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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 $\gamma=e^{it}$?2012-03-26

2 Answers 2

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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 $

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    @Hassan: yes, but $\frac1z\,\mathrm{d}z=i\,\mathrm{d}t$.2012-03-27
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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 $\int_C \frac{e^{\alpha z}}z \,dz = \int_C \frac{f(z)}{z-0}\,dz = 2\pi i f(0) = 2\pi i.$

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    How do you use the result to show that the next integral is $\pi$?2012-03-29