Residue theorem for contour lying in multiple Riemann sheets?

Question

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.

Answer 1

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 $$f(M) \sim \cases { \frac {\pi e^{-b M}} {a b^2 M} & $\epsilon = 0$ \\ \vphantom {\displaystyle \int} \frac {2 e^{-b M}} {b^2 M^2} \! \int_{\mathbb R^+} \hspace {-1px} \frac {e^{-z}} {z^2 + i \epsilon} dz & $\epsilon > 0$}, \quad M \to \infty.$$

Answer 2

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.