Monday Math 18: Gamma Function (Part 1/?) « Twisted One 151’s Weblog

Monday Math 18: Gamma Function (Part 1/?)

The gamma function $\operatorname{\Gamma}(z)$ is defined for all complex z except at z=0,-1,-2,-3,…, by $\operatorname{\Gamma}(z)=\int_{0}^{\infty}t^{z-1}e^{-t}\,dt$
We see first that $\operatorname{\Gamma}(1)=\int_{0}^{\infty}e^{-t}\,dt=1$ . Next, we can apply integration by parts to $\operatorname{\Gamma}(x)$ for real x to get:
$\begin{eqnarray}\operatorname{\Gamma}(x)&=&\int_{0}^{\infty}t^{x-1}e^{-t}\,dt\\&=&[-t^{x-1}e^{-t}]_{0}^{\infty}+\int_{0}^{\infty}(x-1)t^{x-2}e^{-t}\,dt\\&=&(x-1)\int_{0}^{\infty}t^{x-2}e^{-t}\,dt\\&=&(x-1)\operatorname{\Gamma}(x-1)\end{eqnarray}$ .
Thus for an integer n, we see $\operatorname{\Gamma}(n+1)=n\operatorname{\Gamma}(n)=n(n-1)\operatorname{\Gamma}(n-1)=\ldots=n!$
Thus the gamma function is an extension of the factorial to non-integer and non-real values.
We can also compute $\operatorname{\Gamma}(\frac{1}{2})$ :
$\operatorname{\Gamma}(\frac{1}{2})=\int_{0}^{\infty}\frac{e^{-t}}{\sqrt{t}}\,dt$ .
Making the u-substitution $u^2=t\to\,dt=2u\,du$ , we get:
$\begin{eqnarray}\operatorname{\Gamma}(\frac{1}{2})&=&2\int_{0}^{\infty}e^{-u^2}\,du\\&=&\int_{-\infty}^{\infty}e^{-u^2}\,du\\&=&\sqrt{\pi}\end{eqnarray}$ .
And so
$\operatorname{\Gamma}(\frac{3}{2})=\frac{\sqrt{\pi}}{2}$
$\operatorname{\Gamma}(\frac{5}{2})=\frac{3}{2}\frac{\sqrt{\pi}}{2}=\frac{3\sqrt{\pi}}{4}$
and, thus for integer n:
$\begin{eqnarray}\operatorname{\Gamma}(\frac{1}{2}+n)&=&\frac{1\cdot3\cdot\ldots\cdot(2n-1)}{2^n}\sqrt{\pi}\\&=&\frac{(2n)!}{2^{2n}n!}\sqrt{\pi}\end{eqnarray}$
We also have
$\operatorname{\Gamma}(-\frac{1}{2})=\frac{\operatorname{\Gamma}(\frac{1}{2})}{-\frac{1}{2}}=-2\sqrt{\pi}$
$\operatorname{\Gamma}(-\frac{3}{2})=\frac{\operatorname{\Gamma}(-\frac{1}{2})}{-\frac{3}{2}}=\frac{4}{3}\sqrt{\pi}$
and so on.

If we take the original integral and make the substitution $u=e^{-t}$ , we get:
$\operatorname{\Gamma}(z)=\int_{1}^{0}(-\ln{u})^{z-1}u\frac{-du}{u}$
$\operatorname{\Gamma}(z)=\int_{0}^{1}(-\ln{u})^{z-1}\,du$
And thus for an integer n>0,
$\int_{0}^{1}(\ln{u})^{n}\,du=(-1)^{n}\operatorname{\Gamma}(n+1)=(-1)^{n}n!$

Also, according to the Bohr-Mollerup theorem, the gamma function is the only function on x>0 for which the following three properties hold:
I. $f(1)=1$
II. $f(x+1)=xf(x)$
III. $\ln(f(x))$ is convex.

Tags: Math, Monday Math, Factorial, Gamma Function

This entry was posted on May 5, 2008 at 1:00 am and is filed under Math/Science. You can follow any responses to this entry through the RSS 2.0 feed. You can leave a response, or trackback from your own site.

6 Responses to “Monday Math 18: Gamma Function (Part 1/?)”

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Monday Math 18: Gamma Function (Part 1/?)

6 Responses to “Monday Math 18: Gamma Function (Part 1/?)”

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