Integrating a Function with Multiple Simple Poles

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    We want to derive the product theorem of Karl Theodor Wilhelm Weierstrass (1815 Oct 31 B 1897 Feb 19). That theorem solves a problem that comes from the fundamental theorem of algebra.

    The fundamental theorem of algebra tells us that if some polynomial function of a complex number, f(z), has different values for different values of the complex number z=x+iy, then there must exist at least one value of z for which f(z)=0; that is, every variable polynomial function of the complex numbers has roots. That fact leads directly to the fact that any n-th order polynomial function of a complex number has the form of an n-fold product,

(Eq=n 1)

in which A represents an arbitrary constant and zk represents the k-th point in the complex plane where the function takes the value of zero.

    Look at the converse of that proposition. If we have a set of points in the complex plane, {zk}, then Equation 1 tells us how to create a polynomial function valid over the whole complex plane with those points representing its roots. That function satisfies the criterion that makes it what mathematicians call an entire function: it is analytic over the entire complex plane, which means that we can differentiate it anywhere and everywhere on the complex plane. Weierstrass asked whether that fact remains true to mathematics if we let n go to infinity. The answer he devised says that it doesn=t, because the infinite product does not converge to a finite value everywhere on the complex plane. If, however, each factor approaches one as n goes to infinity, then the infinite product will converge properly. Weierstrass devised just the factors whose infinite product does, indeed, converge properly everywhere on the complex plane. We can find those factors ourselves by way of Cauchy=s integral formula and the residue theorem.

    We take g(z) to represent a function that is analytic everywhere in the complex plane except at the points zk occupied by isolated, first-order (or simple) poles. Mathematicians call that a meromorphic function. I use g(z) to represent that meromorphic function, the one with poles, because I want to find f(z) as a function that has zeroes at the points zk and no singularities anywhere. For the indices on the poles we have 1# k<4 with the proviso that we apply the indices to the poles in the order in which the absolute magnitudes of the poles= location vectors increase, thereby ensuring that the contour that we use to cut them from the complex plane does not cross itself. In this situation, then, we have

(Eq=n 2)

In that equation Cn represents a circle, centered on the origin of the complex plane, that contains within it the first n poles of the function. Because, by hypothesis, all of the poles are of first order, we calculate their residues as

(Eq=n 3)

We can thus rewrite Equation 2 as

(Eq=n 4)

We now eliminate the integral from that equation by exploiting the fact that we can multiply the integrand by 1=Z/Z without affecting the integration. We then subtract and add the same term proportional to z and get

(Eq=n 5)

But Equation 2 with z=0 gives us

(Eq=n 6)

which allows us to replace the first integral on the right side of Equation 5. As for the second integral on the right side of Equation 5, if the magnitude of g(Z) approaches a finite value as the absolute value of Z goes to infinity, then g(Z)/Z(Z-z) decreases faster than 2πZ increases as Z goes to infinity, which means that the integral goes to zero as a limit as the contour Cn grows to include the entire complex plane (and, by the way, all of the poles we have specified). With those facts in mind we can rewrite Equation 4 as

(Eq=n 7)

which we call the Mittag-Leffler expansion of the meromorphic function g(z).

    That analysis doesn=t seem to do us much good, until we remember that an analytic function with first-order poles stands in direct proportion to the reciprocal of an analytic function that has simple zeroes at all of the points zk. Further, g(z) is not a constant, so in accordance with Liouville=s theorem there must be at least one point on the complex plane, not coinciding with any of the points zk, at which it takes the value of zero. But f(z) has no singularities, so we must have

(Eq=n 8)

in which f(z) provides the poles of g(z) by way of its zeroes and φ(z) gives g(z) zeroes that don=t put singularities into f(z). We must, of course, ensure that φ(z) does not equal zero at any of the points zk. We meet that criterion most expediently by making

(Eq=n 9)

That equation also makes g(z) uniquely dependent upon f(z) and vice versa, which means that we can use Equation 7 to determine the form of f(z) without encountering any need to introduce other considerations.

    It also simplifies our calculation of the residues in Equation 7. Recall that for g(z)=1/z we have

(Eq=n 10)

In light of Equations 8 and 9 we then have in the present case

(Eq=n 11)

We know that equation stands true to mathematics because we know, from our basic calculus, that any segment of any curve representing an analytic function becomes indistinguishable from a straight line as its length goes to zero. Thus, in any of the minuscule patches that we use to isolate the poles of g(z) we have, as the patch=s radius approaches zero, f(z)=Bz, in which B represents some constant, with arbitrary precision. If we incorporate that fact into Equation 11, then B appears in the integrand in both the dividend and the divisor, thereby canceling out, and we get Equation 10.

    Making the appropriate substitution from Equations 8, 9, and 11, we can now rewrite Equation 7 as

(Eq=n 12)

If we multiply that equation by dz and integrate it, we get

(Eq=n 13)

which gives us, in turn,

(Eq=n 14)

in which

(Eq=n 15)

Equation 14 expresses the Weierstrass product theorem.

Example Application:

The Sine as an Infinite Product

    We start with the infinite-series expansion of

(Eq=n 16)

From that equation we get f(0)=a and [df/dz]z=0=0 along with the fact that f(z) has no singularities anywhere on the complex plane. We also know that the function has zeroes at all the points zk=" kπ. Thus we have Equation 14 as

(Eq=n 17)

But Sinz gives us a pure sine function only for purely real values of z, so we must eliminate the imaginary part, replacing z with x. Then multiplying the result by x gives us

(Eq=n 18)

Thus we obtain the infinite-product representation of the sine.

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