3.6 Analytic continuation for gamma function, 5

A First Course in String Theory

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

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

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

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

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Since both $\displaystyle{e^{-t}}$ and $\displaystyle{\sum_{n=0}^N \frac{(-t)^n}{n!}}$ are finite, $\displaystyle{ \begin{aligned} \sum_{n=N+1}^\infty \frac{(-t)^n}{n!} dt \end{aligned} \\ }$ is also finite.

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

$\displaystyle{ \begin{aligned} &= \sum_{n=N+1}^\infty \frac{(-1)^n}{n!} \frac{1}{z+n} \left[ 1 - 0^{z+n} \right] \\ \end{aligned} \\ }$

This term converges as long as $\displaystyle{ \Re (z) > - N - 1 }$ .

So this condition is not only necessary but also sufficient for the convergence of the analytic continuation of Gamma function,

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

— Me@2023-09-07 07:38:17 PM

Physics Town

continues

3.6 Analytic continuation for gamma function, 5

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