trying to catch something about the gamma function.

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trying to catch something about the gamma function.

Mathematics
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\(\large\color{black}{ \displaystyle \Gamma(-x);~~{\rm where}~x\in {\bf Z},~x>0}\) this does not have a value, but rather an asymptote. (based on the graph and wolf) Why is that?
well, x≥0, really...
is there anything i can try to dig, like to write a power series representation for it and see what happens at particular values of x, or what should i try?

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well, recall that \(\Gamma\) is analytically continued from the factorial: $$\Gamma(n+1)=n!$$and satisfies a similar function equation: $$\Gamma(t+1)=t\Gamma(t)$$
yes, i know these two facts....
http://www.iiste.org/Journals/index.php/MTM/article/viewFile/6988/7036 I think this article has some good points.
Page 3, equation (6).
I have a little thought..... Factorial, working forward 1!=1 2!=1•2=2 3!=1•2•3=6 4!=1•2•3•4=24 Now working backwards 3!=(1•2•3•4)÷4=24÷4=6 2!=(1•2•3)÷3=6÷3=2 1!=(1•2)÷2=2÷2=1 going back further 0!=1÷1=1 (-1)! (really a gamma of -2, but....) = 1÷0 ... bingo?
Chill Out, I wll read. tnx
oh it hits that idea, you end up dividing by 0! for negative x values, and this is why it diverges....
i mean by 0, and exclamation mark of mine is not to denote factorial
will see how much i can understand from that..... tnx for googling for me..... If any sugestions, everyone is welcome to say anything.... (even to study more before posting this question), of course....
I'm bad with analysis, just learning with this as much as I can :D
it always has references.... the author didn't want to make it easy, no, rather professional:O
lol, that identity is all you need: $$\Gamma(t+1)=t\Gamma(t)\implies \Gamma(t)=\frac1t\Gamma(t+1)\\\implies \Gamma(t)=\frac1{t(t+1)(t+2)\dots (t+k)}\Gamma(t+k+1)$$ so if are interested in extending \(\Gamma\) to negative numbers \(t\), we just need to pick a \(k\) big enough so that \(t+k+1\) is in the domain of the already defined \(\Gamma\)
now you'll notice that if we try to do this for negative integers, we'll have trouble -- the only way to make \(t+k+1>0\) for negative integer \(t\), positive integer \(k\) will necessarily mean that one of those factors in the denominator *will* be zero, so our definition will blow up and does not apply to negative integers
and the integral definition of the gamma function just follows from observations of integration by parts with sufficiently 'small' growth: $$\begin{align*}\int_1^\infty e^{-t} f(t)\, dt&=-e^{-t}f(t)\big|_1^\infty-\int_1^\infty (-e^{-t}) f'(t)\,dt\\&=\frac1e f(1)+\int_1^\infty e^{-t} f'(t)\,dt\\&=\frac1e\left(f(1)+f'(1)+f''(1)+\dots \right)\end{align*}$$now imagine if we took \(f(t)=t^n\) we know the derivatives look like: $$f(1)=1\\f'(1)=n\\f'(2)=n(n-1)\\\dots\\f^{(n)}(1)=n(n-1)(n-2)\cdots 1\\f^{(n+1)}(1)=0\\f^{(n+2)}(1)=0\\\dots$$ so consider \(f(t)=et^n\):$$\int e^{-t}\cdot et^n\, dt=1+n+n(n-1)+\dots+n(n-1)(n-2)\cdots1$$
oops, that last integral should be on \((1,\infty)\). anyways, now that is more or less like the Laplace transform: $$\int_0^\infty e^{-st} t^n\, dt =\frac{n!}{s^{n+1}}$$let \(s=1\) so $$\int_0^\infty e^{-t} t^n\, dt=n!$$

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