# 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 $\gamma(t)=e^\alpha t$?

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Residue theorem? | Do you mean $\gamma=e^{it}$? – anon Mar 26 '12 at 23:47

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 $$\int_C \frac{e^{\alpha z}}z \,dz = \int_C \frac{f(z)}{z-0}\,dz = 2\pi i f(0) = 2\pi i.$$
How do you use the result to show that the next integral is $\pi$? – Hassan Muhammad Mar 29 '12 at 6:14
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 $$\int_0^\pi e^{\alpha\cos(t)}\cos(\alpha\sin(t))\,\mathrm{d}t=\pi$$
If $z=e^{it}$ then $\frac{1}{z}=e^{-it}$ not $i$ – Hassan Muhammad Mar 27 '12 at 5:54
@Hassan: yes, but $\frac1z\,\mathrm{d}z=i\,\mathrm{d}t$. – robjohn Mar 27 '12 at 9:05