Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

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

2 Answers 2

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 $$\int_C \frac{e^{\alpha z}}z \,dz = \int_C \frac{f(z)}{z-0}\,dz = 2\pi i f(0) = 2\pi i.$$

share|improve this answer
    
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 $$ \int_0^\pi e^{\alpha\cos(t)}\cos(\alpha\sin(t))\,\mathrm{d}t=\pi $$

share|improve this answer
    
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.