Vanishing integral and lower bound of cosh

Question

I want to show $$\lim_{R\to\infty}\int_\gamma f(z)\ \text{d}z=\lim_{R\to\infty}\int_\gamma\frac{e^{-ikz}}{\cosh(z)}\text{d}z = 0$$ for $$ \gamma(t) = R + it \qquad t \in [0, \pi] $$ and some $k \in \mathbb{R}$
My idea is to prove that $$ \lim_{R\to\infty}\left|\int_\gamma\frac{e^{-ikz}}{\cosh(z)}\text{d}z\right|\le\lim_{R\to\infty}\int_\gamma\frac{|e^{-ikz}|}{|\cosh(z)|}\text{d}z\le 0 $$ What I already got is $$ |e^{-ikz}| \le |e^{ky}| \le |e^{k\pi}| \le C $$ My attempt to estimate a useful lower bound of $|\cosh(z)|$ wasn't successful.

Do I have to evaluate the integral more properly to find the upper bound of $f(z)$ or is there another solution?

Answer 1

Note that using the addition angle formula

$$\cosh(R+it)=\cosh(R)\cos(t)+i\sinh(R)\sin(t)$$

we have $|\cosh(R+it)|=\sqrt{\cosh^2(R)\cos^2(t)+\sinh^2(R)\sin^2(t)}$.

Then, using the identities $\cosh^2(z)-\sinh^2(z)=1$ and $\cos^2(z)+\sin^2(z)=1$, we can write

$$\begin{align} \left|\int_\gamma \frac{e^{-ikz}}{\cosh(z)}\,dz\right|&=\left|\int_0^\pi \frac{e^{-ik(R+it)}}{\cosh(R+it)}\,i\,dt\right|\\\\ &\le \int_0^\pi \frac{e^{kt}}{|\cosh(R)\cos(t)+i\sinh(R)\sin(t)|}\,dt\\\\ &=\int_0^\pi \frac{e^{kt}}{\sqrt{\cosh^2(R)-\sin^2(t)}}\,dt\\\\ &\le \frac{\pi e^{k\pi}}{\sqrt{\cosh^2(R)-1}}\\\\ &\to 0\,\,\text{as}\,\,R\to \infty \end{align}$$

And we are done!