Monday Math 21: The Gamma Function (Part 4/?)

By twistedone151

The Gamma Function Part 4: Euler Reflection Formula and the Sine Function:
(Part 1)
(Part 2)
(Part 3)

Let us now define the function $\phi(x)$ , for non-integer x, as $\phi(x)\equiv\operatorname{\Gamma}(x)\operatorname{\Gamma}(1-x)\sin{\pi{x}}$ . Now, we remember that $\operatorname{\Gamma}(x+1)=x\operatorname{\Gamma}(x)$ , and if we replace x with -x, we see that $\operatorname{\Gamma}(1-x)=-x\operatorname{\Gamma}(-x)$ , and so $\operatorname{\Gamma}(-x)=-\frac{\operatorname{\Gamma}(1-x)}{x}$ . Thus, when we examine $\phi(x+1)$ , we see:
$\begin{eqnarray}\phi(x+1)&=&\operatorname{\Gamma}(x+1)\operatorname{\Gamma}(-x)\sin(\pi(x+1))\\&=&\left(x\operatorname{\Gamma}(x)\right)\left(-\frac{\operatorname{\Gamma}(1-x)}{x}\right)\sin(\pi{x}+\pi)\\&=&-\operatorname{\Gamma}(x)\operatorname{\Gamma}(1-x)\left(-\sin\pi{x}\right)\\&=&\operatorname{\Gamma}(x)\operatorname{\Gamma}(1-x)\sin\pi{x}\\&=&\phi(x)\end{eqnarray}$
So $\phi(x)$ is periodic with a period of one.

Now, we examine the behavior of φ near zero (and thus near all integers due to periodicity). We can write $\phi(x)$ as:
$\phi(x)=\frac{\operatorname{\Gamma}(1+x)}{x}\operatorname{\Gamma}(1-x)\sin{\pi{x}}$
or thus
$\phi(x)=\operatorname{\Gamma}(1+x)\operatorname{\Gamma}(1-x)\frac{\sin{\pi{x}}}{x}$
The gamma functions both have values of unity at x=0, and the singularity at the origin for the fraction $\frac{\sin{\pi{x}}}{x}$ is removable; expanding the sine in its Maclaurin series, we have:
$\begin{eqnarray}\phi(x)&=&\operatorname{\Gamma}(1+x)\operatorname{\Gamma}(1-x)\frac{1}{x}\left[\pi{x}-\frac{(\pi{x})^3}{3!}+\frac{(\pi{x})^5}{5!}-\ldots\right]\\&=&\operatorname{\Gamma}(1+x)\operatorname{\Gamma}(1-x)\left[\pi-\frac{\pi^3x^2}{3!}+\frac{\pi^5x^4}{5!}-\ldots\right]\end{eqnarray}$
Which gives $\phi(0)=\pi$ , and thus we have $\phi(x)$ for x an integer as well.

From part 3, we had the Legendre duplication formula:
$\operatorname{\Gamma}\left(\frac{x}{2}\right)\operatorname{\Gamma}\left(\frac{x+1}{2}\right)=\frac{\sqrt{\pi}}{2^{x-1}}\operatorname{\Gamma}(x)$ .
If we replace x with 1-x, we obtain:
$\operatorname{\Gamma}\left(\frac{1-x}{2}\right)\operatorname{\Gamma}\left(1-\frac{x}{2}\right)=\sqrt{\pi}{2^{x}}\operatorname{\Gamma}(1-x)$ .
Using these, we see:
$\begin{eqnarray}\phi\left(\frac{x}{2}\right)\phi\left(\frac{x+1}{2}\right)&=&\operatorname{\Gamma}\left(\frac{x}{2}\right)\operatorname{\Gamma}\left(1-\frac{x}{2}\right)\sin{\frac{\pi{x}}{2}}\operatorname{\Gamma}\left(\frac{x+1}{2}\right)\operatorname{\Gamma}\left(\frac{1-x}{2}\right)\cos{\frac{\pi{x}}{2}}\\&=&\operatorname{\Gamma}\left(\frac{x}{2}\right)\operatorname{\Gamma}\left(\frac{x+1}{2}\right)\operatorname{\Gamma}\left(\frac{1-x}{2}\right)\operatorname{\Gamma}\left(1-\frac{x}{2}\right)\sin{\frac{\pi{x}}{2}}\cos{\frac{\pi{x}}{2}}\\&=&\left[\frac{\sqrt{\pi}}{2^{x-1}}\operatorname{\Gamma}(x)\right]\left[\sqrt{\pi}{2^{x}}\operatorname{\Gamma}(1-x)\right]\left[\frac{1}{2}\sin{\pi{x}}\right]\\&=&\pi\operatorname{\Gamma}(x)\operatorname{\Gamma}(1-x)\sin\pi{x}\\&=&\pi\phi(x)\end{eqnarray}$ .

Now, define $g(x)\equiv\frac{d^2}{dx^2}\ln\phi(x)$ . Then g(x) is periodic, since φ is periodic. Now, we can apply the duplication formula for φ to see:
$\begin{eqnarray}g(x)&=&\frac{d^2}{dx^2}\ln\phi(x)\\&=&\frac{d^2}{dx^2}\ln\left(\frac{1}{\pi}\phi\left(\frac{x}{2}\right)\phi\left(\frac{x+1}{2}\right)\right)\\&=&\frac{d^2}{dx^2}\left[\ln\phi\left(\frac{x}{2}\right)+\ln\phi\left(\frac{x+1}{2}\right)-\ln\pi\right]\\&=&\frac{d^2}{dx^2}\ln\phi\left(\frac{x}{2}\right)+\frac{d^2}{dx^2}\ln\phi\left(\frac{x+1}{2}\right)\\&=&\frac{1}{4}\left(g\left(\frac{x}{2}\right)+g\left(\frac{x+1}{2}\right)\right)\end{eqnarray}$

Now, since g(x) is continuous on the closed interval [0,1], it is bounded; there is a constant M such that $|g(x)|\le{M}$ for $0\le{x}\le1$ (see the ‘boundedness theorem’ here). As g(x) is periodic, we see it is thus bounded by M for all x. Applying our previous result for g(x):
$\begin{eqnarray}|g(x)|&=&\frac{1}{4}\left|g\left(\frac{x}{2}\right)+g\left(\frac{x+1}{2}\right)\right|\\&\le&\frac{1}{4}\left(\left|g\left(\frac{x}{2}\right)\right|+\left|g\left(\frac{x+1}{2}\right)\right|\right)\\&\le&\frac{1}{4}\left(M+M\right)\\&\le&\frac{M}{2}\end{eqnarray}$
And thus g(x) can be bounded by $\frac{M}{2}$ . Repeating this process, we obtain bound $\frac{M}{4}$ . We can continue repeating; the result is that the bound of g(x) goes to zero. Thus $g(x)=0$ , which tells us that $\ln\phi(x)$ is linear. As it is periodic and continuous, it must then be a constant; and thus $\phi(x)$ is a constant. Since $\phi(0)=\pi$ , we thus see that $\phi(x)=\pi$ for all x. Using the definition of $\phi(x)$ , we have the Euler reflection formula:
$\operatorname{\Gamma}(x)\operatorname{\Gamma}(1-x)=\frac{\pi}{\sin\pi{x}}$

Solving the reflection formula for $\sin\pi{x}$ , we have:
$\sin\pi{x}=\frac{\pi}{\operatorname{\Gamma}(x)\operatorname{\Gamma}(1-x)}$ .
Now, using $\operatorname{\Gamma}(1-x)=-x\operatorname{\Gamma}(-x)$ , we find that
$\sin\pi{x}=-\frac{\pi}{x\operatorname{\Gamma}(x)\operatorname{\Gamma}(-x)}$
In part 2, we found the Weierstrass form of the gamma function:
$\operatorname{\Gamma}(z)=e^{-\gamma{z}}\frac{1}{z}\prod_{k=1}^{n}\frac{e^{z/k}}{1+\frac{z}{k}}$
Plugging in -x, we obtain:
$\operatorname{\Gamma}(-x)=-e^{\gamma{x}}\frac{1}{x}\prod_{k=1}^{n}\frac{e^{-x/k}}{1-\frac{x}{k}}$
And thus
$\begin{eqnarray}\operatorname{\Gamma}(x)\operatorname{\Gamma}(-x)&=&\left(e^{-\gamma{x}}\frac{1}{x}\prod_{k=1}^{n}\frac{e^{x/k}}{1+\frac{x}{k}}\right)\left(-e^{\gamma{x}}\frac{1}{x}\prod_{k=1}^{n}\frac{e^{-x/k}}{1-\frac{x}{k}}\right)\\&=&-\frac{1}{x^2}\prod_{k=1}^{n}\frac{1}{(1+\frac{x}{k})(1-\frac{x}{k})}\\&=&-\frac{1}{x^2}\prod_{k=1}^{n}\frac{1}{1-\frac{x^2}{k^2}}\end{eqnarray}$
Plugging this into our formula for $\sin\pi{x}$ :
$\begin{eqnarray}\sin\pi{x}&=&-\frac{\pi}{x\cdot\left(-\frac{1}{x^2}\right)\prod_{k=1}^{n}\frac{1}{1-\frac{x^2}{k^2}}}\\&=&\pi{x}\prod_{k=1}^{n}\left(1-\frac{x^2}{k^2}\right)\end{eqnarray}$ .

Thus, we have an expression for the sine function as an infinite product.

Tags: Euler Reflection Formula, Gamma Function, Math, Monday Math, Sine Product Formula

This entry was posted on May 26, 2008 at 12:01 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.

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Monday Math 21: The Gamma Function (Part 4/?)

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