08 | December | 2023 | Physics Town

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.

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

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

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

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

Therefore, for the integral to converge, $\displaystyle{ \Re (z) + n > 0 }$ is required. So for the overall integral $\displaystyle{ \int_{0}^{1} t^{z-1} \sum_{n=0}^\infty \frac{(-t)^n}{n!} dt}$ to converge, a necessary condition is

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

Therefore,

$\displaystyle{ \Re (z) > 0 }$

Under this condition,

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