gamma function question for x < 0

[dts5044]

so I know the gamma function defined:
T(x) = [integral from 0 to infinity] t^(x-1)e^(-t)dt

converges for x > 0

I also know that T(x+1) = xT(x)

does this make it possible to extend the T(x) function for x <0? (except the negative integers), and if so, why is it okay to apply this identity even though T(x) is not defined (divergent) for x <= 0?

This is more of a conceptual question, so if anyone understands this I would love to hear an explanation. Thanks!

 

[galactus]

\(\displaystyle {\Gamma}=\int_{0}^{\infty}t^{x-1}e^{-t}dt\)

For 0<x<1, it is an improper integral because \(\displaystyle t^{x-1}\) becomes infinite at the lower limit.

But, it is convergent for x>0. For \(\displaystyle x\leq 0\) it is not defined because the integral diverges.

For \(\displaystyle x\leq 0\), \(\displaystyle {\Gamma}(x)\) has not so far been defined. BUT, let's define it by the recursion relation

\(\displaystyle {\Gamma}(x+1)=x{\Gamma}x\). Which is what you're onto.

\(\displaystyle {\Gamma}(x)=\frac{1}{x}{\Gamma}(x+1)\) defines \(\displaystyle {\Gamma}(x)\) for x<0. .........[1]

Here is an example:

\(\displaystyle {\Gamma}(-0.5)=\frac{1}{-0.5}{\Gamma}(0.5)\approx -3.545....\)

\(\displaystyle {\Gamma}(-1.5)=\frac{1}{-1.5}\cdot \frac{1}{{\Gamma}(0.5)}\approx 2.363....\) and so on.

Since \(\displaystyle {\Gamma}(1)=1\), we see that \(\displaystyle {\Gamma}(x)=\frac{{\Gamma}(x+1)}{x}\rightarrow{\infty} \;\ as \;\ x\rightarrow 0\)

From this and successive uses of [1], we see that \(\displaystyle {\Gamma}(x)\) becomes infinite not only at 0 but also at all negative integers.

For positive x, \(\displaystyle {\Gamma}(x)\) is a continuous function passing through the points \(\displaystyle x=n, \;\ {\Gamma}(x)=(n-1)!\).

For negative x, \(\displaystyle {\Gamma}(x)\) is discontinuous at the negative integers. In the intervals between the integers it alternates between positive and negative: negative from 0 to -1, positive from -1 to -2, and so on.

We can also use the identity, which I will not bother deriving, \(\displaystyle {\Gamma}(-x)=\frac{{-\pi}}{x\cdot {\Gamma}(x)sin({\pi}x)}\)

As can be seen, if \(\displaystyle x\in Z\), then the denominator is 0 because sin is 0 for integer multiples of Pi.

Was that a nice tutorial?. I have always been interested in the gamma and beta functions. They are cool and can be used to evaluate integrals which are otherwise difficult using elementary methods.

 

[galactus]

I have studied Gamma quite a bit over the years. Here is something I thought I would impart because I think it is very cool.

Like a lot of things in math, we find connections between seemingly unrelated things. Are you familiar with the zeta function?.

How about the Digamma function?.

The Digamma function, a.k.a Psi function is \(\displaystyle {\psi}(x)=\frac{{\Gamma}'(x)}{{\Gamma}(x)}\)

Here is a connection with zeta and gamma:

\(\displaystyle {\Gamma}(x)=\int_{0}^{\infty}t^{x-1}e^{-t}dt\)

make a change of variables t=ru:

\(\displaystyle {\Gamma}(x)=\int_{0}^{\infty}(ru)^{x-1}e^{-ru}rdu=r^{x}\int_{0}^{\infty}u^{x-1}e^{-ru}du\)

Therefore, \(\displaystyle \frac{1}{r^{x}}=\frac{1}{{\Gamma}(x)}\int_{0}^{\infty}u^{x-1}e^{-ru}du\)

\(\displaystyle {\zeta}(x)=\sum_{r=1}^{\infty}\frac{1}{r^{x}}=\frac{1}{{\Gamma}(x)}\sum_{r=1}^{\infty}\int_{0}^{\infty}u^{x-1}e^{-ru}du\)

\(\displaystyle =\frac{1}{{\Gamma}(x)}\int_{0}^{\infty}u^{x-1}\sum_{r=1}^{\infty}e^{-ru}du\)

There, we pushed the sigma through the integral, and summing the infinite geometric series gives us:

\(\displaystyle {\zeta}(x)=\frac{1}{{\Gamma}(x)}\int_{0}^{\infty}u^{x-1}\frac{e^{-u}}{1-e^{-u}}du\)

And we get a wonderful formula:

\(\displaystyle \Large\boxed{{\zeta}(x){\Gamma}(x)=\int_{0}^{\infty}\frac{u^{x-1}}{e^{u}-1}du}\)

Which is valid for \(\displaystyle x\notin {1,0,-1,-2,.....}\)

That may be more than you wanted to know, but you touched on a topic I find interesting.

Another cool thing to investigate is the gamma constant, \(\displaystyle {\gamma}\)

\(\displaystyle {\gamma}={\displaystyle}\lim_{n\to {\infty}}\left(\sum_{k=1}^{n}\frac{1}{k}-ln(n)}\right)\)

\(\displaystyle {\gamma}\approx \sum_{k=1}^{n}\frac{1}{k}-\int_{1}^{n}\frac{1}{x}dx\)

 

[dts5044]

haha don't worry, nothing is ever more than I want to know when it comes to math! I love it! Thanks for the help both of you, I don't think I could have asked for better answers!