# Residue theorem for contour lying in multiple Riemann sheets?

Suppose you have a contour that passes over multiple Riemann sheets, but eventually comes back to itself (it's a closed contour). For example, the function $$\sqrt{z}$$ is analytic on the Riemann surface $$\Omega$$ consisting of two Riemann sheets (each identical to $$\mathbb{C}$$) which are glued along the positive real axis $$[0,\infty)$$. If I were to have a function $$f(z)=g(z)\sqrt{z}$$, where $$g(z)$$ has isolated poles at some $$\{z_i\}\in\Omega$$, how would Cauchy's residue theorem change if I consider contours which wind around both Riemann sheets? Now it seems a little difficult for me to define "inside" and "outside" the contour.

The reason I'm interested in this is because I'm trying to evaluate an integral by contour integration where the integrand actually has 4 branch cuts and 4 Riemann sheets.

$$\lim_{M\rightarrow\infty}\int_{-\infty}^{\infty} \frac{1}{(z^2+\frac{a^2}{M^2}+i\epsilon)(z^2+b^2M^2+i\epsilon)}e^{-\sqrt{z^2+\frac{a^2}{M^2}}-\sqrt{z^2+b^2M^2}} dz$$

I see no way to get a simple answer for this by constructing a contour which dodges the branch cuts. Therefore I'm thinking that, maybe, I can construct a contour which actually jumps through multiple sheets, and picks up multiple poles (on different sheets) via a 'modified' residue theorem.

• I don't know how to compute your integral using residues - the contour enters an essential singularity at $z=\infty$ and I don't see how to modify it. But anyway, residue theorem for Riemann surfaces says that if a contour is the boundary of a (compact) part of the surface (i.e. if it cuts the surface to 2 parts) and if your function (or rather the 1-form $f\,dz$) has only isolated singularities in that part, then the integral is given by the sum of the residues as usual. – user8268 Oct 24 at 19:15
• @user8268 The contour doesn't need to enter the essential singularity at $\infty$. You could just take a contour which gets asymptotically close to it. And for the residue theorem - it's that simple? I guess I just have a hard time visualizing what the "inside" and "outside" of my contours (which pass through multiple Riemann sheets) are. Would you be able to point me to literally any sources, even discussions, on this topic? Specifically for simple Riemann surfaces like what I'm considering, i.e. a (finite) bunch of $\mathbb{C}$'s glued together along branch cuts. – Arturo don Juan Oct 24 at 21:06

## 2 Answers

I'll assume $$a > 0, \, b > 0, \, \epsilon \geq 0$$. Let $$A = z^2 + (a/M)^2$$ and $$B = z^2 + (b M)^2$$. Since $$\left| \frac {e^{-\sqrt A - \sqrt B}} {(A + i \epsilon) (B + i \epsilon)} \right| \leq \frac {e^{-|z| - b M}} {(a/M)^2 (b M)^2},$$ the limit of the integral is zero. If the question is about the asymptotic behaviour of the integral, we have to consider the cases $$\epsilon > 0$$ and $$\epsilon = 0$$ separately. The non-exponential part is $$\frac 1 {(A + i \epsilon) (B + i \epsilon)} = \frac {M^4} {b^2 M^4 - a^2} \left( \frac 1 {M^2 (A + i \epsilon)} - \frac 1 {M^2 (B + i \epsilon)} \right).$$ Consider the $$e^{-\sqrt A - \sqrt B}/(M^2 (A + i \epsilon))$$ term first. If $$\epsilon = 0$$, then, setting $$z = x/M$$, we obtain $$\lim_{M \to \infty} M e^{b M} \int_{\mathbb R} \frac {e^{-\sqrt A - \sqrt B}} {M^2 A} dz = \int_{\mathbb R} \frac {dx} {x^2 + a^2},$$ since $$b M -\sqrt A - \sqrt B \rvert_{z = x/M}$$ is negative and tends to zero when $$M \to \infty$$. If $$\epsilon > 0$$, then $$\lim_{M \to \infty} M^2 e^{b M} \int_{\mathbb R} \frac {e^{-\sqrt A - \sqrt B}} {M^2 (A + i \epsilon)} dz = \int_{\mathbb R} \frac {e^{-|z|}} {z^2 + i \epsilon} dz.$$ It can be shown in the same way that the integral of the $$e^{-\sqrt A - \sqrt B}/(M^2 (B + i \epsilon))$$ term is asymptotically smaller. Therefore $$f(M) \sim \cases { \frac {\pi e^{-b M}} {a b^2 M} & \epsilon = 0 \\ \vphantom {\displaystyle \int} \frac {2 e^{-b M}} {b^2 M^2} \! \int_{\mathbb R^+} \hspace {-1px} \frac {e^{-z}} {z^2 + i \epsilon} dz & \epsilon > 0}, \quad M \to \infty.$$

Putting aside the issue of the residue theorem on more complicated Riemann surfaces, I can actually solve my original integral for the leading order behavior in $$M$$ using Laplace's method. First, make the substitution $$z=Mx$$

\begin{align} f(M)&=\int_{-\infty}^{\infty}\frac{1}{\left(z^2+\frac{a^2}{M^2}\right)\left(z^2+b^2M^2\right)}e^{-\sqrt{z^2+\frac{a^2}{M^2}}-\sqrt{z^2+b^2M^2}}\,dz\\ &=M^{-3}\int_{-\infty}^{\infty}\frac{1}{\left(x^2+\frac{a^2}{M^4}\right)\left(x^2+b^2\right)}e^{-M\left[\sqrt{x^2+\frac{a^2}{M^4}}+\sqrt{x^2+b^2}\right]}\,dx \\ &=M^{-3}\int_{-\infty}^{\infty}h(x;M)\,e^{Mg(x;M)}\,dx \end{align}

where

$$h(x;M)=\frac{1}{\left(x^2+\frac{a^2}{M^4}\right)(x^2+b^2)}$$

$$g(x;M)=-\left[\sqrt{x^2+\frac{a^2}{M^4}}+\sqrt{x^2+b^2}\right]$$

Since $$g(x;M)$$ has a unique maximum value for all $$M$$, namely $$x_0=0$$, we are in a perfect position to apply Laplace's method directly. Applying it gives us:

$$\boxed{f(M)\underset{M\rightarrow \infty}{\approx}\frac{1}{ab}\sqrt{\frac{2\pi}{ab^2M}}e^{-bM}}$$

One can see that $$f(M\rightarrow\infty)\rightarrow 0$$, in agreement with @Maxim's answer.

• I assume that you mean $z = M x$, then $f(M) = M^{-3} \int_{-\infty}^\infty h e^{M g} dx$ with your $h$ and $g$. Then expanding $h$ and $g$ around zero gives $f(M) \sim a^{-3/2} b^{-2} \sqrt {2 \pi/M} e^{-b M}$. This does not check out numerically though. You cannot apply Laplace's method directly, both $h$ and $g$ depend on $M$ in a non-trivial way. You have to prove that the method is still applicable, and my claim is that it isn't (not directly). – Maxim Oct 25 at 21:20
• @Maxim Yes sorry I did mean that. Also I made a few typos. I've corrected and edited my answer. – Arturo don Juan Oct 25 at 22:10
• @Maxim I do see that Laplace's method is not technically directly applicable, since $h(x)$ and $g(x)$ do still depend on $M$. Moreover, I see that if this non-trivial dependence directly violates one of the base assumptions going into Laplace's method, it will invalidate its use. I'm trying to come up with a reason for why I can, or can't, use it. Do you have any ideas? – Arturo don Juan Oct 25 at 22:18
• @Maxim I feel like it should be alright. The $M$-dependence of $h$ shouldn't really matter, since it's a polynomial sitting out from (the exponential will totally dominate). The $M$-dependence of $g$ however is quite dramatic - it directly affects the second derivative near the point $x_0=0$. However, past $x=0$, $g$ falls of at rate pretty much proportional to $3$, i.e. the exponential dies off as $\exp (-3Mx)$, plus corrections proportional to powers of $1/M$. So I feel like we should be safe here. – Arturo don Juan Oct 25 at 22:23
• $h$ is not a polynomial; consider that $$\int_{\mathbb R} \frac {e^{-M z^2}} {\sqrt {z^2 + \frac 1 M}} dz \not \sim \int_{\mathbb R} \frac {e^{-M z^2}} {\sqrt {\frac 1 M}} dz.$$ I've updated my answer. – Maxim Oct 27 at 17:24