Define the semicircular arc $\gamma_R$ by $\gamma_R(t)=Re^{it}$, where $0\leq t\leq\pi$ and $R>1$ is a real constant. Let $\gamma$ be the join of $\gamma_R$ and the line segment from $-R$ to $R$. Also, $z\in\text{Range}(\gamma_R)$.
Evaluate, $$\int_\gamma \frac{e^{iz}}{z^2+1} \ dz. \ \ \ \ (1)$$
I have already shown the following results: \begin{align} \left|\frac{e^{iz}}{z^2+1}\right|&\leq\frac{1}{R^2-1}, \\ \left|\int_{\gamma_R}\frac{e^{iz}}{z^2+1}\right|&\leq\frac{R\pi}{R^2-1} \ \ \text{(by the ML Lemma)}. \end{align}
I have tried to solve $(1)$, by parametrising the line segment $\gamma$ and using the contour integral formula, $$\int_{0}^{1} f(\gamma(t))\gamma'(t) \ dt.$$ But the working became to convoluted. Is there an easier way?
I misread your question initially. This is in fact quite easy to do via the Residue Calculus: note that $\gamma$ is not just a line segment, but a semi-circular contour. Notice that the integrand has poles at $\pm i$. Only the pole at $i$ could matter, and so we have three cases:
$R<1:$ we don't include any poles so the integrand is zero by Cauchy's Theorem.
$R>1:$ we include the pole at $i$ so our integral is $2\pi i \operatorname{Res(i)} = (2\pi i)\left(\frac{-i}{2e}\right) = \pi/e$
$R = 1:$ Things become a bit complicated because the pole at $i$ is on the boundary of the circle, but we can use the Cauchy Principle Value to associate a value with the integral
Since in your case $R>1$ we can safely say the answer is $\pi/e.$