\mathrm dx =\dfrac{2\pi i}{1-e^{2 p\pi i}}\cdot \dfrac{1}{(m-1)!}\prod_{j=1}^{m-1} (p-j).
\]

As a corollary, $\Gamma(s)\Gamma(1-s)=\dfrac{\pi}{\sin \pi s}$. $\Gamma(1/2)=?$ follows directly.