3.6 Analytic continuation for gamma function, 1

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$\displaystyle{ \begin{aligned} \Gamma (z) = &\int_{0}^{1}dt~t^{z-1} \left( e^{-t} - \sum_{n=0}^N \frac{(-t)^n}{n!} \right) \\ &+ \sum_{n=0}^N \frac{(-t)^n}{n!} \frac{1}{z+n} + \int_1^\infty dt~e^{-t}t^{z-1} \\ \end{aligned} \\ }$

Explain why the above right-hand side is well defined for $\displaystyle{ \Re (z) > - N - 1 }$ .

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$\displaystyle{ \begin{aligned} \Gamma (z) &= \int_{0}^{\infty}t^{z-1}e^{-t}dt \\ &= - \int_{0}^{\infty}t^{z-1}de^{-t} \\ &= - \left[ t^{z-1} e^{-t} \right]_{0}^{\infty} + (z-1) \int_{0}^{\infty} t^{z-2}e^{-t} dt \\ &= - \left[ \lim_{t \to \infty} t^{z-1} e^{-t} - 0^{z-1} e^{-0} \right] + (z-1) \int_{0}^{\infty} t^{z-2}e^{-t} dt \\ \end{aligned} \\ }$

Note that this can only prove that for convergence, it may be necessary to have $\displaystyle{ \Re (z) > 1 }$ , but not $\displaystyle{ \Re (z) > 0 }$ .

Also, the condition $\displaystyle{ \Re (z) > 1 }$ may be not necessary at all, because an infinity in one term may be cancelled out by another infinity in a latter term.

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

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 $t > 0$ :