Difference between revisions of "Contour integral representation of reciprocal gamma"

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==Theorem==
<strong>Theorem:</strong> The following formula holds for a [[positively orientation|positively oriented]] [[contour]] $C$ is a path encircling $0$ beginning at and returning to $+\infty$:
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The following formula holds for a [[positively orientation|positively oriented]] [[contour]] $C$ is a path encircling $0$ beginning at and returning to $+\infty$:
 
$$\dfrac{1}{\Gamma(z)} = \dfrac{i}{2\pi} \displaystyle\oint_C (-t)^{-z}e^{-t} \mathrm{d}t,$$
 
$$\dfrac{1}{\Gamma(z)} = \dfrac{i}{2\pi} \displaystyle\oint_C (-t)^{-z}e^{-t} \mathrm{d}t,$$
 
where $\dfrac{1}{\Gamma}$ denotes the [[reciprocal gamma]] function, $\pi$ denotes [[pi]], and $e^{-t}$ denotes the [[exponential]] function.
 
where $\dfrac{1}{\Gamma}$ denotes the [[reciprocal gamma]] function, $\pi$ denotes [[pi]], and $e^{-t}$ denotes the [[exponential]] function.
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<strong>Proof:</strong> █
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==Proof==
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==References==
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[[Category:Theorem]]
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[[Category:Unproven]]

Latest revision as of 07:17, 16 June 2016

Theorem

The following formula holds for a positively oriented contour $C$ is a path encircling $0$ beginning at and returning to $+\infty$: $$\dfrac{1}{\Gamma(z)} = \dfrac{i}{2\pi} \displaystyle\oint_C (-t)^{-z}e^{-t} \mathrm{d}t,$$ where $\dfrac{1}{\Gamma}$ denotes the reciprocal gamma function, $\pi$ denotes pi, and $e^{-t}$ denotes the exponential function.

Proof

References