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\begin{center}

{\bf The Gamma Function}

\end{center}

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{\bf I.  Integral Definition}

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

This is convergent since $t^me^{-t}<t^{m-n}n!$ thus proving
convergence at the upper limit and also at the lower limit if
$z>0$ since $\ds \lim_{\epsilon\to0}\int_\epsilon t^{z-1}dt$
exists.

For $z$ complex the convergence holds for $R(z)>0$.

We have

\begin{itemize}
\item[i)]
$\ds\Gamma(1)=\int_0^\infty e^{-t}dt=1$

\item[ii)]
$z\Gamma(z)=\Gamma(z+1)$

$\ds\int_0^\infty
t^{z-1}e^{-t}dt=\left[\frac{t^z}{z}e^{-t}\right]_0^\infty+
\int_0^\infty\frac{t^z}{z}e^{-t}dt$

$\ds\Gamma(z)=0+\frac{1}{z}\Gamma(z+1)$

\item[iii)]
$\Gamma(n+1)=n! \hspace{0.5in} (n\geq0$, an integer)
\end{itemize}

${}$

{\bf II.  Alternative Integral Definition}

Substitute $t=u^2$, this gives $\ds\Gamma(z)=2\int_0^\infty
u^{2z-1}e^{-u^2}du$

${}$

{\bf III.  Limit Definition (Euler)}

$\ds\Gamma(z)=\lim_{n\to\infty}\frac{n!n^z}{z(z+1)\cdots(z+n)}$

This holds for all $z$ except $0,-1,-2,\cdots$

We can derive III from I in two ways

${}$
\begin{itemize}
\item[a)]
$\ds e^{-t}=\lim_{n\to\infty}\left(1-\frac{t}{n}\right)^n$

${}$

Define $\ds\Gamma_n(z)=\int_0^n
t^{z-1}\left(1-\frac{t}{n}\right)^n dt$

$\ds\int_0^n t^{z-1}e^{-t}dt-\Gamma_n(z)=\int_0^n
t^{z-1}\left\{e^{-t}-\left(1-\frac{t}{n}\right)^n\right\}dt$

$\ds\int_0^n
t^{z-1}e^{-t}\left\{1-e^t\left(1-\frac{t}{n}\right)^n\right\}dt$

\newpage

We have

\begin{itemize}
\item[i)]
$e^t\geq t+1 \hspace{0.5in}$ for all $t$

\item[ii)]
$e^{-t}\geq 1-t \hspace{0.4in}$ for all $t$
\end{itemize}

From ii)  $1\geq e^t(1-t) \hspace{0.2in}$ for all $t$

From i) multiplying by $(1-t), \,\,\, t\leq1$

$e^t(1-t)\geq(1-t^2)$

Therefore $1\geq e^t(1-t)\geq 1-t^2 \hspace{0.2in}$ for $t\leq1$

Taking the nth power $(0\leq t\leq1)$

$1\geq e^{nt}(1-t)^n\geq(1-t^2)^n$

Replace $t$ by $\frac{t}{n}$

$\ds 1\geq
e^t\left(1-\frac{t}{n}\right)^n\geq\left(1-\frac{t^2}{n^2}\right)^n
\hspace{0.3in} 0\leq t\leq n$

Hence $\ds 1-\left(1-\frac{t^2}{n^2}\right)^n\geq
1-e^t\left(1-\frac{t}{n}\right)^n\geq0$

For $0\leq x\leq1 \hspace{0.5in} 0\leq1-x^n\leq n(1-x)$

Therefore $\ds
\frac{t^2}{n}\geq1-e^t\left(1-\frac{t}{n}\right)^n\geq0$

Therefore $\ds\left|\int_0^n
t^{z-1}e^{-t}\left(1-e^t\left(1-\frac{t}{n}\right)^n\right)dt\right|$

$\ds
\leq\int_0^n|t^{z-1}|e^{-t}\left|1-e^t\left(1-\frac{t}{n}\right)^n\right|dt$

$\ds \leq\int_0^n|t^{z-1}|e^{-t}\frac{t^2}{n}dt$

${}$

$|t^{z-1}|=e^{(x-1)\log t}=t^{x-1}=t^{Re(z)-1}$

Therefore $\ds\left|\int_0^n
t^{z-1}e^{-t}\left(1-e^t\left(1-\frac{t}{n}\right)^n\right)dt\right|\leq
\frac{1}{n}\int_0^nt^{Re(z)+1}e^{-t}dt$

$\ds\leq\frac{1}{n}\int_0^\infty
t^{Re(z+1)}e^{-t}dt=\frac{\Gamma(Re(z+2))}{n}=\frac{const}{n}$

Hence $\ds\lim_{n\to\infty}\int_0^n
t^{z-1}e^{-t}\left(1-e^t\left(1-\frac{t}{n}\right)^n\right)dt=0$

Hence $\ds\lim_{n\to\infty}\left\{\int_0^n
t^{z-1}e^{-t}dt-\Gamma_n(z)\right\}=0$

Therefore $\ds\Gamma(z)=\lim_{n\to\infty}\Gamma_n(z)$

\begin{eqnarray*}\Gamma_n(z)&=&\int_0^n
t^{z-1}\left(1-\frac{t}{n}\right)^ndt\\ &=&n^z\int_0^1
s^{z-1}(1-s)^nds\\
&=&n^z\frac{n}{z}\frac{n-1}{z+1}\cdots\frac{1}{z+n}.1\end{eqnarray*}

Therefore
$\ds\Gamma(z)=\lim_{n\to\infty}\frac{n!n^z}{z(z+1)\cdots(z+n)}$

${}$

\item[b)]
Define $f_n(t)=\left\{\begin{array}{cr} (1-\frac{t}{n})^nt^{z-1} &
0<t<n\\ 0 & t\geq n\end{array}\right.$

then $0\leq|f_n(t)|\leq |e^{-t}t^{z-1}|$ for every $n$, and
$f_n(t)\to e^{-t}t^{z-1}$ as $n\to\infty$ for every $t$, also
$\ds\exists\int_0^\infty e^{-t}t^{z-1}dt<\infty$

Hence by Lebesgue's theorem on dominated cgce.

$\ds\int_0^\infty\lim_{n\to\infty}f_n(t)=\lim_{n\to\infty}\int_0^\infty
f_n(t)dt$

i.e. $\ds\int_0^\infty e^{-t}t^{z-1}dt=\lim_{n\to\infty}\int_o^n
f_n(t)dt$

and the result follows as before.
\end{itemize}


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{\bf IV.  Infinite Product Definition (Weierstrass)}

${}$

Infinite Products

We have a sequence $1+a_1,\, 1+a_2,\, \cdots$ none of which are
zero.  We form the product defined by

$\prod_m=(1+a_1)(1+a_2)\cdots(1+a_m)$

If $\prod_m$ tends to a limit other than zero as $m\to\infty$ then
the infinite product $(1+a_1)(1+a_2)\cdots$ is said to converge
and is written $\ds\prod_{n=1}^{\infty}(1+a_n)$.

A necessary condition for convergence is $\ds
\lim{n\to\infty}a_n=0$, for $\ds
1+a_n=\frac{\prod_n}{\prod_{n-1}}$ and we have $\ds
\lim{n\to\infty}\prod_n=\lim_{n\to\infty}\prod_{n-1}\not=0$

A sufficient condition for convergence is that the series
$\ds\sum_{n=1}^\infty\log(1+a_n)$ is convergent.

(Here we take the principal value of the log.

i.e. such that $-\pi<\arg(1+a_n)<\pi$ and $\log(1+a_n)\to0$ as
$n\to\infty$ and $a_n\to0$.)

for $\ds\prod_m=\exp\left\{\log\prod_1^m(1+a_n)\right\}$

$\ds\hspace{0.5in}=\exp\left\{\sum_{n+1}^m\log(1+a_n)\right\}$

Hence, since the exponential function is continuous, $\lim\exp
s_m=\exp\lim s_m$, and the result follows when we take $\ds
s_m=\sum_{n=1}^m\log(1+a_n)$.

If $\sum\log(1+a_n)$ is absolutely convergent then $\prod(1+a_n)$
is said to be absolutely convergent.

${}$

Theorem

A necessary and sufficient condition for absolute convergence is
that the series $\sum a_n$ is absolutely convergent.

Proof

Since $\ds\lim_{n\to\infty}a_n=0$ we can find $m$ where
$|a_n|<\frac{1}{2}$ for $n\geq m$.

Then
\begin{eqnarray*}\log(1+a_n)&=&a_n-\frac{a_n^2}{2}+\frac{a_n^3}{3}-\cdots\\
\frac{\log(1+a_n)}{a_n}-1&=&-\frac{a_n}{2}+\frac{a_n^2}{3}-\cdots\\
\left|\frac{\log(1+a_n)}{a_n}-1\right|&\leq&
\frac{|a_n|}{2}+\frac{|a_n|^2}{3}+\cdots\\
&\leq&\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+\cdots
\hspace{0.5in} n\geq m\\ &\leq&\frac{1}{2}\end{eqnarray*}

Therefore
$\ds\frac{1}{2}\leq\left|\frac{\log(1+a_n)}{a_n}\right|\leq\frac{3}{2}$

Therefore
$\ds\frac{1}{2}|a_n|\leq|\log(1+a_n)|\leq\frac{3}{2}|a_n|
\hspace{0.5in} n\geq m$

Hence by the comparison test $\sum|\log(1+a_n)|$ converges or
diverges as $\sum|a_n|$ converges or diverges.

N.B.  If a finite number of factors $(1+a_1),\cdots$ vanish and if
the product omitting these factors is convergent the product is
said to converge to zero.  If no factor vanishes but
$\ds\lim_{m\to\infty}\prod_m=0$ then the product is said to
diverge to zero.

${}$

Returning to the Gamma-Function we have:

$\ds\frac{1}{\Gamma(z)}=\lim_{n\to\infty}zn^z\left(1+\frac{z}{1}\right)
\left(1+\frac{z}{2}\right)\cdots\left(1+\frac{z}{n}\right)$

The product $\ds\prod\left(1+\frac{z}{n}\right)$ is divergent
$(z\not=0)$ for

$\ds\log\left(1+\frac{z}{n}\right)=\frac{z}{n}+O\left(\frac{z^2}{n^2}\right)
\hspace{0.5in} \left\{\frac{|z|}{n}<1\right\}$

and the series $\sum\frac{1}{n}$ is divergent and $\sum
O(\frac{1}{n^2})$ is convergent.

Hence $\sum\log(1+a_n)$ is divergent $(z\not=0)$.

Now $\ds\prod_{n=1}^m\left(1+\frac{z}{n}\right)=
\prod_{n=1}^m\left\{\left(1+\frac{z}{n}\right)e^{-\frac{z}{n}}\right\}
e^{z\left(1+\frac{1}{2}+\cdots+\frac{1}{m}\right)}$

Also $\ds\log\left(1+\frac{z}{n}\right)e^{-\frac{z}{n}}=
O\left(\frac{z^2}{n^2}\right)$

Hence the product
$\ds\prod_{n=1}^\infty\left(1+\frac{z}{n}\right)e^{-\frac{z}{n}}$
is convergent.

Therefore $\ds\frac{1}{\Gamma(z)}=
z\lim_{m\to\infty}m^{-z}e^{z\left(1+\frac{1}{2}+\cdots+\frac{1}{m}\right)}
\prod_{n=1}^m\left(1+\frac{z}{n}\right)e^{-\frac{z}{n}}$

$\ds\hspace{1.15in}=
z\lim_{m\to\infty}e^{z\left(1+\frac{1}{2}+\cdots+\frac{1}{m}-\log
m\right)}\prod_{n=1}^m\left(1+\frac{z}{n}\right)e^{-\frac{z}{n}}$

Now
$\ds\lim_{m\to\infty}\left(1+\frac{1}{2}+\cdots+\frac{1}{m}-log
m\right)=\gamma \hspace{0.2in}$ (Euler's constant)

Therefore $\ds\frac{1}{\Gamma(z)}=ze^{\gamma
z}\prod_{n=1}^\infty\left(1+\frac{z}{n}\right)e^{-\frac{z}{n}}$

This is Weierstrass' definition.

\newpage

{\bf Euler's Constant} - Proof that $\gamma$ exists.

$\ds 1+\frac{1}{2}+\cdots+\frac{1}{m}-\log(m+1)=
\sum_{n=1}^m\frac{1}{n}-\sum_{n=1}^m\log\frac{n+1}{n}$

$\ds=\sum_{n=1}^m u_n$ where $u_n=\frac{1}{n}-\log\frac{n+1}{n}$

$\ds=\frac{1}{n}\int_0^1dt-\int_0^1\frac{dt}{t+n}=
\int_0^1\left(\frac{1}{n}-\frac{1}{t+n}\right)dt$

$\ds=\int_0^1\frac{tdt}{n(t+n)}\leq\int_0^1\frac{t}{n^2}dt=
\frac{1}{2n^2}$

Therefore $\sum u_n$ is convergent by comparison with
$\sum\frac{1}{2n^2}$

Therefore
$\ds\lim_{m\to\infty}\left[1+\frac{1}{2}+\cdots+\frac{1}{m}-\log(m+1)\right]=
\sum_{n=1}^\infty u_n$

Now $\ds 1+\frac{1}{2}+\cdots+\frac{1}{m}-\log m=
\left[1+\frac{1}{2}\cdots\frac{1}{m}-\log(m+1)\right]+
\log\frac{m+1}{m}$

and $\ds \lim_{m\to\infty}\log\frac{m+1}{m}=\log 1=0$.  Therefore
$\ds\gamma=\sum_{n=1}^\infty u_n$.

${}$

{\bf Properties of $\Gamma(z)$} (Weierstrass Form)

\begin{itemize}
\item[i)]
The RHS is convergent for all $z<\infty$.  Hence $\Gamma(z)$ has
no zeros.

\item[ii)]
The RHS has simple zeros at $z=0,-1,\cdots$.  Hence $\Gamma(z)$
has simple poles at these points.

\item[iii)]
$z\Gamma(z)=\Gamma(z+1)$

\item[iv)]
$\ds\Gamma(z)\Gamma(1-z)=\pi\csc\pi z\hspace{0.5in}
\left(\frac{\sin
z}{z}=\prod\left(1-\frac{z^2}{n^2\pi^2}\right)\right)$

\item[v)]
$2^{2z-1}\Gamma(z)\Gamma(z+\frac{1}{2})=
\Gamma(\frac{1}{2})\Gamma(2z) \hspace{0.5in}$ duplication formula

\item[vi)]
$\Gamma(\frac{1}{2})=\pi^{\frac{1}{2}} \hspace{0.5in}$ (same as
iv) but with $z=\frac{1}{2}$)
\end{itemize}

${}$

{\bf Behaviour of $\Gamma(x)$ for real $x$}

$\Gamma(x)>0$ for $x>0 \hspace{0.5in} \Gamma(n)=(n-1)!$

$z(z+1)\cdots(z+n)\Gamma(z)=\Gamma(n+1+z)$

Therefore
$\ds(z+n)\Gamma(z)=\frac{\Gamma(n+1+z)}{z(z+1)\cdots(z+n+1)}$

$\ds\lim_{z\to-n}(z+n)\Gamma(z)=
\frac{\Gamma(1)}{-n(-n+1)\cdots(-1)}= \frac{(-1)^n}{n!}$

This is the residue at the simple pole $z=-n$.


DIAGRAM


${}$

${}$

{\bf The Beta Function}

$\ds B(m,n)=\int_0^1 u^{m-1}(1-u)^{n-1}du \hspace{0.5in} (m,n>0)$

We shall show that $\ds
B(m,n)=\frac{\Gamma(m)\Gamma(n)}{\Gamma(m+n)}$

We note that $B(m,n)=B(n,m)$ (putting $u=1-v$).

Also putting $u=\cos^2\theta$

$\ds
B(m,n)=2\int_0^\frac{\pi}{2}\cos^{2m-1}\theta\sin^{2n-1}\theta
d\theta$

Define $\ds\Gamma(m;R)=2\int_0^R x^{2m-1}e^{-x^2}dx$

then $\ds\lim_{R\to\infty}\Gamma(m;R)=\Gamma(m)$

$\ds\Gamma(m;R)\Gamma(n;R)=4\int_0^R x^{2m-1}e^{-x^2}dx\int_0^R
y^{2n-1}e^{-y^2}dy$

Assume for the moment that $m,n\geq\frac{1}{2}$.  Then
$x^{2m-1}e^{-x^2}y^{2n-1}e^{-y^2}$ is a continuous function of $x$
and $y$ in $x\geq0, \,\,\, y\geq0$.

Hence $\ds 4\int\int x^{2m-1}e^{-x^2}y^{2n-1}e^{-y^2}dxdy=
4\int_0^R x^{2m-1}e^{-x^2}dx\int_0^R y^{2n-1}e^{-y^2}dy$

$\hspace{0.5in} _{{\rm \ square}}$

$\hspace{0.5in} _{0\leq x\leq R}$

$\hspace{0.5in} _{0\leq y\leq R}$

${}$

DIAGRAM

${}$

$\ds 4\int\int=4\int\int+4\int\int$

$_{{\rm \ square}} \hspace{0.2in} _{{\rm quadrant}} \hspace{0.3in}
_{\sum}$

$\hspace{0.6in} _{0\leq\theta\leq\frac{\pi}{2}}$

$\hspace{0.6in} _{0\leq r\leq R}$

$\ds 4\int\int=4\int\int x^{2m-1}y^{2n-1}e^{-r^2}rdrd\theta$

$_{{\rm quadrant}}$

$\ds \hspace{0.3in} =4\int_0^R r^{2m+2n-1}e^{-r^2}dr
\int_0^\frac{\pi}{2}\cos^{2m-1}\theta\sin^{2n-1}\theta d\theta$

$\hspace{0.3in} =\Gamma(m+n;R)B(m,n)$

Therefore
$\ds\Gamma(m;R)\Gamma(n;R)=\Gamma(m+n;R)B(m,n)+4\int\int$

$\hspace{3.85in} _{\sum}$

$\ds\left|4\int\int\right|\leq4\int\int
x^{2m-1}y^{2n-1}e^{-x^2-y^2}dxdy \hspace{0.3in} (m,n$ are real)

$\hspace{0.2in} _{\sum} \hspace{0.55in} _{\sum}$

${}$

$\ds\hspace{0.6in} \leq 4\int\int
r^{2m+2n-1}e^{-r^2}\cos^{2m-1}\theta\sin^{2n-1}\theta drd\theta$

$\hspace{0.8in} _{0\leq\theta\leq\frac{\pi}{2}}$

$\hspace{0.8in} _{R\leq r\leq R\sqrt2}$

$\ds\hspace{0.6in} =4\int_R^{R\sqrt2}r^{2m+2n-1}e^{-r^2}dr
\int_0^\frac{\pi}{2}\cos^{2m-1}\theta\sin^{2n-1}\theta d\theta$

$\ds\hspace{0.6in} =[\Gamma(m+n;R\sqrt2)-\Gamma(m+n;R)]B(m,n)$

Therefore $\ds\lim_R\to\infty\int\int=0$

$\hspace{1.45in} _{\sum}$

Therefore as $R\to\infty\hspace{0.5in}
\Gamma(m)\Gamma(n)=\Gamma(m+n)B(m,n)$

This proof holds for $m,n\geq \frac{1}{2}$.

${}$

Extension to $m,n>0$

We have, for $m>0$, $\ds\Gamma(m)=\int_0^\infty t^{m-1}e^{-t}dt$
where the integral exists as an improper integral when $m>0$.

Also $m\Gamma(m)=\Gamma(m+1) \hspace{0.3in} (m>0)$

$\ds B(m,n)=
\lim_{\alpha\to0+,\,\beta\to0-}\int_\alpha^{1-\beta}u^{m-1}(1-u)^{n-1}du$

$\ds B(m,n+1)=
\lim_{\alpha\to0+,\,\beta\to0-}\int_\alpha^{1-\beta}u^{m-1}(1-u)^ndu$

$\ds\hspace{0.9in}=\lim_{\alpha,\beta\to0}
\left\{\left[\frac{u^m}{m}(1-u)^n\right]_\alpha^{1-\beta}+
\frac{n}{m}\int_\alpha^{1-\beta}u^m(1-u)^{n-1}du\right\}$

$\ds\hspace{0.9in}=\lim_{\alpha,\beta\to0}
\left[\frac{-\alpha^m(1-\alpha)^n+(1-\beta)^m\beta^n}{m}\right]
+\frac{n}{m}B(m+1,n)$

Therefore $\ds\frac{B(m,n+1)}{n}=\frac{B(m+1,n)}{m} \hspace{1in}
(i)$

$\ds\int_0^1u^{m-1}(1-u)^{n-1}=\int_0^1u^{m-1}(1-u)^{n-1}\{u+(1-u)\}du$

Therefore $B(m,n)=B(m+1,n)+B(m,n+1) \hspace{0.3in} (ii)$

${}$

Hence from $(i)$ and $(ii)$

$\ds\frac{B(m,n+1)}{n}=\frac{B(m+1,n)}{m}=
\frac{B(m+1,n)+B(m,n+1)}{m+n}=\frac{B(m,n)}{m+n}$

Therefore $\ds B(m+1,n)=\frac{m}{m+n}B(m,n)$

$\ds B(m,n+1)=\frac{n}{m+n}B(m,n)$

Therefore

$\ds
B(m+1,n+1)=\frac{n}{m+n+1}B(m+1,n)=\frac{mn}{(m+n)(m+n+1)}B(m,n)$

Therefore $\ds
B(m,n)=\frac{(m+n)(m+n+1)}{mn}\frac{\Gamma(m+1)\Gamma(n+1)}{\Gamma(m+n+2)}$

$\ds \hspace{1.4in} =\frac{\Gamma(m)\Gamma(n)}{\Gamma(m+n)}
\hspace{0.5in} (m,n>0)$

${}$

Example on Fubini's Theorem

$\ds\Gamma(x)\Gamma(y)=\int_0^\infty e^{-t}t^{x-1}dt\int_0^\infty
e^{-u}u^{y-1}du$

$\ds=\int_0^\infty t^{x-1}dt\int_0^\infty e^{-(t+u)}u^{y-1}du$

(Using Fubini's theorem and regarding the integral as a multiple
integral.)

$\ds\int_0^\infty t^{x-1}dt\int_t^\infty e^{-u}(u-t)^{y-1}du$

$\ds=\int_0^\infty t^{x-1}dt\int_0^\infty
e^{-u}(u-t)^{y-1}X(ut)dt$

(Note $X(ut)=1$ if $u>t$, but $X(ut)=0$ if $u\leq t$)

$\ds=\int_0^\infty e^{-u}du\int_0^u t^{x-1}(u-t)^{y-1}dt \hfill$
(Fubini's theorem)

$\ds=\int_0^\infty e^{-u}du\int_0^1 u^{x+y-1}w^{x-1}(1-w)^{y-1}dw
\hfill (t=uw)$

$\ds=\int_0^\infty e^{-u}u^{x+y-1}du\int_0^1 w^{x-1}(1-w)^{y-1}dw$

$\ds=\Gamma(x+y)B(x;y)$

(All valid for $x>0, \,\, y>0$)

${}$

{\bf Contour Integral}

DIAGRAM


Consider $\int t^{z-1}e^{-t}dt$ around the contour $ABA'C'DCA$
where $CDC'$ is the circle $|t|=\epsilon$ and $AC, \,C'A'$ are the
upper and lower sides of the real axis from $t=\epsilon$ to $t=R$,
and $ABA'$ is a simple loop.  If $z$ is not an integer
$t^{z-1}=e^{(z-1)\log t}$ is not one-valued (as a function of
$t$).  Choose that branch of $\log t$ which is real when $t$ is at
$A$.  Hence along $ABA' \,\, \arg t$ increases from 0 to $2\pi$;
along $A'C' \,\, \arg t=2\pi$; along $C'DC \,\, \arg t$ decreases
from $2\pi$ to 0; along $CA \,\, \arg t=0$ i.e. $\log t$ returns
to its initial value ($\log R$) and hence so does $t^{z-1}$, and
also $t^{z-1}e^{-t}$ (since $e^{-t}$ is one-valued).  Hence
$t^{z-1}e^{-t}$ is one-valued inside and on the whole contour.  It
is also regular inside and on the contour.

By Cauchy's Theorem $\ds \int_{ABA'C'DCA}t^{z-1}e^{-t}=0$

i.e. $\ds \int_{ABA'}=-\int_{CA}+\int_{C'A'}-\int_{C'DC}$

On $CA \,\,\, t=v$ (real and positive)

On$C'A' \,\,\, t=ve^{2\pi i} \,\, (v$ real and positive)

Hence $\ds\int_{CA}=\int_\epsilon^R u^{z-1}e^{-u}du$

$\ds \hspace{0.5in}\int_{C'A'}=\int_\epsilon^R (ue^{2\pi
i})^{z-1}e^{-u}du$

On $C'DC \,\,\, t=\epsilon e^{i\theta} \,\,\, 0\leq\theta\leq2\pi$

Hence if $z=x+iy$

$t^{z-1}=e^{(x-1+iy)\log t}=e^{(x-1+iy)(\log\epsilon+i\theta)}$

$|t^{z-1}|=e^{Re(x-1+iy)(\log\epsilon+i\theta)}=e^{(x-1)\log\epsilon-y\theta)}=
\epsilon^{x-1}e^{-y\theta}$

Hence $|t^{z-1}|\leq\epsilon^{x-1}e^{2\pi|y|}$

Also $|e^{-t}|=e^{-Re(t)}\leq e^\epsilon$

Hence $|t^{z-1}e^{-t}|\leq\epsilon^{x-1}e^{2\pi|y|+\epsilon}$

Therefore $\ds\left|\int_{C'DC}t^{z-1}e^{-t}dt\right|
\leq\epsilon^{x-1}e^{2\pi|y|+\epsilon}2\pi\epsilon=2\pi\epsilon^x
e^{2\pi|y|+\epsilon}$

Hence $\ds\lim_{\epsilon\to0}\int_{C'DC}=0\,\,\,$ if $x=Re(z)>0$

Hence when $\epsilon\to0$

$\ds\int_{ABA'}t^{z-1}e^{-t}dt=(e^{2\pi iz}-1)\int_0^R
u^{z-1}e^{-u}du$

Now let $R\to\infty$ so that the loops $ABA'$ takes a limiting
form as shown.

We write $\ds\int_\infty^{0+}$ for this loop integral.

DIAGRAM

Since $\ds\lim_{R\to\infty}\int_0^R u^{z-1}e^{-u}du=\Gamma(z)
\,\,\,\, (Re(z)>0)$

$\ds\Gamma(z)=\frac{1}{e^{2\pi
iz}-1}\int_\infty^{0+}t^{z-1}e^{-t}dt \,\,\,\, (Re(z)>0, \,\,
z\not=$integer)

We can now dispense with the condition $Re(z)>0$ since the path of
integration does not pass through the origin.  We can show by
integration by parts that this new definition
$z\Gamma(z)=\Gamma(z+1)$.  For $Re(z)>0$ Euler's definite integral
definition is equivalent to Weierstrass's product, and also to the
contour integral definition.  Both the product and the contour
integral satisfy $z\Gamma(z)=\Gamma(z+1)$.  So both define the
whole function now in the whole plane where both have a meaning.
From Weierstrass's definition $\Gamma(z)\Gamma(1-z)=\pi\csc\pi z$.
Hence the contour integral satisfies this equation.

So

\begin{eqnarray*}\frac{1}{\Gamma(1-z)}&=&\Gamma(z)\frac{\sin\pi
z}{\pi}=\frac{\sin\pi z}{\pi}\frac{1}{e^{2\pi
iz}-1}\int_\infty^{0+}t^{z-1}e^{-t}dt\\ &=&\frac{e^{\pi
iz}-e^{-\pi iz}}{2\pi i}\frac{1}{e^{2\pi
iz}-1}\int_\infty^{0+}t^{z-1}e^{-t}dt\\ &=&\frac{1}{2\pi ie^{\pi
iz}}\int_\infty^{0+}t^{z-1}e^{-t}dt\end{eqnarray*}

Replacing $z$ by $1-z$ we have

\begin{eqnarray*}\frac{1}{\Gamma(z)} &=&\frac{1}{2\pi ie^{\pi
i(1-z)}}\int_\infty^{0+}t^{-z}e^{-t}dt\\ &=&\frac{-1}{2\pi
i}\int_\infty^{0+}(te^{-\pi i})^{-z}e^{-t}dt\end{eqnarray*}

Put $s=te^{-\pi i}$ then $ds=e^{-\pi i}dt=-dt$

$\ds\frac{1}{\Gamma(z)}=\frac{1}{2\pi
i}\int_{-\infty}^{0+}s^{-z}e^sds$


DIAGRAM

${}$

{\bf Asymptotic Behaviour of $\Gamma(z)$ as $z\to\infty$}

We shall show that $\log\Gamma(z)=(z-\frac{1}{2})\log
z-z+\log(2\pi)^\frac{1}{2}+\epsilon(z)$ where
$\epsilon(z)=O(\frac{1}{z})$ in a sector $-\pi+\delta\leq\arg
z\leq\pi-\delta$.  For $z=x$ we show that
$\epsilon(x)=\frac{\theta}{12x} \,\,\, 0<\theta<1$

i.e.
$\Gamma(z)=z^{z-\frac{1}{2}}e^{-z}(2\pi)^{\frac{1}{2}}\exp\{\epsilon(z)\}$

$\Gamma(z)\sim z^{z-\frac{1}{2}}e^{-z}(2\pi)^{\frac{1}{2}}$

(Stirling's formula $n!\sim
n^{n+\frac{1}{2}}e^{-n}(2\pi)^{\frac{1}{2}})$

We have by Euler's Limit Formula.

$\ds\log\Gamma(z)=\lim_{n\to\infty}[z\log n+\log n!-\{\log
z+\log(z+1)+\cdots+\log(z+n)\}] \hfill (1)$

Now $\ds\int_r^{r+1}f(t)dt=\int_0^1
1.f(r+t)dt=\left[(t-\frac{1}{2})f(r+t)\right]_0^1
-\int_0^1(t-\frac{1}{2}f'(r+t)dt$

$\ds\hspace{0.5in}
=\frac{1}{2}\left[f(r+1)+f(r)\right]-\int_0^1(t-\frac{1}{2})f'(r+t)dt
\hfill (2)$

(N.B.  This is a starting point for obtaining the Euler-Maclaurin
formula for approximate integration.)

Summing from $r=o$ to $r=n-1$ we have

$\ds\int_0^n f(t)dt$

$\ds\hspace{0.2in}=\frac{1}{2}f(0)+f(1)+\cdots+f(n-1)+\frac{1}{2}f(n)
-\int_0^1(t-\frac{1}{2}\sum_{r=0}^{n-1}f'(r+t)dt \hfill (3)$

Define

$\begin{array}{cccr}\phi(t)&=&t-\frac{1}{2}&0\leq t<1\\
\phi(t+1)&=&\phi(t)&t\geq0\end{array}$


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Then $\ds\int_0^1(t-\frac{1}{2})\sum_{r=0}^{n-1}f'(r+t)dt=
\int_0^1\sum_{r=0}^{n-1}\phi(r+t)f'(r+t)dt$

$\ds=\int_0^n\phi(t)f'(t)dt$

Hence we have

$\ds\int_0^n\log(t+z)dt$

$\ds=\frac{1}{2}\log
z+\log(z+1)+\cdots+\log(z+n-1)+\frac{1}{2}\log(z+n)
-\int_0^n\frac{\phi(t)}{t+z}dt$

Therefore $z\log n-\log z-\cdots-\log(z+n)+\log n!$

$\ds\hspace{0.3in}=z\log
n-\left[\int_0^n\log(t+z)dt+\frac{1}{2}\log
z+\frac{1}{2}\log(z+n)\right.$

$\ds\hspace{2in}+\left.\int_0^n\frac{\phi(t)}{t+z}dt+\log
n!\right] \hfill (4)$

Now
$\ds\int_0^n\log(t+z)dt=[(t+z)\log(t+z)-t]_0^n=(n+z)\log(n+z)-n-z\log
z$

RHS of (4)

$\ds=z\log
n-\left(n+z+\frac{1}{2}\right)log(n+z)+\left(z-\frac{1}{2}\right)\log
z+\log n!+n-\int_0^n\frac{\phi(t)}{t+z}dt$

${}$

$\ds z\log n-\left(z+n+\frac{1}{2}\right)\log(n+z)$

$\ds=z\log n-\left(z+n+\frac{1}{2}\right)\left[\log
n+\log\left(1+\frac{z}{n}\right)\right]$

$\ds=-\left(n+\frac{1}{2}\right)\log
n-\left(z+n+\frac{1}{2}\right)\log\left(1+\frac{z}{n}\right)$

$\ds=-\left(n+\frac{1}{2}\right)\log
n-\left(z+n+\frac{1}{2}\right)\left[\frac{z}{n}+O\left(\frac{z^2}{n^2}\right)\right]
\hspace{0.3in} n>|z|$

$\ds=-\left(n+\frac{1}{2}\right)\log
n-z+O\left(\frac{1}{n}\right)$

where $O\left(\frac{1}{n}\right)$ involves $z$.

Therefore RHS of (4)

$\ds=\left(z-\frac{1}{2}\right)\log z-z+\left[\log
n!-\left(n+\frac{1}{2}\right)\log
n+n\right]-\int_0^n\frac{\phi(t)}{t+z}dt+O\left(\frac{1}{n}\right)$

Now $\ds\lim_{n\to\infty}$LHS of (4)$=\log\Gamma(z)$

We show later that

$\ds-\epsilon(z)=\lim_{n\to\infty}\int_0^n\frac{\phi(t)}{t+z}dt=
\int_0^\infty\frac{\phi(t)}{t+z}dt \hfill$ (A)

and also $\ds\lim_{z\to\infty}\epsilon(z)=0$ where
$-\pi+\delta<\arg z<\pi-\delta \hfill$ (B)

Assuming (A) we have that $\ds\lim_{n\to\infty}\left\{\log
n!-\left(n+\frac{1}{2}\right)\log n+n\right\}=c$

(This result can be proved independently by a rather simpler
method.)

Also assuming (B) we can evaluate $c$, we have

$\ds\log\Gamma(z)+\log\Gamma\left(z+\frac{1}{2}\right)
+(2z-1)\log2-\log\Gamma(2z)=\log\Gamma\left(\frac{1}{2}\right)$

When (A) and (B) are assumed we have

$\ds\log\Gamma(z)=\left(z-\frac{1}{2}\right)\log
z-z+c+\epsilon(z)$

where $\ds\lim_{z\to\infty}\epsilon(z)=0$

Applying this to the previous equation

$\ds\left[\left(z-\frac{1}{2}\right)\log z-z+c+\epsilon(z)\right]+
\left[z\log\left(z+\frac{1}{2}\right)-\left(z+\frac{1}{2}\right)+c+\epsilon
\left(z+\frac{1}{2}\right)\right]$

$\ds\hspace{0.5in}
+(2z-1)\log2-\left[\left(2z-\frac{1}{2}\right)\log2z-2z+c+\epsilon(2z)\right]
=\log\Gamma\left(\frac{1}{2}\right)$

Therefore

$\ds\log z\left[z-\frac{1}{2}+z-\left(2z-\frac{1}{2}\right)\right]
+z\log\left(1+\frac{1}{2z}\right)+\left(-z-z-\frac{1}{2}+2z\right)$

$\ds\hspace{0.3in}+c+\epsilon(z)+\epsilon\left(z+\frac{1}{2}\right)-\epsilon(2z)
+(\log2)\left[-\left(2z-\frac{1}{2}\right)+2z-1\right]
=\log\Gamma\left(\frac{1}{2}\right)$

Taking the limit as $z\to\infty$ gives

$\ds
c+\frac{1}{2}-\frac{1}{2}-\frac{1}{2}\log2=\log\Gamma\left(\frac{1}{2}\right)$

Therefore $\ds
c=\log2^\frac{1}{2}\Gamma\left(\frac{1}{2}\right)=\log(2\pi)^\frac{1}{2}$

$\ds\left[{\rm since \
}\lim_{z\to\infty}z\log\left(1+\frac{1}{2z}\right)=
\lim_{z\to\infty}\log\left(1+\frac{1}{2z}\right)^z\right.$

$\ds\left.=\log\lim_{z\to\infty}\left(1+\frac{1}{2z}\right)^z=
\log\left(e^\frac{1}{2}\right)=\frac{1}{2}\right]$

Hence we have

$\ds\log\Gamma(z)=\left(z-\frac{1}{2}\right)\log
z-z+\log(2\pi)^\frac{1}{2}+\epsilon(z) \hfill$ (5)

where $\ds\epsilon(z)=-\int_0^\infty\frac{\phi(t)}{t+z}dt \hfill$
(6)

${}$

{\bf To prove (A)}

We consider $\ds\int_0^n\frac{\phi(t)}{t+z}dt$

Define $\phi_1(t)$ by $\phi_1'(t)=\phi(t) \,\,\,\, 0<t<1$

$\phi_1(0)=0 \hspace{0.5in} \phi_1(t+1)=\phi_1(t) \,\,\,\, t\geq0$

Therefore $\ds\phi_1(t)=\frac{1}{2}t^2-\frac{1}{2}t \,\,\,\,
(0\leq t<1)$

Then
$\ds\int_0^n\frac{\phi(t)}{t+z}dt=\int_0^n\frac{\phi_1'(t)}{t+z}dt
=\left[\frac{\phi_1(T)}{(t+z)}\right]_0^n
+\int_0^n\frac{\phi_1(t)}{(t+z)^2}dt$

$\ds\hspace{0.5in} =0+\int_0^n\frac{\phi_1(t)}{(t+z)^2}dt$

$\phi_1(t)$ is bounded, $\hspace{0.3in} (0\geq\phi_1(t)\geq
-\frac{1}{8})$

and also for $t>2|z|$ we have

$|t+z|\geq t-|z|>\frac{1}{2}t$

Therefore $\ds\frac{1}{|t+z|^2}<\frac{4}{t^2}$

So $\ds\int_0^n\frac{\phi_1(t)}{(t+z)^2}dt$ converges absolutely
as $n\to\infty$.

This proves (A) and we have

$\ds\epsilon(z)=-\int_0^\infty\frac{\phi(t)}{t+z}dt=
-\int_0^\infty\frac{\phi_1(t)}{(t+z)^2}dt \hfill$ (7)

${}$

{\bf To prove B}

\begin{itemize}
\item[I)]
Behaviour of $\epsilon(x)$ for real positive $x$.

$\ds\epsilon(x)=-\int_0^\infty\frac{\phi_1(t)}{(t+x)^2}dt$

Firstly we see that $\epsilon(x)>0$.  Also
$0\leq-\phi(t)\leq\frac{1}{8}$

Therefore $\ds 0\leq\epsilon(x)\leq
\frac{1}{8}\int_0^\infty\frac{dt}{(t+x)^2}= \frac{1}{8x}$

This is sufficient to prove B for real positive $x$.  The bound
can however be improved.

Let $\phi_2(t)=\frac{1}{6}t^3-\frac{1}{4}t^2+\frac{1}{12}t
\,\,\,\, 0\leq t<1$

$\hspace{0.5in}\phi_2(t+1)=\phi_2(t) \,\,\,\, t\geq0$

Then $\phi_2(0)=\phi_2(1)=0$

and $\phi_2'(t)=\phi_1(t)+\frac{1}{12}$

Therefore

$\ds\epsilon(x)=
-\int_0^\infty\frac{-\frac{1}{12}+\phi_2'(t)}{(t+x)^2}dt=
\frac{1}{12x}-\left[\frac{\phi_2(t)}{(t+x)^2}\right]_0\infty-
2\int_0^\infty\frac{\phi_2(t)dt}{(t+x)^3}dt$

$\ds\hspace{0.5in}=
\frac{1}{12x}-2\int_0^\infty\frac{\phi_2(t)}{(t+x)^3}dt$

$\ds\int_0^\infty\frac{\phi_2(t)}{(t+x)^3}dt>0$

Therefore $0\leq\epsilon(x)\leq\frac{1}{12x} \,\,\,\,
\epsilon(x)=\frac{\theta}{12x} \,\,\,\, 0\leq\theta\leq1$

${}$

\item[II]
Behaviour of $\epsilon(z)$ for complex $z$

$\ds\epsilon(z)=-\int_0^\infty\frac{\phi_1(t)}{(t+z)^2}dt$

We suppose $-\pi+\delta\leq\arg z\leq\pi-\delta, \,\,\,\,\,\,
0<\delta<\pi$

$\ds|\epsilon(z)|\leq
\int_0^\infty\left|\frac{\phi_1(t)}{(t+z)^2}\right|dt\leq
\frac{1}{8}\int_0^\infty\frac{dt}{|t+z|^2}$

$|t+z|^2=R^2=(r\sin\theta)^2\csc^2X$

$t=r\sin\theta\cot X-r\cos\theta$

$dt=-r\sin\theta\csc^2XdX$

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Therefore $\ds\frac{dt}{|t+z|^2}=-\frac{dX}{r\sin\theta}$

Therefore $\ds\int_0^\infty\frac{dt}{|t+z|^2}=
\int_\theta^0\frac{-dX}{r\sin\theta}=
\frac{\theta}{\sin\theta}\frac{1}{r}=
\frac{\theta}{\sin\theta}\frac{1}{|z|}$

The same result is obtained when $\theta<0$

$\ds0\leq\left|\frac{\theta}{\sin\theta}\right|\leq
\frac{\pi-\delta}{\sin\delta}$

Therefore $\ds\epsilon(z)|\leq
\frac{1}{8}\frac{\pi-\delta}{\sin\delta}\frac{1}{|z|}$

Therefore $\ds\lim_{z\to\infty}\epsilon(z)=0$ uniformly with
respect to $\arg z$ for

$|\arg z|\leq\pi-\delta$
\end{itemize}

${}$

{\bf Some Applications of the asymptotic formula}

$\ds\lim_{z\to\infty}
\left[\log\Gamma(z)-\left(z-\frac{1}{2}\right)\log
z+z-\log(2\pi)^\frac{1}{2}\right]=0$

\begin{itemize}
\item[1)]
$\ds\lim_{z\to\infty}\frac{\Gamma(z+a)}{\Gamma(z)}z^{-a}=1$

In fact
$\ds\frac{\Gamma(z+a)}{\Gamma(z)}z^{-a}=1+O\left(\frac{1}{z}\right)$

${}$

$\ds\log\frac{\Gamma(z+a)}{\Gamma(z)}z^{-a}=-a\log
z+\left(z+a-\frac{1}{2}\right)\log(z+a)-(z+a)$

$\ds\hspace{1in} -\left(z-\frac{1}{2}\right)\log
z+z+\epsilon(z+a)-\epsilon(z)$

$\ds=\log z\left\{-a+z+a-\frac{1}{2}-z+\frac{1}{2}\right\}+
\left(z+a-\frac{1}{2}\right)\log\left(1+\frac{a}{z}\right)$

$\ds\hspace{0.3in} -a+\epsilon(z+a)-\epsilon(z)$

$\ds=\left(z+a-\frac{1}{2}\right)
\left(\frac{a}{2}+O\left(\frac{1}{z^2}\right)\right)-
a+\epsilon(z+a)-\epsilon(z)$

$\ds=O\left(\frac{1}{2}\right)+\epsilon(z+a)-\epsilon(z)$

Hence the result.

${}$

\item[2)]
Behaviour of $\Gamma(z)$ as $T(z)\to\pm\infty$

If $z=x+iy \,\,\,\,\, y\to\pm\infty (x$ fixed)

then $\ds|\Gamma(x+iy)|\sim
|y|^{x-\frac{1}{2}}e^{-\frac{1}{2}\pi|y|}(2\pi)^\frac{1}{2}$

$\Gamma(z)\to0$ as $I(z)\to\pm\infty$ for any $R(z)$
\end{itemize}

${}$

{\bf Bernoulli Polynomials and Numbers}

Define a sequence of polynomials $P_n(t) \,\,\,\, n=1,2\cdots$ by

$\begin{array}{lcll} P_1(t)&=&t-\frac{1}{2}\\
P_2'(t)&=&P_1(t)&P_2(0)=0\\ P_3'(t)&=&P_2(t)-\bar{P_2}&P_3(0)=0\\
P_4'(t)&=&P_3(t)&P_4(0)=0\\
P_5'(t)&=&P_4(t)-\bar{P_4}&P_5(0)=0\end{array}$

Where $\bar{P_2}=$ mean value of $P_2(t)$ in $(0,1)$.  i.e.
$\bar{P_2}=\int_0^1 P_2(t)dt$

We show that

\begin{itemize}
\item[a)]
$P_1(t), \, P_3(t),\, \cdots$ are anti-symmetric about
$t=\frac{1}{2}$

$P_2(t), \, P_4(t),\, \cdots$ are symmetric about $t=\frac{1}{2}$

\item[b)]
$P_2(t), \, P_4(t),\, \cdots$ have zeros at $t=0, \, t=1$ and no
others in $[0,1]$.

$P_3(t), \, P_5(t),\, \cdots$ have zeros at $t=0, \,
t=\frac{1}{2}, \, t=1$ and no others in $[0,1]$.
\end{itemize}


DIAGRAM


\begin{eqnarray*} P_1(t)&=&t-\frac{1}{2}\\
P_2(t)&=&\frac{1}{2}t^2-\frac{1}{2}t\\
P_3(t)&=&\frac{1}{6}t^3-\frac{1}{4}t^2+\frac{1}{12}t\\
P_4(t)&=&\frac{1}{24}t^4-\frac{1}{12}t^3+\frac{1}{24}t^2\\
P_5(t)&=&\frac{1}{120}t^5-\frac{1}{48}t^4+\frac{1}{72}t^3-\frac{1}{720}t\\
\bar{P_2}&=&-\frac{1}{12} \hspace{0.5in}
\bar{P_4}=\frac{1}{720}\end{eqnarray*}

The Bernoulli Polynomial $\phi_n(t)$ is defined by
$\phi_n(t)=n!P_n(t)$.

The Bernoulli number $B_n$ is defined by
$B_n=(-1)^n(2n)!\bar{P_{2n}}$.

$B_1=\frac{1}{6} \hspace{0.3in} B_2=\frac{1}{30} \hspace{0.3in}
B_3=\frac{1}{42}$

Also $\ds\frac{B_n}{(2n)!}=\frac{2S_{2n}}{(2\pi)^{2n}}
\hspace{0.5in} S_k=\sum_{r=0}^\infty\frac{1}{r^k}$

${}$

{\bf Proof of a)}

$\ds\frac{d}{dt}\left[P_{2k}(t)-P_{2k}(1-t)\right]=P'_{2k}(t)+P'_{2k}(1-t)$

$\ds\hspace{1.7in}=P_{2k-1}(t)+P_{2k-1}(1-t)\hfill$ (i)

$\ds\frac{d}{dt}\left[P_{2k+1}(t)-P_{2k+1}(1-t)\right]=P'_{2k+1}(t)+P'_{2k+1}(1-t)$

$\ds\hspace{2in}=P_{2k}(t)+P_{2k}(1-t)\hfill$ (ii)

Assume $P_{2k-1}(t)$ is anti-symmetric about $t=\frac{1}{2}
\hfill$ (A)

then the RHS of (i)=0.

Therefore $P_{2k}(t)-P_{2k}(1-t)=const=P_{2k}(1)-P_{2k}(0)$

$\ds=\int_0^1P'_{2k}(t)dt=\int_0^1P_{2k-1}(t)=0\hfill$ by (A)

Therefore A$\Rightarrow$B: $P_{2k}$ is symmetric about
$t=\frac{1}{2}$.

Hence RHS of (ii)=0, and we have

$P_{2k+1}(t)+P_{2k+1}(1-t)=const=P_{2k+1}(1)+P_{2k+1}(0)$

$P_{2k+1}(0)=0$ by the construction of $P_n(t)$

Therefore $\ds
P_{2k+1}(1)=\int_0^1P'_{2k+1}(t)dt=\int_0^1(P_{2k}(t)-\bar{P_{2k}})dt=0$

Therefore $P_{2k+1}(t)$ is anti-symmetric about $t=\frac{1}{2}$.

i.e. $A(k)\Rightarrow A(k+1)$ and $B(k)$

$P_1(t)=t-\frac{1}{2}$ therefore $A(1)$ is true, hence (a) is
proven by induction.

${}$

{\bf Proof of (b)}

We know that $P_{2k-1}(t)$ vanishes at $t=0$ (by construction) and
at $t=\frac{1}{2}$ and $t=1$ (by anti-symmetry about
$t=\frac{1}{2}$).  Also $P_{2k}(t)$ vanishes at $t=0$ by
construction and at $t=1$ (by symmetry about $t=\frac{1}{2}$).  We
use the fact that if $f(t)$ is continuous and $f'(t)$ exists then
$f'(t)$ has at least one zero between consecutive zeros of $f(t)$.

(A)  Assume that $P_{2k-1}(t)$ has no zero other than
$t=\frac{1}{2}$ in $0<t<1$.  Now $P_{2k-1}(t)\not\equiv0$ and is
anti-symmetric about $t=\frac{1}{2}$.  Hence either $P_{2k-1}(t)$
is positive in $0<t<\frac{1}{2}$ and negative in
$\frac{1}{2}<t<1$, or vice versa.

Therefore $P'_{2k}(t)=P_{2k-1}(t)$ either steadily increases in
$0<t<\frac{1}{2}$ and steadily decreases in $\frac{1}{2}<t<1$ or
vice versa, and $P'_{2k}(\frac{1}{2})=0$.

$P_{2k}(0)=P_{2k}(1)=0$, therefore $P_{2k}(t)$ has no zeros in
$0<t<1 \hfill$ (B)

Also $P_{2k}(t)-c$ has at most two zeros in $0\leq t\leq1$ for any
$c$.  In particular $P'_{2k+1}(t)=P_{2k}(t)-\bar{P_{2k}}$ has at
most 2 zeros in $0\leq t\leq 1$, therefore $P_{2k+1}(t)$ has no
zeros in $(0,\frac{1}{2})$ or $(\frac{1}{2},1)$.

Therefore $A(k)\Rightarrow A(k+1)$ and $B(k)$.

A(1) is true, hence the result by induction.

N.B.  The zeros of $P_3(t), \, P_5(t)\cdots$ at $0,\frac{1}{2},1$
are all simple.

The zeros of $P_4(t), \, P_6(t)\cdots$ at $0,1$ are all double.

It can be proved that

$\ds\frac{he^{ht}-1}{e^h-1}=\sum_{n=0}^\infty P_n(t)h^n$

${}$

{\bf Asymptotic Expansion of log}${\bf\Gamma(z)}$

$\log\Gamma(z)=(z-\frac{1}{2})\log
z-z+\log(2\pi)^\frac{1}{2}+\epsilon(z)$

where $\ds \epsilon(z)=-\int_0^\infty\frac{\psi_1(t)}{t+z}dt$

Where $\psi_1(t)$ (formerly $\phi_1(t)$) is defined by

$\psi_1(t)=(t-\frac{1}{2}) \hspace{0.5in} (0\leq t\leq1)$

$\psi_1(t+1)=\psi_1(t) \hspace{0.5in} t\geq0$

Let $\psi_n(t)=P_n(t)=\frac{\phi_n(t)}{n!} \hspace{0.5in} 0\leq
t<1 \hspace{0.5in} n=2,3\cdots$

$\psi_n(t+1)=\psi_n(t) \hspace{0.5in} t\geq0$

Then $\psi'_{2n}(t)=\psi_{2n-1}(t)$

$\psi'_{2n+1}(t)=\psi_{2n}(t)-\bar{P_{2n}}$

Now $\ds\int_0^\infty\frac{\psi_1(t)}{t+z}dt=
\int_0^\infty\frac{\psi'_2(t)}{t+z}dt=
\left[\frac{\psi_2(t)}{t+z}\right]_0^\infty+
\int_0^\infty\frac{\psi_2(t)}{(t+z)^2}dt$

$\ds\int_0^\infty\frac{\psi'_3(t)+\bar{P_2}}{(t+z)^2}dt=
\frac{\bar{P_2}}{z}+\left[\frac{\psi_3(t)}{(t+z)^2}\right]_0^\infty
+2!\int_0^\infty\frac{\psi_3(t)}{(t+z)^3}dt$

$\ds=\frac{\bar{P_2}}{z}+2!\int_0^\infty\frac{\psi_3(t)}{(t+z)^3}dt$

Continuing this process we find

$\ds\int_0^\infty\frac{\psi_1(t)}{t+z}dt=
\frac{\bar{P_2}}{z}+2!\frac{\bar{P_4}}{z^3}+ \cdots+
\frac{(2n-2)!\bar{P_{2n}}}{z^{2n-1}}+
(2n)!\int_0^\infty\frac{\psi_{2n+1}(t)}{(t+z)^{2n+1}}dt$

Therefore
$\ds\epsilon(z)=-\sum_{r=1}^n\frac{(2r-2)!\bar{P_{2r}}}{z^{2r-1}}+
R_n(z)$

$\ds
R_n(z)=-(2n)!\int_0^\infty\frac{\psi_{2n+1}(t)}{(t+z)^{2n+1}}dt$

$\ds=-(2n+1)!\int_0^\infty\frac{\psi_{2n+2}(t)}{(t+z)^{2n+2}}dt$

after one further integration by parts with
$\psi_{2n+1}(t)=\psi_{2n+2}(t)$.

Substituting $B_n=(-1)^n(2n)!\bar{P_{2n}}$ we get

$\ds\epsilon(z)= \sum_{n=1}^m\frac{(-1)^{n-1}B_n}{(2n-1)^{2n}}
\frac{1}{z^{2n-1}}+R_m(z)$

${}$

${}$

Magnitude of $R_n(z)$

First note that $\psi_2(t)\leq0, \,\,\, \psi_2(t)\leq0$ etc.

Let $z=x$, real and positive.  In this case $R_0(x)>0, \,\,\,
R_1(x)<0$ etc.  i.e. the $R_n(x)$ are alternately positive and
negative.  Hence $\epsilon(x)$ lies between the sums to $n$ and
$n+1$ terms of the series

$\ds\frac{B_1}{1.2}\frac{1}{x}-\frac{B_2}{3.4}\frac{1}{x^3}+\cdots$

and $R_n(x)$ is numerically less than the term
$\ds\frac{(-1)^nB_{n+1}}{(2n+1)(2n+2)}\frac{1}{x^{2n+1}}$

In particular $\ds\lim_{x\to\infty}x^{2n-1}R_n(x)=0$ ($n$ fixed)
\hfill (1)

In fact $\ds\lim_{x\to\infty}x^{2n+1}R_n(x)$ exists and
$\ds=\frac{(-1)^nB_{n+1}}{(2n+1)(2n+2)}$ \hfill (2)

The property (1) characterises the asymptotic nature of the series
$\ds\frac{B_1}{1.2}\frac{1}{x}-\cdots$

${}$

Divergence of the series

The above series taken to $\infty$ is divergent for all $x$.  i.e.
$\ds\lim_{n\to\infty}R_n(x)$ does not exist.

This follows the result (here quoted but not proved) that

$\ds(-1)^n\bar{P_{2n}}=\frac{B_n}{(2n)!}=\frac{2S_{2n}}{(2\pi)^{2n}}$

$\ds\frac{B_n}{(2n)!}\sim \frac{2}{(2\pi)^{2n}}$

In that case the nth term of the series is

$\ds
(-1)^n(2n-2)!\frac{2S_{2n}}{(2\pi)^{2n}}\frac{1}{z^{2n-1}}\sim
\frac{2(-1)^n}{2\pi}\frac{(2n-2)!}{(2\pi z)^{2n-1}}$

And $\ds\sum_{n=1}^\infty\frac{(-1)^n(2n-2)!}{z^{2n-1}}$ diverges
for all $z$.

Hence the result.

We write $\ds\epsilon(z)=\frac{B_1}{1.2}\frac{1}{z}-
\frac{B_2}{3.4}\frac{1}{z^3}+\cdots$ and

$\ds\log\Gamma(z)\sim \left(z-\frac{1}{2}\right)\log
z-z+\log(2\pi)\frac{1}{2}+\frac{B_1}{1.2}\frac{1}{z}-
\frac{B_2}{3.4}\frac{1}{z^3}+\cdots$

${}$

Behaviour of $R_n(z)-z$ complex -as $z\to\infty$

We have $\ds R_n(z)=
-(2n+1)!\int_0^\infty\frac{\psi_{2n+2}(t)}{(t+z)^{2n+2}}dt$

We suppose as before that $|\arg z|\leq\pi-\delta, \,\,\,
(0<\delta<\pi)$ necessary for the existence of $R_n(z)$.

\begin{itemize}
\item[a)]
when $x=Re(z)>0$

$|t+z|\geq t+x$ for all $t\geq0$ and hence

$\ds|R_n(z)|\leq
(2n+1)!\int_0^\infty\frac{|\psi_{2n+2}(t)|}{(t+x)^{2n+2}}dt$


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Either $\psi_{2n+2}(t)\leq0 \hspace{0.5in} n=0,2,4,\cdots$

or $\hspace{0.3in} \psi_{2n+2}(t)\geq0 \hspace{0.5in}
n=1,3,5,\cdots$

$|R_n(z)|\leq|R_n(x)|$

If $m_n$ is the upper bound of $|\psi_{2n}(t)|$

$\ds|R_n(z)|\leq\frac{(2n+1)!m_{n+2}}{(2n+1)x^{2n+1}}=
\frac{K_n}{x^{2n+1}}= \frac{K_n\sec^{2n+1}\theta}{|z|^{2n+1}}$

Hence $|R_n(z)|=O(|z|^{-(2n+1)}) \hspace{0.5in} |\arg
z|<\frac{\pi}{2}$

${}$

\item[b)]
when $|\arg z|\leq\pi-\delta$

$\ds|R_n(z)|\leq
(2n+1)!\int_0^\infty\frac{|\psi_{2n+2}(t)|}{|t+z|^{2n+2}}dt$

$|t+z|=y\csc X \hspace{0.5in} t=y\cot X-x$

$\ds|R_n(z)|\leq(2n+1)!\int_0^\pi
\frac{|\psi_{2n+2}(t)|y\csc^2XdX}{y^{2n+2}\csc^{2n+2}X}$

$\ds\hspace{0.5in}\leq
\frac{(2n+1)!m_{n+2}}{y^{2n+1}}\int_0^\pi\sin^{2n}XdX$

$\ds\hspace{0.5in}=\frac{K_n'}{y^{2n+1}}=
\frac{K_n'\csc^{2n+1}\theta}{|z|^{2n+1}}$

$\ds\hspace{0.5in}\leq\frac{K_n'\csc^{2n+1}\delta}{|z|^{2n+1}}
\hspace{0.5in} |\theta|\leq\pi-\delta$

$|R_n(z)|=O(z^{-(2n+1)})$

\end{itemize}

\end{document}