3.6 Analytic continuation for gamma function, 4

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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Under this condition,

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

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

So for $\displaystyle{\Re (z) > 0}$ ,

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

If we define this as the analytic continuation of Gamma function:

$\displaystyle{ \begin{aligned} \Gamma_c (z) \end{aligned} \\ }$

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

its convergence depends on the convergence of the middle term:

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

By the argument above, a necessary condition for this term to converge is

$\displaystyle{ \Re (z) + n_{\text{min}} > 0 }$

$\displaystyle{ \Re (z) + N + 1 > 0 }$

$\displaystyle{ \Re (z) > - N - 1 }$

However, besides the integral convergence, we have to consider also the summation convergence.

To check whether the order of integration and summation can be interchanged, we rewrite the term as

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

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.