I was reading Gamma by Julian Havil and I got hung up on one of the alternate definitions it gave for the Gamma function. It then used that definition to derive an extension of Gamma that converges for all C sans negative integers.

It says that Euler (hey, BTW, what is the proper way to pronounce his name?) proposed this convoluted definition in 1729 in a letter:

[math]\Gamma _{r}(x)=\frac{r!r^{x}}{x(x+1)(x+2)...(x+r)}[/math]

[math]=\frac{r^{x}}{x(x+\frac{x}{1})(x+\frac{x}{2})...(x+\frac{x}{r})}[/math]

and related it to Gamma in this way:

[math]\Gamma(x) =\lim_{r \to \infty }\Gamma _{r}(x)[/math]

It justified this by showing that [imath]\lim_{r \to \infty }\Gamma _{r}(x)[/imath] satisfied the identity [imath]\lim_{r \to \infty }\Gamma _{r}(x+1) = x\lim_{r \to \infty }\Gamma _{r}(x)[/imath] just like the Gamma function, and then it showed that [imath]\Gamma (1) = \lim_{r \to \infty }\Gamma _{r}(1)[/imath].

This is sufficient proof that the two definitions are equal for natural numbers, but it doesn't prove to me that they are the same for all x > 0. Is anyone familiar with this definition? Could anyone point me in the direction of a proof or offer insight so I could prove it myself? Thanks

## Alternate Definitions of the Gamma Function

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### Alternate Definitions of the Gamma Function

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- jestingrabbit
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### Re: Alternate Definitions of the Gamma Function

Does your text anywhere mention that Gamma is the unique analytic function which satisfies the recurrence formula and [imath]\Gamma(1)=1[/imath]? Its true, even if it doesn't prove it, though I suspect it does.

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### Re: Alternate Definitions of the Gamma Function

Eastwinn wrote:Euler (hey, BTW, what is the proper way to pronounce his name?)

Oil-er

### Re: Alternate Definitions of the Gamma Function

^ I've been pronouncing it wrong then

Thank for the response. Yes, it does. It names a theorem and mentions that it's log convex. I'll look into that. If I could show that the limit is log convex then all matters would be settled.

I found a paper that proves it using the fact that Gamma is log convex and increasing for input greater than zero but I didn't quite understand it because I'm not too hip on this log convex stuff (yet).

jestingrabbit wrote:Does your text anywhere mention that Gamma is the unique analytic function which satisfies the recurrence formula and [imath]\Gamma(1)=1[/imath]? Its true, even if it doesn't prove it, though I suspect it does.

Thank for the response. Yes, it does. It names a theorem and mentions that it's log convex. I'll look into that. If I could show that the limit is log convex then all matters would be settled.

I found a paper that proves it using the fact that Gamma is log convex and increasing for input greater than zero but I didn't quite understand it because I'm not too hip on this log convex stuff (yet).

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### Re: Alternate Definitions of the Gamma Function

jestingrabbit wrote:Does your text anywhere mention that Gamma is the unique analytic function which satisfies the recurrence formula and [imath]\Gamma(1)=1[/imath]? Its true, even if it doesn't prove it, though I suspect it does.

Strictly speaking, is this correct as written, if you don't also say something about log-convexity? Couldn't you add an analytic function that's zero at the positive integers?

### Re: Alternate Definitions of the Gamma Function

Blatm wrote:Eastwinn wrote:Euler (hey, BTW, what is the proper way to pronounce his name?)

Oil-er

Actually, I would say it's 'Oi-ler'.

### Re: Alternate Definitions of the Gamma Function

Tyl. wrote:Blatm wrote:Eastwinn wrote:Euler (hey, BTW, what is the proper way to pronounce his name?)

Oil-er

Actually, I would say it's 'Oi-ler'.

You're right, that's probably a bit closer.

### Re: Alternate Definitions of the Gamma Function

skullturf wrote:jestingrabbit wrote:Does your text anywhere mention that Gamma is the unique analytic function which satisfies the recurrence formula and [imath]\Gamma(1)=1[/imath]? Its true, even if it doesn't prove it, though I suspect it does.

Strictly speaking, is this correct as written, if you don't also say something about log-convexity? Couldn't you add an analytic function that's zero at the positive integers?

I don't know whether it's true or not, but the theorem that says the gamma function is the unique function that's log-convex that satisfies the recurrence doesn't require analyticity (analyticness?), and the gamma function certainly is analytic, so that may be an equivalent classification.

The recurrence is true for all x, so you can't just add an arbitrary analytic function that's zero on the positive integers, so the statement is actually equivalent to saying that if an analytic function satisfies the recurrence and is 0 on the positive integers (or equivalently at 1, by the recurrence) then it is 0. Is this true? (JR suggests it is)

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### Re: Alternate Definitions of the Gamma Function

The Bohr-Mollerup Theorem says that there is only one function that satisfies f(x+1) = x f(x), f(1) = 1, and log-convexity, and that function is Gamma. I'm yet to understand the proof, however, but still if I can show that the alternative definition in the OP is log-convex then that will be sufficient proof.

Edit: It was simple to find that

[math]\frac{\mathrm{d}^{2} ln\Gamma_{r} (x)}{\mathrm{d}^{2} x} = \sum_{k=0}^{r}\frac{1}{(x+k)^{2}}[/math]

which is positive when x is positive, so that settles the issue -- the alternate definition in my original post is equivalent to the integral definition where convergence allows.

Here's something interesting though. Take the limit as r->infinity...

[math]\frac{\mathrm{d}^{2} ln\Gamma (x)}{\mathrm{d}^{2} x} = \sum_{k=0}^{\infty }\frac{1}{(x+k)^{2}}[/math]

And plug in 1 for x...

[math]\frac{\mathrm{d}^{2} ln\Gamma (1)}{\mathrm{d}^{2} x} = \sum_{k=0}^{\infty }\frac{1}{(1+k)^{2}} = \sum_{k=1}^{\infty }\frac{1}{k^{2}} = \frac{\pi ^{2}}{6}[/math]

Cool!

Edit: It was simple to find that

[math]\frac{\mathrm{d}^{2} ln\Gamma_{r} (x)}{\mathrm{d}^{2} x} = \sum_{k=0}^{r}\frac{1}{(x+k)^{2}}[/math]

which is positive when x is positive, so that settles the issue -- the alternate definition in my original post is equivalent to the integral definition where convergence allows.

Here's something interesting though. Take the limit as r->infinity...

[math]\frac{\mathrm{d}^{2} ln\Gamma (x)}{\mathrm{d}^{2} x} = \sum_{k=0}^{\infty }\frac{1}{(x+k)^{2}}[/math]

And plug in 1 for x...

[math]\frac{\mathrm{d}^{2} ln\Gamma (1)}{\mathrm{d}^{2} x} = \sum_{k=0}^{\infty }\frac{1}{(1+k)^{2}} = \sum_{k=1}^{\infty }\frac{1}{k^{2}} = \frac{\pi ^{2}}{6}[/math]

Cool!

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- jestingrabbit
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### Re: Alternate Definitions of the Gamma Function

skullturf wrote:

Strictly speaking, is this correct as written, if you don't also say something about log-convexity? Couldn't you add an analytic function that's zero at the positive integers?

You're right, you do need the convexity condition, or something like it. Something as weak as being defined for all positive reals might be enough, but you do need more than just analytic, the recurrence and the value. My mistake.

Edit: after thinking about this for a while, I think that you need something as strong as the log convexity, and nothing weaker will get you close. if you have some periodic function that's 1 on the integers, like [imath]f(z) = \exp(\sin(\pi z))[/imath], then [imath]g(z)= \Gamma(z)f(z)[/imath] will still be analytic everywhere that Gamma is and in this case zero nowhere just like Gamma. You need the log convexity, for sure.

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