Evaluation of $\int_{0}^{\infty} \cos(x)/(x^2+1)$ using complex analysis. Evaluate:
$$\int_{0}^{\infty} \frac{\cos(x)}{x^2 + 1} dx$$
Using only complex analysis. 
$$I = \int_{0}^{\infty} \frac{\cos(x)}{x^2 + 1} dx = (\frac{1}{2})\int_{-\infty}^{\infty} \frac{\cos(x)}{x^2 + 1} dx$$
Consider a contour $C$ with a upper-axis semi-circle $B$ and the axis running from $-R \to R$
We will compute: 
$$\oint_{C} \frac{\cos(z)}{z^2 + 1} dz$$
First:
$$z^2 + 1 = 0 \implies Z \in \{-i, i\}$$
Only, $z = i$ is in the semi circle region.
$$\text{Res}_{z=i} = \lim_{z \to i} (z-i)(f(i)) = \lim_{z \to i} \frac{\cos(z)}{z + i} = \frac{\cosh(1)}{2i}$$
Applying the residue theorem:
$$\oint_{C} \frac{\cos(z)}{z^2 + 1} dz = (2\pi i)\cdot \frac{\cosh(1)}{2i} = \pi\cdot\cosh(1)$$
$$\oint_{C} \frac{\cos(z)}{z^2 + 1} dz = \int_{-\infty}^{\infty} \frac{\cos(x)}{x^2 + 1} dx$$
But that is wrong, the answer for the full improper is:
$$\int_{-\infty}^{\infty} \frac{\cos(x)}{x^2 + 1} dx = \frac{\pi}{e}$$
What am I doing wrong?
 A: The cosine function does not vanish on the semicircle as $R \to \infty$; in fact, it does the opposite.  You need to either 1) take the real part of $e^{i x}$ in the upper half plane, or 2) use $\cos{x} = (e^{i x}+e^{-i x})/2$ and use both the upper and lower half planes, respectively.
A: we use  $$f\left( z \right) = \frac{{e^{iz} }}{{1 + z^2 }}$$
A: we use  $$f\left( z \right) = \frac{{e^{iz} }}{{1 + z^2 }}$$
then take real parts of the resulting integral .using the same contour $C$ 

\begin{array}{l}
 i \in C; - i \notin C \\ 
 {\mathop{\rm Re}\nolimits} s\left( {f;i} \right) = \mathop {\lim }\limits_{z \to i} \left\{ {\left( {z - i} \right)\frac{{e^{iz} }}{{1 + z^2 }}} \right\} = \frac{{e^{ - 1} }}{{2i}} \\ 
 \end{array}
\begin{array}{l}
 \oint\limits_C {\frac{{e^{iz} }}{{1 + z^2 }}dz}  = \int\limits_{ - R}^{ + R} {\frac{{e^{ix} dx}}{{1 + x^2 }}}  + \int\limits_{S_R}  {\frac{{e^{iz} dz}}{{1 + z^2 }}}  = 
2\pi i\frac{{e^{ - 1} }}{{2i}} = \frac{\pi }{e} \\ 
  \Rightarrow \int\limits_{ - R}^{ + R} {\frac{{\cos \left( x \right)dx}}{{1 + x^2 }}}  + i\int\limits_{ - R}^{ + R} {\frac{{\sin \left( x \right)dx}}{{1 + x^2 }} + } \int\limits_{S_R}  {\frac{{e^{iz} dz}}{{1 + z^2 }}}  = \pi e^{ - 1}  \\ 
 2\int\limits_0^{ + R} {\frac{{\cos \left( x \right)dx}}{{1 + x^2 }}}  = \pi e^{ - 1} ;\quad \left( {\int\limits_{S_R}  {\frac{{e^{ix} dx}}{{1 + x^2 }} = 0} ;R \to  + \infty \int\limits_{ - R}^{ + R} {\frac{{\sin \left( x \right)dx}}{{1 + x^2 }} = 0} } \right) \\ 
 \mathop {\lim }\limits_{R \to  + \infty } \int\limits_0^{ + R} {\frac{{\cos \left( x \right)dx}}{{1 + x^2 }}}  = \frac{\pi }{2}e^{ - 1}  \\ 
 \end{array}
note:
but if $y = {\mathop{\rm Im}\nolimits} \left( z \right)$ then $\cos \left( z \right) \approx \frac{{e^{\left| y \right|} }}{{\left| z \right|^2 }}$ for large $\left| z \right|$
we have the estimate 
$
\left| {\int\limits_{C_R } {\frac{{e^{iz} }}{{z^2  + 1}}dz} } \right| \le \int\limits_{C_R } {\frac{{e^{ - y} }}{{R^2  - 1}}\left| {dz} \right|}  \le \frac{{\pi R}}{{R^2  - 1}} \to 0
$ as  $R \to \infty $ 
where $
y = {\mathop{\rm Im}\nolimits} \left( z \right) > 0
$by the residue theorem
$
\int\limits_{ - \infty }^{ + \infty } {\frac{{e^{ix} }}{{x^2  + 1}}dx = } 2\pi i\sum\limits_{{\mathop{\rm Im}\nolimits} a > 0} {{\mathop{\rm Re}\nolimits} s_a } \frac{{e^{iz} }}{{z^2  + 1}} = \frac{\pi }{e}
$
