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Calculating some contour integral, I have to prove that $\int^{R+i}_{R}\frac{cosh(az)}{cosh(\pi z)}dz$ goes to zero if R goes to infinity. And we know that $\left|a\right|<\pi$. I want to use the ML-estimation:

$\left|\int^{R+i}_{R}\frac{cosh(az)}{cosh(\pi z)}dz\right|\leq max_{z=R+iy, 0\leq y\leq 1}\left|\frac{cosh(az)}{cosh(\pi z)}\right|$.

But I can't find a good inequality for that maximum, does anyone know?

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Assume without loss of generality that $a \gt 0$. Then

$$\int_R^{R+i} dz \frac{\cosh{a z}}{\cosh{\pi z}} = i\int_0^1 dy \, \frac{\cosh{a (R+i y)}}{\cosh{\pi (R+i y)}}$$

Now,

$$\frac{\cosh{a (R+i y)}}{\cosh{\pi (R+i y)}} = \frac{e^{a (R+i y)}+e^{-a(R+i y)}}{e^{\pi (R+i y)}+e^{-\pi(R+i y)}} = \frac{e^{-(\pi-a)(R+i y)}+e^{-(\pi+a)(R+i y)}}{1+e^{-2 \pi(R+i y)}}$$

$$\left |\frac{\cosh{a (R+i y)}}{\cosh{\pi (R+i y)}} \right | \frac{\le e^{-(\pi-a)R}+e^{-(\pi+a)R}}{1-e^{-2 \pi R}} $$

Clearly, as $R \to \infty$, the above expression goes to zero.

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