15.5 - The Gamma Function

Provided \(t>1\):

\(\Gamma(t)=(t-1) \times \Gamma(t-1) \)

Proof

We'll useintegration by parts with:

\(u=y^{t-1}\) and \(dv=e^{-y}dy\)

to get:

\(du=(t-1)y^{t-2}\) and \(v=-e^{-y}\)

Then, the integration by parts gives us:

\(\Gamma(t)=\lim\limits_{b \to \infty} \left[-y^{t-1}e^{-y}\right]^{y=b}_{y=0} + (t-1)\int_0^\infty y^{t-2}e^{-y}dy\)

Evaluating at \(y=b\)and \(y=0\)for the first term, and using the definition of the gamma function (provided \(t-1>0\)) for the second term, we have:

\(\Gamma(t)=-\lim\limits_{b \to \infty} \left[\dfrac{b^{t-1}}{e^b}\right]+(t-1)\Gamma(t-1)\)

Now, if we were to be lazy, we would just wave our hands, and say that the first term goes to 0, and therefore:

\(\Gamma(t)=(t-1) \times \Gamma(t-1)\)

provided \(t>1\), as was to be proved.

Let's not be too lazy though! Taking the limit as \(b\)goes to infinity for that first term, we get infinity over infinity. Ugh! Maybe we should have left well enough alone! We can take the exponent and the natural log of the numerator without changing the limit. Doing so, we get:

\(-\lim\limits_{b \to \infty} \left[\dfrac{b^{t-1}}{e^b}\right] =-\lim\limits_{b \to \infty} \left\{\dfrac{\text{exp}[(t-1) \ln b]}{\text{exp}(b)}\right\}\)

Then, because both the numerator and denominator are exponents, we can write the limit as:

\(-\lim\limits_{b \to \infty} \left[\dfrac{b^{t-1}}{e^b}\right] =-\lim\limits_{b \to \infty}\{\text{exp}[(t-1) \ln b-b]\}\)

Manipulating the limit a bit more, so that we can easily apply L'Hôpital's Rule, we get:

\(-\lim\limits_{b \to \infty} \left[\dfrac{b^{t-1}}{e^b}\right] =-\lim\limits_{b \to \infty} \left\{\text{exp}\left[(t-1)b\left(\dfrac{ \ln b}{b}-1\right)\right]\right\}\)

Now, let's take the limit as \(b\)goes to infinity:

Okay, our proof is now officially complete! We have shown what we set out to show. Maybe next time, I'll just wave my hands when I need a limit to go to 0.

If \(t=n\), a positive integer, then:

\(\Gamma(n)=(n-1)!\)

Proof

Using the previous theorem:

\begin{align} \Gamma(n) &= (n-1)\Gamma(n-1)\\ &= (n-1)(n-2)\Gamma(n-2)\\ &= (n-1)(n-2)(n-3)\cdots (2)(1)\Gamma(1) \end{align}

And, since by the definition of the gamma function:

\(\Gamma(1)=\int_0^\infty y^{1-1}e^{-y} dy=\int_0^\infty e^{-y} dy=1\)

we have:

\(\Gamma(n)=(n-1)!\)

as was to be proved.