**Saddle Tails Pruning.**

We show that the integral is small for *k*−2/5 < |*t* − *tj*| < *π* log *p*/*q* . We define

$$S\_j^{(1)}(z) = -\frac{1}{2\pi} \int\_{k^{-2/5} < |t - t\_j| < \frac{\pi}{\log p / q}} z^{-r\_0 + it} \Gamma(r\_0 + it) (p^{-r\_0 - it} + q^{-r\_0 - it})^k dt. \tag{121}$$

Note that for |*t* − *tj*| ≤ *π* log *p*/*q* , we have

$$\begin{split} |p^{-r\_0-it} + q^{-r\_0-it}| &= (p^{-r\_0} + q^{-r\_0}) \sqrt{1 - \frac{2p^{-r\_0}q^{-r\_0}}{(p^{-r\_0}+q^{-r\_0})^2}(1 - \cos\left(t \log p/q\right))} \\ &\leq (p^{-r\_0} + q^{-r\_0}) \left(1 - \frac{p^{-r\_0}q^{-r\_0}}{(p^{-r\_0}+q^{-r\_0})^2}(1 - \cos\left(t - t\_j\right) \log p/q)\right) \\ &\qquad \text{since} \sqrt{1-x} \leq 1 - \frac{x}{2} \text{ for } x \in [0, 1] \\ &\leq (p^{-r\_0} + q^{-r\_0}) \left(1 - \frac{2p^{-r\_0}q^{-r\_0}}{\pi^2 (p^{-r\_0} + q^{-r\_0})^2}((t - t\_j)\log p/q)^2\right) \\ &\qquad \text{since} \quad 1 - \cos x \geq \frac{2x^2}{\pi^2} \text{ for } |\mathbf{x}| \leq \pi \end{split} \tag{12}$$

where *γ* = <sup>2</sup>*p*<sup>−</sup>*r*<sup>0</sup> *q*<sup>−</sup>*r*<sup>0</sup> log<sup>2</sup> *p*/*q π*<sup>2</sup>(*p*−*r*<sup>0</sup> + *q*−*r*<sup>0</sup> )2 . Thus,

$$S\_j^{(1)}(z) = O\left(z^{-r\_0} |\Gamma(r\_0 + it)| \int\_{k^{-2/5} < |t - t\_j| < \frac{\pi}{6\log t / i}} |p^{-r\_0 - it} + q^{-r\_0 - it}| dt\right)$$

$$= O\left(z^{-r\_0} (p^{-r\_0} + q^{-r\_0})^k \int\_{k^{-2/5}}^{\infty} e^{-\gamma k x^2} du\right)$$

$$= O\left(z^{-r\_0} (p^{-r\_0} + q^{-r\_0})^k k^{-3/5} e^{-\gamma k^{1/5}}\right), \text{since } \text{erf}(\mathbf{x}) = O\left(\varepsilon^{-\chi^2}/\mathbf{x}\right). \tag{123}$$
