Solution: We will use the method of residue to compute the given integral. Look at the residue theorem. Here the function \[ f(z) = \frac{\sin \pi z}{z^2 + 1} = \frac{\sin \pi z}{(z + \iota )(z - \iota )} \] has simple pole at $z = \pm \iota $. Therefore, using the residue theorem (b), we have \begin{align*} \mathrm{Res} (f, \iota ) & = \lim_{z \to i} (z - \iota ) \frac{\sin \pi z}{z^2 + 1} \\ & = \lim_{z \to \iota} \frac{\sin \pi z}{z + \iota } \\ & = \frac{\sin \pi \iota }{2\iota }; \\[2ex] \mathrm{Res} (f, -\iota ) & = \lim_{z \to -i} (z + \iota ) \frac{\sin \pi z}{z^2 + 1} \\ & = \lim_{z \to -\iota} \frac{\sin \pi z}{z - \iota } \\ & = \frac{\sin (-\pi \iota) }{-2\iota } = \frac{\sin \pi \iota }{2\iota }. \end{align*} Thus, \begin{align*} \int_\gamma \frac{\sin \pi z}{z^2 + 1}\mathrm{d} z & = 2\pi \iota \left( \mathrm{Res} (f,\iota ) + \mathrm{Res} (f,-\iota ) \right) \\ & = 2\pi \iota \left( \frac{\sin \pi \iota }{2\iota } + \frac{\sin \pi \iota }{2\iota } \right) \\ & = 2\pi \sin \pi \iota . \end{align*}