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I sent them an email about a week or so ago with the images of the following from the solutions to exercises section 1 of my Insight Article A Path to Fractional Integral Representations of Some Special Functions:

1.9) Use partial fraction decomposition and the Euler limit definition of the Gamma function ##\Gamma ( z ) := \mathop {\lim }\limits_{\lambda \to \infty } \frac{\lambda ! \lambda ^{z - 1}}{z ( z + 1 ) \cdots ( z + \lambda - 1 ) }## to show that ##\forall z \notin {\mathbb{Z}^ - } \cup \left\{ 0 \right\} ,##

$$\Gamma ( z ) = \mathop {\lim }\limits_{\lambda \to \infty } \, \lambda ^{z - 1} \sum\limits_{k = 0}^{\lambda - 1} ( -1) ^k ( \lambda - k ) \left. _{\lambda} C_{k} \right. ( z + k )^{ - 1}$$

where ##\left. _{\lambda} C_{k} \right. ## is a binomial coefficient and ##:=## means "defined to be".

Start by finding the partial fraction decomposition of ##\tfrac{1}{{z\left( {z + 1} \right) \cdots \left( {z + \lambda - 1} \right)}}## . This is just algebra, set

$$\begin{gathered} \tfrac{1}{{z\left( {z + 1} \right) \cdots \left( {z + \lambda - 1} \right)}} = \tfrac{{{a_0}}}{z} + \tfrac{{{a_1}}}{{z + 1}} + \cdots + \tfrac{a_{\lambda - 1}}{{z + \lambda - 1}} \\ = \sum\limits_{k = 0}^{\lambda - 1} {\tfrac{{{a_k}}}{{z + k}}} \\ \end{gathered}$$

(where the second equality is just more compact notation). Multiply by the common denominator to arrive at

$$1 = \sum\limits_{k = 0}^{\lambda - 1} {{a_k}\prod\limits_{\begin{subarray}{l} m = 0 \\ m \ne k \end{subarray}} ^{\lambda - 1} {\left( {z + m} \right)} }$$

To solve this equation for the unknown coefficients ##{a_k}##, set ##z = - n##, for each ##n = 0,1, \ldots ,\lambda - 1##. Then

$$1 = \sum\limits_{k = 0}^{\lambda - 1} {{a_k}\prod\limits_{\begin{subarray}{l} m = 0 \\ m \ne k \end{subarray} }^{\lambda - 1} {\left( {m - n} \right)} } \Rightarrow {a_n} = {\left\{ {\prod\limits_{\begin{subarray}{l} m = 0 \\ m \ne n \end{subarray}} ^{\lambda - 1} {\left( {m - n} \right)} } \right\}^{ - 1}}$$

from which ##{a_n} = \tfrac{1}{{\left( { - n} \right)\left( {1 - n} \right) \cdots \left( { - 1} \right) \cdot 1 \cdot 2 \cdots \left( {\lambda - n - 1} \right)}} = \tfrac{{{{\left( { - 1} \right)}^n}}}{{n!\left( {\lambda - n - 1} \right)!}}##. Therefore the desired partial fraction decomposition is

$$\boxed{\frac{1}{{z\left( {z + 1} \right) \cdots \left( {z + \lambda - 1} \right)}} = \sum\limits_{k = 0}^{\lambda - 1} {\left[ {\frac{{{{\left( { - 1} \right)}^k}}}{{k!\left( {\lambda - k - 1} \right)!}} \cdot \frac{1}{{z + k}}} \right]} }$$

Hence from the Euler limit form of the Gamma function we have that

$$\Gamma \left( z \right): = \mathop {\lim }\limits_{\lambda \to \infty } \frac{{\lambda !{\lambda ^{z - 1}}}}{{z\left( {z + 1} \right) \cdots \left( {z + \lambda - 1} \right)}} = \mathop {\lim }\limits_{\lambda \to \infty } \left[ {\sum\limits_{k = 0}^{\lambda - 1} {\tfrac{{{{\left( { - 1} \right)}^k}\lambda !{\lambda ^{z - 1}}}}{{k!\left( {\lambda - k - 1} \right)!}} \cdot \tfrac{1}{{z + k}}} } \right]$$

which I simplified to

$$\Gamma \left( z \right) = \mathop {\lim }\limits_{\lambda \to \infty } \,{\lambda ^{z - 1}}\sum\limits_{k = 0}^{\lambda - 1} {{{\left( { - 1} \right)}^k}\left( {\lambda - k} \right)\left( {\begin{array}{*{20}{c}} \lambda \\ k \end{array}} \right){{\left( {z + k} \right)}^{ - 1}}} $$

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I checked my email today and they said my formula was published to

https://functions.wolfram.com/GammaBetaErf/Gamma/09/0003/

1.9) Use partial fraction decomposition and the Euler limit definition of the Gamma function ##\Gamma ( z ) := \mathop {\lim }\limits_{\lambda \to \infty } \frac{\lambda ! \lambda ^{z - 1}}{z ( z + 1 ) \cdots ( z + \lambda - 1 ) }## to show that ##\forall z \notin {\mathbb{Z}^ - } \cup \left\{ 0 \right\} ,##

$$\Gamma ( z ) = \mathop {\lim }\limits_{\lambda \to \infty } \, \lambda ^{z - 1} \sum\limits_{k = 0}^{\lambda - 1} ( -1) ^k ( \lambda - k ) \left. _{\lambda} C_{k} \right. ( z + k )^{ - 1}$$

where ##\left. _{\lambda} C_{k} \right. ## is a binomial coefficient and ##:=## means "defined to be".

Start by finding the partial fraction decomposition of ##\tfrac{1}{{z\left( {z + 1} \right) \cdots \left( {z + \lambda - 1} \right)}}## . This is just algebra, set

$$\begin{gathered} \tfrac{1}{{z\left( {z + 1} \right) \cdots \left( {z + \lambda - 1} \right)}} = \tfrac{{{a_0}}}{z} + \tfrac{{{a_1}}}{{z + 1}} + \cdots + \tfrac{a_{\lambda - 1}}{{z + \lambda - 1}} \\ = \sum\limits_{k = 0}^{\lambda - 1} {\tfrac{{{a_k}}}{{z + k}}} \\ \end{gathered}$$

(where the second equality is just more compact notation). Multiply by the common denominator to arrive at

$$1 = \sum\limits_{k = 0}^{\lambda - 1} {{a_k}\prod\limits_{\begin{subarray}{l} m = 0 \\ m \ne k \end{subarray}} ^{\lambda - 1} {\left( {z + m} \right)} }$$

To solve this equation for the unknown coefficients ##{a_k}##, set ##z = - n##, for each ##n = 0,1, \ldots ,\lambda - 1##. Then

$$1 = \sum\limits_{k = 0}^{\lambda - 1} {{a_k}\prod\limits_{\begin{subarray}{l} m = 0 \\ m \ne k \end{subarray} }^{\lambda - 1} {\left( {m - n} \right)} } \Rightarrow {a_n} = {\left\{ {\prod\limits_{\begin{subarray}{l} m = 0 \\ m \ne n \end{subarray}} ^{\lambda - 1} {\left( {m - n} \right)} } \right\}^{ - 1}}$$

from which ##{a_n} = \tfrac{1}{{\left( { - n} \right)\left( {1 - n} \right) \cdots \left( { - 1} \right) \cdot 1 \cdot 2 \cdots \left( {\lambda - n - 1} \right)}} = \tfrac{{{{\left( { - 1} \right)}^n}}}{{n!\left( {\lambda - n - 1} \right)!}}##. Therefore the desired partial fraction decomposition is

$$\boxed{\frac{1}{{z\left( {z + 1} \right) \cdots \left( {z + \lambda - 1} \right)}} = \sum\limits_{k = 0}^{\lambda - 1} {\left[ {\frac{{{{\left( { - 1} \right)}^k}}}{{k!\left( {\lambda - k - 1} \right)!}} \cdot \frac{1}{{z + k}}} \right]} }$$

Hence from the Euler limit form of the Gamma function we have that

$$\Gamma \left( z \right): = \mathop {\lim }\limits_{\lambda \to \infty } \frac{{\lambda !{\lambda ^{z - 1}}}}{{z\left( {z + 1} \right) \cdots \left( {z + \lambda - 1} \right)}} = \mathop {\lim }\limits_{\lambda \to \infty } \left[ {\sum\limits_{k = 0}^{\lambda - 1} {\tfrac{{{{\left( { - 1} \right)}^k}\lambda !{\lambda ^{z - 1}}}}{{k!\left( {\lambda - k - 1} \right)!}} \cdot \tfrac{1}{{z + k}}} } \right]$$

which I simplified to

$$\Gamma \left( z \right) = \mathop {\lim }\limits_{\lambda \to \infty } \,{\lambda ^{z - 1}}\sum\limits_{k = 0}^{\lambda - 1} {{{\left( { - 1} \right)}^k}\left( {\lambda - k} \right)\left( {\begin{array}{*{20}{c}} \lambda \\ k \end{array}} \right){{\left( {z + k} \right)}^{ - 1}}} $$

*********************End*********************

I checked my email today and they said my formula was published to

https://functions.wolfram.com/GammaBetaErf/Gamma/09/0003/

**Edit:**there's some kind of issue with the link I was emailed it doesn't appear to be the correct formula, I have emailed back and awaiting a response.
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