Complex Contour Integration - Complex Analysis I'm just practising for my upcoming exam, and I've come across a question I'm having a bit of difficulty with.
I've been asked to show the following;
$$\int_{0}^{\infty} \frac{dz}{\cosh(z)} = \frac{\pi}{2}$$
Which, we're told, needs to be integrated around an appropriate contour, with the hint that this countour should be a rectangle with vertices at $-R$, $R$, $R + \pi i$, $-R + \pi i$, oriented in the positive sense.
So, I started by noting that the function $f(z) = \frac{1}{\cosh(z)}$ is even, so we can say that, as $R \rightarrow \infty$, $\int_{-R}^{R} f(z)\cdot dz = PV \int_{-\infty}^{\infty} f(z)\cdot dz = \int_{-\infty}^{\infty} f(z)\cdot dz$
So, now, we can evaluate the contour using Cauchy's Residue Theorem, noting that the only singularity to occur within this contour is $z = \frac{\pi i}{2}$, which has the residue $-i$, so the integral simply becomes;
$$\int_C \frac{dz}{\cosh(z)} = 2 \pi i \times -i = 2 \pi$$
Then, I need to deal with the integrals above the real axis. So, let's do the following;
$$\Gamma_{1} = \int_{0}^{R + \pi i} f(z) \cdot dz$$
$$\Gamma_{2} = \int_{R + \pi i}^{-R + \pi i} f(z) \cdot dz$$
$$\Gamma_{3} = \int_{0}^{-R + \pi i} f(z) \cdot dz$$
Then;
$$\int_{\Gamma_{1} + \Gamma_{2} + \Gamma_{3}}f(z) \cdot dz = \int_{\Gamma_1}f(z) \cdot dz + \int_{\Gamma_2}f(z) \cdot dz + \int_{\Gamma_3}f(z) \cdot dz$$
$$ = \int_{0}^{R + \pi i} f(z) \cdot dz + \int_{R + \pi i}^{-R + \pi i} f(z) \cdot dz - \int_{-R + \pi i}^{0} f(z) \cdot dz$$
$$ = \int_{0}^{R + \pi i} f(z) \cdot dz + \int_{0}^{-R + \pi i} f(z) \cdot dz +\int_{R + \pi i}^{0} f(z) \cdot dz - \int_{0}^{-R + \pi i} f(z) \cdot dz$$
$$ = \int_{0}^{R + \pi i} f(z) \cdot dz +\int_{R + \pi i}^{0} f(z) \cdot dz = 0$$
Assuming I've done all of that correctly, I get;
$$Lim(R \rightarrow \infty) \int_{-R}^{R} \frac{dz}{\cosh(z)} = \int_{-\infty}^{\infty} \frac{dz}{\cosh(z)} = 2\pi$$
And, as $f(z)$ is even, we have;
$$\int_{0}^{\infty} \frac{dz}{\cosh(z)} = \pi$$
What have I missed?? Unless those three gamma integrads are meant to equate to give $\pi$, I can't see what I've done incorrectly here. I know it's a lot of working to go through, but any help would be fantastic. :)
 A: $\newcommand{\+}{^{\dagger}}
 \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
 \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack}
 \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,}
 \newcommand{\dd}{{\rm d}}
 \newcommand{\down}{\downarrow}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}
 \newcommand{\fermi}{\,{\rm f}}
 \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}
 \newcommand{\half}{{1 \over 2}}
 \newcommand{\ic}{{\rm i}}
 \newcommand{\iff}{\Longleftrightarrow}
 \newcommand{\imp}{\Longrightarrow}
 \newcommand{\isdiv}{\,\left.\right\vert\,}
 \newcommand{\ket}[1]{\left\vert #1\right\rangle}
 \newcommand{\ol}[1]{\overline{#1}}
 \newcommand{\pars}[1]{\left(\, #1 \,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\pp}{{\cal P}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,}
 \newcommand{\sech}{\,{\rm sech}}
 \newcommand{\sgn}{\,{\rm sgn}}
 \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}}
 \newcommand{\ul}[1]{\underline{#1}}
 \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}
 \newcommand{\wt}[1]{\widetilde{#1}}$
$\ds{\int_{0}^{\infty}{\dd z \over \cosh\pars{z}} = {\pi \over 2}:\ {\large ?}}$

Besides the $\ds{\color{#c00000}{\tt 'tricky' complex\ integration}}$, it's indeed quite simple:

\begin{align}&\color{#66f}{\Large\int_{0}^{\infty}{\dd z \over \cosh\pars{z}}}
=2\
\overbrace{\int_{0}^{\infty}{\expo{z}\,\dd z \over \expo{2z} + 1}}
^{\ds{\mbox{Set}\ \expo{z}\ \equiv t}}\ =\
=2\int_{1}^{\infty}{\dd t \over t^{2} + 1}
=\left.\vphantom{\huge A}2\arctan\pars{t}\,\right\vert_{1}^{\infty}
\\[3mm]&=2\pars{{\pi \over 2} - {\pi \over 4}} = \color{#66f}{\Large{\pi \over 2}}
\end{align}
