3.6 Analytic continuation for gamma function, 2

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$\displaystyle{ \begin{aligned} \Gamma (z) &= \int_{0}^{1}t^{z-1}e^{-t}dt + \int_{1}^{\infty}t^{z-1}e^{-t}dt \\\\ \end{aligned} \\ }$

Prove that $\displaystyle{\int_{1}^{\infty}t^{z-1}e^{-t}dt}$ converges.

Since $\displaystyle{ \begin{aligned} e^t &= \sum_{s=0}^\infty \frac{t^s}{s!} \\ \end{aligned} \\ }$ ,

for any integer $r \ge 0$ and any real $t > 0$ ,

$\displaystyle{ \begin{aligned} e^t &\ge \frac{t^r}{r!} \\ \end{aligned}~~~}$

Then

$\displaystyle{ \begin{aligned} e^{-t} &\le \frac{1}{\frac{t^r}{r!}} \\ \end{aligned} \\ }$

Let $\displaystyle{ \begin{aligned} x &= \Re(z) \\ \end{aligned} \\ }$ :

$\displaystyle{ \begin{aligned} t^{x-1}e^{-t} &\le \frac{t^{x-1}}{\frac{t^r}{r!}} \\ t^{x-1}e^{-t} &\le r! t^{x - r - 1} \\ \end{aligned} \\ }$

By choosing an $r$ such that $r \ge 1$ and $r \ge \Re(z) + 1$ ,

for any $t \ge 1$ ,

$\displaystyle{ \begin{aligned} t^{x-1}e^{-t} &\le r! t^{x - (x+1) - 1} \\ t^{x-1}e^{-t} &\le r! t^{-2} \\ \end{aligned} \\ }$

Therefore,

$\displaystyle{ \begin{aligned} \int_1^\infty t^{x-1}e^{-t} dt &\le r! \int_1^\infty t^{-2} dt = - r! [ t^{-1} ]_1^\infty = r! \\ \end{aligned} \\ }$

In other words, for any complex $z$ ,

$\displaystyle{ \begin{aligned} \left| \int_1^\infty t^{z-1} e^{-t} dt \right| &\le r! \\ \end{aligned} \\ }$

for some $r$ such that $r \ge 1$ and $r \ge \Re(z) + 1$ .

$\displaystyle{ \begin{aligned} \left| \int_1^\infty t^{z-1} e^{-t} dt \right| &\le r! \\ \end{aligned} \\ }$

converges for any $z$ .

Prove that $\displaystyle{\int_{0}^{1}t^{z-1}e^{-t}dt}$ converges.

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— Me@2023-09-07 07:38:17 PM

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