$$\frac{\Gamma(j-q)}{\Gamma (-q) \Gamma (j+1)}$$ a) Does anyone know how this could be the expression for gamma function and how it could be derived from integral definition of gamma function ?

b) And the above definition of gamma function gives the below result

$$\frac{\Gamma (j-q)}{\Gamma (-q) \Gamma (j+1)} = \frac{[-]^{j}}{j!} \sum\limits_{m = 0}^{j} S_{j}^{(m)} q^m$$

where S is the Stirling number and m=0 in summation sign . I'm following the book fractional calculus by KEITH B. OLDHM and JEROME SPANIER this book is really advanced and sometimes I'm unable to understand the things . Your help would be appreciated thanks .

For reference Euler limit definition of gamma function and integral transform definition.

$$\Gamma (x) \equiv \lim_{N \rightarrow \infty } [N!N^x/x[x+1][x+2]...[x+N]]$$

$$ \Gamma\left(x\right)\equiv\int_{0}^{\infty} y^{x-1}\exp\left(-y\right){\text{d}y} $$ where x>0

  • $\begingroup$ How is $\gamma(x)$ defined? Also what is j? $\endgroup$ – Daniel McLaury Nov 17 '16 at 7:12
  • $\begingroup$ @DanielMcLaury I have updated the post , kindly find the it. j is the non negative number and q may take any value $\endgroup$ – Aamir Nov 17 '16 at 7:34

The OP's question is a good one, despite making a mess of the notation. Let me restate it here, along with some work towards a solution.

The OP wants a proof of \begin{equation} \frac{\Gamma(j-q)}{\Gamma(-q)\Gamma(j+1)} = \frac{(-1)^{j}}{j!} \sum\limits_{m=0}^{j} \mathrm{S}_{j}^{(m)} q^{m} \end{equation} for $j \in \mathbb{N}_{0},\,\, q \in \mathbb{C}$.

As per the book referenced (paraphrasing): "For small numerical values of $j$, the left hand side of the above expression is readily simplfied to a polynomial in $q$ by application of the two equations below. This procedure generalizes to give the above expression."

$\Gamma(x+1) = x\Gamma(x)$ and $\Gamma(x-1) = \Gamma(x)/(x-1)$

We can rewrite the sum on the rhs using the falling factorial, $$(q)_{j} = \sum\limits_{m=0}^{j} \mathrm{S}_{j}^{(m)} q^{m} = \frac{\Gamma(q+1)}{\Gamma(q-j+1)}$$

Hopefully, someone can take it from here.


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