Since the only singularity is at $z=0$, we get that $ \frac{e^{\alpha z}}{z}=\frac{1+\alpha z+\frac12\alpha^2z^2+\frac16\alpha^3z^3+\dots}{z}\tag{1} $ Thus, as long as $C$ circles the origin once clockwise, $ \int_{C}\frac{e^{\alpha z}}{z}\mathrm{d}z=2\pi i\tag{2} $ Notice that with $z=e^{it}=\cos(t)+i\sin(t)$, $ \begin{align} \int_C \frac{e^{\alpha z}}{z}\,\mathrm{d}z &=\int_{-\pi}^\pi e^{\alpha(\cos(t)+i\sin(t))}\,i\,\mathrm{d}t\\ &=i\int_{-\pi}^\pi e^{\alpha\cos(t)}(\cos(\alpha\sin(t))+i\sin(\alpha\sin(t)))\,\mathrm{d}t\\ &=i\int_{-\pi}^\pi e^{\alpha\cos(t)}\cos(\alpha\sin(t))\,\mathrm{d}t\\ &=2i\int_0^\pi e^{\alpha\cos(t)}\cos(\alpha\sin(t))\,\mathrm{d}t\tag{3} \end{align} $ Combining $(2)$ and $(3)$ yields $ \int_0^\pi e^{\alpha\cos(t)}\cos(\alpha\sin(t))\,\mathrm{d}t=\pi $