Integrating $\cos z/(z^2+1)$

Question

Let $C_R\subset\mathbb C$ be a directed line segment that goes from $-R+2i$ to $R+2i$. I would like to evaluate $$ \lim_{R\rightarrow\infty} \int_{C_R} \frac{\cos z}{z^2 + 1}\ dz. $$ I obviously tried to do this by residue calculus. I tried expressing $\cos$ in terms of $\exp$, or experimenting with various closed curves that contains $C_R$, but in vain.

I would be grateful if you could give a clue.

Answer 1

Let $\gamma_R^+$ be the curve that goes from $-R+2i$ to $R+2i$ then circles counter-clockwise back above the line to $-R+2i$ and $\gamma_R^-$ be the curve that goes from $-R+2i$ to $R+2i$ then circles clockwise back below the line to $-R+2i$. Then, your limit equals $$ \lim_{R\to\infty}\left(\frac12\int_{\gamma_R^+}\frac{e^{iz}}{z^2+1}\,\mathrm{d}z +\frac12\int_{\gamma_R^-}\frac{e^{-iz}}{z^2+1}\,\mathrm{d}z\right)\tag{1} $$ $\gamma_R^+$ does not contain any singularities, so the first integral is $0$. $\gamma_R^-$ contains the singularities at $i$ and $-i$. The residue at $i$ is $-\frac{ei}{2}$ and the residue at $-i$ is $\frac{i}{2e}$. The integral over $\gamma_R^-$ is $-2\pi i$ times the sum of the residues, which is $-\pi\left(e-\frac1e\right)$. Thus, $(1)$ is $$ -\frac\pi2\left(e-\frac1e\right)\tag{2} $$

Answer 2

$\displaystyle{% {\cal I}\left(R\right) \equiv \int_{C_{R}}\frac{\cos\left(z\right)}{z^{2} + 1}\,{\rm d}z}\,, \quad \lim_{R \to \infty}{\cal I}\left(R\right) = ?$

\begin{align} {\cal I}\left(R\right) &= -2\pi{\rm i}\,{\cos\left({\rm i}\right) \over 2{\rm i}} - \int_{2}^{0} {\cos\left(R + {\rm i}y\right) \over \left(R + {\rm i}y\right)^{2} + 1}\,{\rm i\,d}y - \int_{R}^{-R}{\cos\left(x\right) \over x^{2} + 1}\,{\rm d}x - \int_{0}^{2} {\cos\left(-R + {\rm i}y\right) \over \left(-R + {\rm i}y\right)^{2} + 1}\,{\rm i\,d}y \\[5mm]& \lim_{R \to \infty}{\cal I}\left(R\right) = -\pi\cosh\left(1\right) + \int_{-\infty}^{\infty}{\cos\left(x\right) \over x^{2} + 1}\,{\rm d}x = -\pi\cosh\left(1\right) + \Re\int_{-\infty}^{\infty} {{\rm e}^{{\rm i}x} \over \left(x - {\rm i}\right)\left(x + {\rm i}\right)}\,{\rm d}x \\[3mm]&= -\pi\cosh\left(1\right) + \Re\left\lbrack% 2\pi{\rm i}\,{{\rm e}^{{\rm i}\ {\rm i}} \over {\rm i} + {\rm i}} \right\rbrack = -\pi\cosh\left(1\right) + \pi{\rm e}^{-1} = {\large -\pi\,\sinh\left(1\right)} \end{align}