# Analytic continuation of the Riemann zeta function using contour integration

To find the analytic continuation of the Riemann zeta function using contour integration one can integrate $\displaystyle f(z) = \frac{z^{s-1}}{e^{-z}-1}$ around a contour that consists of rays just above and just below the negative real axis (where the branch cut is placed) and a small circle about the origin that is traversed counterclockwise.

Then the argument is that as long as radius of the circle remains small enough, the value of the contour integral is independent of the circle's radius.

Some books say that this follows from Cauchy's integral theorem. But why is Cauchy's integral theorem applicable if the contour is seemingly not closed?

• You close it by cutting of the infinite rays, and adding sides to make a "square" (avoiding the poles, of course). Then you check that the integrals over the sides of the big square tend to $0$ when you let the side length of the square tend to $\infty$ (and the sum of the residues converges). – Daniel Fischer Dec 1 '13 at 22:12
• I don't quite understand. Whether the integral remains the same when you vary the radius of the circle doesn't depend on the functional equation or anything, just on whether you cross a pole of the integrand. By the way, it's simpler to just consider the contour made up of the two circles and the parts of the straight lines connecting them, didn't think of that immediately. – Daniel Fischer Dec 1 '13 at 22:50

## 1 Answer

The contour is not closed, but the value of the integral is independent (for small radius) of the method you might use to close off the contour near $Re(z)=-\infty$ and take a limit.

Near $-\infty$ means modifying the contour only in a region $Re(z) < a$ for large enough negative $a$. In this problem any $a<0$ will leave unchanged the set of poles and residues inside the contour, and as $a \to -\infty$ any changes in the value of the integral due to the contour modification crossing the branch cut, converge to $0$.

The un-modified contour can be considered as a closed integration path on a compactification of the left half-plane by adding a point at real part $-\infty$, which is justified when integrating a power of $z$ that is suppressed exponentially (in $Re(z)$) when approaching the added point.

• So is it nonsense to invoke Cauchy's integral theorem when referring to the unmodified contour? – Random Variable Dec 2 '13 at 0:02
• It is no more and no less nonsense than talking about the Riemann integral of log(x) or 1/sqrt(x) on [0,1]. The Riemann integral is only defined for bounded functions and the objects in the Cauchy theorem are only defined for closed contours that satisfy some conditions like bounded, compact, maybe piecewise smooth. The conditions are not satisfied, but there is a natural way to take the limit of related computations where the theorems do apply, and in that limit things are well defined. – zyx Dec 2 '13 at 0:28
• You could also forget about the branch cut and start from a closed finite contour, but delete the arc of the contour from height $-h$ to $h$ that crosses the cut, for small enough $h$, then take the limit of such path-contours as they approach the one in the question, and observe that the part crossing the cut has vanishing contribution to the integral (in the limit). – zyx Dec 2 '13 at 0:40
• Is your last comment referring to the contour depicted in the proof of Fact 2 in the following pdf? people.math.gatech.edu/~xchen/teach/comp_analysis/note-zeta.pdf – Random Variable Dec 2 '13 at 1:49
• To the contour in the question, which opens in the other direction, toward $-\infty + 0i$, but the same argument would apply to the contour in the link. The point is that differences in the choice of branch, and effects from the infinite length of the contour, are suppressed by the exponential. – zyx Dec 2 '13 at 19:14