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I want to understand why we care about the $\Re(n) \gt (-1).$
I tried calculating it by myself but plugging in $0$ and $\infty$ didn't really work out well.

$$\Gamma (n+1)=\int _{0}^{\infty }x^{n}e^{-x}\,dx,\ \qquad \Re (n)>-1\,.$$

Piita
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Xposed
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1 Answers1

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There is a notational issue.
The gamma function is defined for all complex numbers except the non-positive integers. For this, the variable $z$ is standard. Variable $n$ is standard for integers.

For positive integers $n$,
$$\Gamma (n)= (n-1)!$$
For complex numbers $z$ with a positive real part, the gamma function is defined via a convergent improper integral: $$\Gamma (z)=\int _{0}^{\infty }t^{z-1}e^{-t}\,dt,\ \qquad \Re (z)>0\,.$$

When you change the argument to $(z+1)$, you adjust to get

$$\Gamma (z+1)=\int _{0}^{\infty }t^{z}e^{-t}\,dt,\ \qquad \Re (z)\gt -1\,.$$

Piita
  • 965