# Functions.wolfram.com accepted and published my formula for Gamma Fn

• I
• benorin

#### benorin

Homework Helper
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/
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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• quasar987, CGandC, DrClaude and 1 other person

Tried to follow through your proof but I can't understand how you define ##\alpha##.

Oh I'm sorry Delta2, my proof got mangled up with the proof above it pre-copy+paste, gimme a few and I'll fix it.

• Delta2
Ehm, now the first section seems deleted, I guess you still editing...

I guess I should 've made the straightforward implication that ##\alpha## is the solution to the equation $$u^{\alpha-1}e^{-u}=u^{z-1}$$

But then again it is not so straightforward if we can treat it as constant with respect to ##u##.

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@Delta2 refresh the page (as I edited the OP).

Sorry @Delta2 Oh nvm, you did. Yes the first section which contained an alpha was part of the previous problem and was removed, there's more (I think) to the beginning of the second part now.

• Delta2
Ok , I think I understand now, seems to me you correctly calculate the ##a_k## and the rest is up to some algebraic processing of the Euler Limit of Gamma function.

I copy+pasted the link from the email but looking at it is it even right? Looking at the date added at the bottom of the page says 2001 so seems off.

When I click on the link, my browser doesn't seem to display all the page correctly, but what I can see clearly is that $$\Gamma(z)=\lim_{n\to\infty}\frac{(1)_nn^{z-1}}{(z)_n}$$. This doesn't look exactly as your formula...

There should be a sum in it, see note at bottom of OP after refresh

No can't see any sum anywhere in that link.