Gamma function/partial fractions (a problem from my Insights article)

[benorin]

I'm typing up answers to the exercises in my Insight article "A Path to Fractional Integral Representations of Some Special Functions", this problem is from section 1 (Gamma/Beta functions). I need a bit of help with this one:

The problem statement:
1.9) Use partial fraction decomposition to show that $\forall z \notin {\mathbb{Z}^ - } \cup \left\{ 0 \right\},\;\Gamma \left( z \right) = \mathop {\lim }\limits_{\lambda \to \infty } \frac{{\lambda !{\lambda ^{z - 1}}}}{{z\left( {z + 1} \right) \cdots \left( {z + \lambda - 1} \right)}} = \mathop {\lim }\limits_{\lambda \to \infty } \,{\lambda ^{z - 1}}\sum\limits_{k = 0}^{\lambda - 1} {{{\left( { - 1} \right)}^k}\left( {\lambda - k} \right)\left( {\begin{array}{*{20}{c}} \lambda \\ k \end{array}} \right){{\left( {z + k} \right)}^{ - 1}}}$, where $\left( {\begin{array}{*{20}{c}} \lambda \\ k \end{array}} \right)$ is a binomial coefficient.

Solution:
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

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

I don’t understand why Wolfram|Alpha says that the expression I had for $\Gamma \left( z \right)$ (pre-simplification) is only true for $z > - 1 \vee z \notin \mathbb{Z}$. I think the inequality treats z as real variable and the second statement treats z as complex but non-integer and the union of those domains would be $\mathbb{C}\backslash {\mathbb{Z}^ - }$, which oddly includes zero? I just wanted to make sure my partial fractions didn’t alter the domain of convergence. Why does Wolfram|Alpha give a different answer for the simplified expression?

 

[benorin]

And this is probably a silly question, but I find myself needing to ask it: when I solved the partial fractions equation for the coefficients in the above, I didn't include the exponential factor $\lambda ^{z-1}$ on the LHS. That is how I solved the problem initially, but upon going to type it up I discovered I lost the paper I hashed it out on. So like how I try to keep simple formulas in my memory by finding an easy way to derive them on the fly, (works for trig and calc at least): I figured I'd just write it up again. That time I included the exponential factor in my partial fractions equation. I did much as that above only upon setting z equal to the nonpositive integers which make all but one of the terms vanish on the RHS (after multiplying by the common denominator) the exponential factor which varies with z changed value of the coefficients in that "solution". Why should I know better? Wouldn't I need a full on power series for that? Yes, I suppose so.
Right, thanks. Good chat!

 

[poonamdeo13]

I tried accessing this blog with a screen reader, it seems to recognize the Math content but the screen reader is not able to recognize the Math characters, whereas it is reading as Math instead of reading the expression.