Evaluate $\int_{C}\frac{e^{\alpha z}}{z}dz$ where $\alpha \in \mathbb R$ and C is the circle $\gamma(t)=e^{it}$... 
Let $\alpha \in \mathbb R$ and C be the circle $\gamma(t)=e^\alpha t$, $-\pi\le t \le \pi$ 
Evaluate $$\int_{C}\frac{e^{\alpha z}}{z}dz.$$
Use the above, to show that $$\int_{0}^{\pi}e^{\alpha \cos t}\cos(\alpha \sin t)dt= \pi.$$

I want to use cauchy integral formula for this problem, but I do not know how to start. Can I use the circle $\gamma(t)=e^\alpha t$?  
 A: 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
$$
\int_0^\pi e^{\alpha\cos(t)}\cos(\alpha\sin(t))\,\mathrm{d}t=\pi
$$
A: There's no need for the sledgehammer that is the residue theorem here. Cacuhy's integral formula (as the poster asked for) is enough. Let $f(z) = e^{\alpha z}$. Then
$$\int_C \frac{e^{\alpha z}}z \,dz = \int_C \frac{f(z)}{z-0}\,dz = 2\pi i f(0) = 2\pi i.$$
