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

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

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

 

[blue_raver22]

To transform this integral to a contour, we can use the substitution $$z=e^{i\theta}$$, which transforms the path of integration from a circle to a unit circle. This means that we can rewrite the integral as $$\int_{0}^{2\pi}f(e^{i\theta})ie^{i\theta} \mathrm{d}\theta$$

Next, we can use the fact that $$\cos(\theta+\sin(\theta)) = \mathrm{Re}(e^{i(\theta+\sin(\theta))})$$ to rewrite the integral as $$\mathrm{Re}\left(\int_{0}^{2\pi}f(e^{i\theta})ie^{i\theta} \mathrm{d}\theta\right)$$

Since $$f(z)=\exp(z)$$ is holomorphic on the unit circle, we can use Cauchy's integral formula to evaluate this integral as $$\mathrm{Re}\left(\frac{1}{2\pi i}\int_{\gamma(0;1)}\frac{f(z)}{z} \mathrm{d}z\right)$$

But since $$\int_{\gamma(0;1)}\frac{f(z)}{z} \mathrm{d}z = 0$$, we can conclude that $$\int_0^{2\pi}\exp(\theta)\cos(\theta+\sin(\theta)) \mathrm{d}\theta = 0$$

Therefore, we have successfully transformed the original integral into a contour integral that we can use to evaluate the given integral.