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Euler-Heisenberg Lagrangian & heat kernel: Why do we need to substitute $s \to -is$ to continue calculating?

Starting from the general effective lagrangian for a constant background electromagnetic field, $$\mathcal{L}_{\mathrm{EH}}=-\frac{1}{4}F_{\mu\nu}^{2}-\frac{e^{2}}{32\pi^{2}}\int_{0}^{\infty}\frac{ds}{s}e^{-s\varepsilon}e^{-ism^{2}}\frac{\mathrm{Re}\cos(esX)}{\mathrm{Im}\cos(esX)}\frac{1}{2}\varepsilon^{\mu\nu\alpha\beta}F_{\mu\nu}F_{\alpha\beta},\tag{33.69}$$ where $$X\equiv\sqrt{\frac{1}{2}F_{\mu\nu}^{2}+\frac{i}{4}\varepsilon^{\mu\nu\alpha\beta}F_{\mu\nu}F_{\alpha\beta}}.\tag{33.70}$$ Why do we need to substitute $s \to -is$ before continuing to calculate any interesting quantities like pair creation? What does this substituting entail for the underlying physics? And why are we allowed to make this substitution? After the substituting, the Euler-Heisenberg Lagrangian reads,

$$\mathcal{L}_{\mathrm{EH}}=-\frac{1}{4}F_{\mu\nu}^{2}-\frac{e^{2}}{32\pi^{2}}\int_{0}^{\infty}\frac{ds}{s}e^{is\varepsilon}e^{-sm^{2}}\frac{\mathrm{Re}\cosh(esX)}{\mathrm{Im}\cosh(esX)}\frac{1}{2}\varepsilon^{\mu\nu\alpha\beta}F_{\mu\nu}F_{\alpha\beta}.\tag{33.71}$$

Reference: Quantum Field Theory and the Standard Model, Matthew D. Schwartz, page 715.