**The Gamma Function Part 4: Euler Reflection Formula and the Sine Function:**

(Part 1)

(Part 2)

(Part 3)

Let us now define the function , for non-integer x, as . Now, we remember that , and if we replace *x* with –*x*, we see that , and so . Thus, when we examine , we see:

So 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 as:

or thus

The gamma functions both have values of unity at x=0, and the singularity at the origin for the fraction is removable; expanding the sine in its Maclaurin series, we have:

Which gives , and thus we have for x an integer as well.

From part 3, we had the Legendre duplication formula:

.

If we replace *x* with 1-*x*, we obtain:

.

Using these, we see:

.

Now, define . Then *g*(*x*) is periodic, since *φ* is periodic. Now, we can apply the duplication formula for *φ* to see:

Now, since *g*(*x*) is continuous on the closed interval [0,1], it is bounded; there is a constant *M* such that for (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*):

And thus *g*(*x*) can be bounded by . Repeating this process, we obtain bound . We can continue repeating; the result is that the bound of *g*(*x*) goes to zero. Thus , which tells us that is linear. As it is periodic and continuous, it must then be a constant; and thus is a constant. Since , we thus see that for all *x*. Using the definition of , we have the Euler reflection formula:

Solving the reflection formula for , we have:

.

Now, using , we find that

In part 2, we found the Weierstrass form of the gamma function:

Plugging in –*x*, we obtain:

And thus

Plugging this into our formula for :

.

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

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

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