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Transforming an integral of Exponential to a Contour

shen07

Member
Aug 14, 2013
54
Hi friends, i need some help for this number:

By considering the integral $$\int_{\gamma(0;1)}\exp(z) \mathrm{d}z$$,show that

$$\int_0^{2\pi}\exp(\theta)\cos(\theta+\sin(\theta)) \mathrm{d}\theta = 0$$

i know that since $$f(z)=\exp(z)$$ is holomorphic on and inside $$\gamma(0;1)$$,$$\int_{\gamma(0;1)}\exp(z) \mathrm{d}z = 0$$

But now how do i transform it to a contour so that i can use that integral?
 

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
Use $z=e^{i\theta }\,\,\,\, 0\leq \theta \leq 2\pi $

Now use the well-known formula for smooth curves and continuous function $f$

$$\int_{\gamma} f(z)\, dz = \int^b_a f(\gamma(t))\, \gamma'(t)\, dt $$

for $a\leq t \leq b $
 

shen07

Member
Aug 14, 2013
54
Use $z=e^{i\theta }\,\,\,\, 0\leq \theta \leq 2\pi $

Now use the well-known formula for smooth curves and continuous function $f$

$$\int_{\gamma} f(z)\, dz = \int^b_a f(\gamma(t))\, \gamma'(t)\, dt $$

for $a\leq t \leq b $
I have worked out and simplify, can anyone tell me if it is correct and how can i continue with it.

$$on \gamma(0;1)\,\,\,\,\, cos(\theta+sin(\theta))= \frac{z^2+1}{2z} cos(\frac{z^2-1}{2iz})+i \frac{z^2-1}{2z} sin(\frac{z^2-1}{2iz})$$

i express it in terms of exponential, i.e

$$\frac{\sqrt{2z^4+2}}{2z}exp(\frac{z^2-1}{2iz})$$

Now this implies that my integral has been tranformed on $$\gamma(0;1)$$

$$\int_0^{2\pi}\exp(cos(\theta))\cos(\theta+sin( \theta)) \mathrm{d}\theta=\frac{-i}{2}\int_{\gamma(0;1)} exp(\frac{z^2+1}{2z})\,\,\frac{\sqrt{2z^4+2}}{z^2}\,\,exp(\frac{z^2-1}{2iz})\,\,\mathrm{d}z$$

Now im stuck, how do i show that this integral is ZERO?
 

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
You are working in reverse which made things difficult .

We are given

\(\displaystyle \int_{\gamma(0,1)}\text{exp}(z)\, dz \)

Now we can paramatrize the circle \(\displaystyle \gamma(0,1)\) as \(\displaystyle e^{it},\,\, 0\leq t \leq 2\pi \)

so applying the rule we get


\(\displaystyle i\int^{2\pi}_0 \text{exp}(e^{it}) e^{it}\, dt=0\)