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I am trying to evaluate $\lim_{x \to -N} (x+N) \Gamma(x) $ =$(-1)^N/N!$, for N = 0, 1, 2, ... but I keep getting zero.

I tried using the definition of $\Gamma(x) = \Gamma(x+1)/x $ with no luck.

Is there any identity for gamma I could use to get the desired result?

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Using some of these ideas, and trying to formulate my own answer based off of what I actually know, I got $lim_{x \rightarrow -N} (x+N) \Gamma(x)$ = $lim_{x \rightarrow -N} (x+N) \frac{\Gamma(x+N)}{(x)_{N}}$ = $lim_{x \rightarrow -N} \frac {\Gamma(x+N+1)}{x(x+1)(x+2)...(x+N+1)}$ =$ \frac{\Gamma(1)}{(-1)[N(N-1)(N-2)...1]}$ =$\frac{1}{(-1)(N!)}$ which is still wrong I think. –  renagade629 Oct 7 '12 at 6:10
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3 Answers

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Assuming you are exposed to the Euler's reflection formula: $$ \Gamma(x) \Gamma(1-x) = \frac{\pi}{\sin(\pi x)} \tag{1} $$ we get: $$ \lim_{x \to -n} (x+n)\Gamma(x) = \lim_{x\to-n} \frac{\pi (x+n)}{\sin(\pi x)}\frac{1}{\Gamma(1-x)} \stackrel{y=x+n}{=} \lim_{y \to 0} \frac{\pi y}{\sin(\pi y - \pi n)} \frac{1}{\Gamma(1 + n -y )} $$ Using identity $\sin(\pi y - \pi n) =(-1)^n \sin(\pi y)$, valid for integer $n$, we conclude $$ \lim_{x \to -n} (x+n)\Gamma(x) = \frac{(-1)^n}{\Gamma(1+n)} \, \underbrace{ \lim_{y \to 0} \frac{\pi y}{\sin(\pi y)} }_{ \lim_{z\to 0} \frac{z}{\sin(z)}=1}= \frac{(-1)^n}{n!} $$

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I can follow your work easiest since this formula is in my reference sheet. I don't understand the purpose of making the y = x+n substitution. Is it necessary? Can it be done without it? For $\Gamma (1+n-y)$ it disappeared from your second line when you transitioned to the third line to $\Gamma (1+n)$. Wasn't the $sin(\pi y - \pi n) = (-1)^n sin(\pi y)$ in the denominator? How did the $(-1)^n$ get moved to the numerator? –  renagade629 Oct 7 '12 at 5:47
    
@renagade629 There is no need to introduce $y$. I thought it would be easier. Anyway, moving $(-1)^n$ from the denominator to the numerator is done as follows: $$ \frac{1}{(-1)^n} = \frac{(-1)^n}{(-1)^{2n}}$$ but for integer $n$, $2n$ is even, and thus $(-1)^{2n} = 1$. –  Sasha Oct 7 '12 at 12:41
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Let $L$ denote the desired limit. Write $(x+N)\Gamma(x)=\dfrac{\Gamma(x)}{(x+N)^{-1}}$.

Recall that $\Gamma(x)$ has simple poles at the nonpositive integers, with residue at $x=-N$ equal to $l=\frac{(-1)^N}{N!}.$ Thus, near $x=-N,$ we can express $\Gamma(x)=\frac{l}{x+N}+h(x)$ with $h(x)$ holomorphic. We obtain

$$L=\lim_{x\to-N}\dfrac{\Gamma(x)}{(x+N)^{-1}}=\lim_{x\to-N}\dfrac{\frac{l}{x+N}+h(x)}{(x+N)^{-1}}=l+0=l,$$

as desired. Of course, the real crux of the question is to prove that the residue of the gamma function at $x=-N$ is $l.$ I'll leave this to you, because it really is a nice exercise on the idea of analytic continuation.

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Thanks I will look into it. I am taking an undergraduate course in differential equations and we are using the gamma function for bessel ODE and we are using this notation. So I know nothing of the terminology that you used. I just have a sheet with gamma properties. –  renagade629 Oct 7 '12 at 2:48
    
You're welcome! The important part is that $\Gamma(x)$ can be extended to a function on the complex numbers, minus the points $x=0,-1,-2,-3,\ldots.$ However, near each of these there is a Laurent expansion (like a Taylor series except allowing negative powers) of $\Gamma(x).$ So, in some small neighbourhood of $x=-N,$ we know that $\Gamma(x)=\frac{(-1)^N}{N!(x+N)}+a_0+a_1(x+N)+a_2(x+N)^2+\cdots,$ where the $a_i$ are constants. (In my answer, $h(x)=a_0+\cdots.$) This is extremely useful, used to prove identities such as Euler's reflection mentioned in another answer. –  Andrew Oct 7 '12 at 3:18
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Recalling the rising and falling factorials

$$ x^{(N)}=\frac{\Gamma(x+N)}{\Gamma(x)}\,,\quad (x)_N=\frac{\Gamma(x+1)}{\Gamma(x-N+1)}\,, \quad {(-a)}^{(N)} = {(-1)}^N {(a)}_{{N}}\,,$$

where $ \Gamma(x+1)= x! \,.$

$$ (x+N)\Gamma(x)= \frac{\Gamma(x)(x+N)\Gamma(x+N)}{\Gamma(x+N)}=\frac{\Gamma(x)}{\Gamma(x+N)}(x+N)!=\frac{(x+N)!}{x^{(N)}}\,,$$

$$\Rightarrow (x+N)\Gamma(x) = \frac{(x+N)!}{x^{(N)}} $$

Taking the limit of the above equation gives

$$ \lim_{n\to \infty} (x+N)\Gamma(x) = \frac{0!}{(-N)^{(N)}} = \frac{1}{(-1)^N(N)_N}= \frac{1}{(-1)^N N!}\,.$$

Note that, $(N)_N = N!\,,$ by the second identity of the falling factorial.

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