MHB Transforming an integral of Exponential to a Contour

[shen07]

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?

 

[alyafey22]

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]

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 I am stuck, how do i show that this integral is ZERO?

 

[alyafey22]

You are working in reverse which made things difficult .

We are given

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

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

so applying the rule we get $$i\int^{2\pi}_0 \text{exp}(e^{it}) e^{it}\, dt=0$$