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How to integrate
$ 1)\displaystyle \int_0^{2\pi} e^{\cos \theta} \cos( \sin \theta) d\theta$
$ 2)\displaystyle \int_0^{2\pi} e^{\cos \theta} \sin ( \sin \theta) d\theta$

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    $\begingroup$ Sir, yes, sir! How about you show your progress so it doesn't sound so imperative? $\endgroup$ – Lazar Ljubenović Apr 11 '13 at 17:42
  • $\begingroup$ @LazarLjubenović stuck at evaluating $\int_0^{2\pi} e^{e^{i \theta }} d\theta$ $\endgroup$ – Mula Ko Saag Apr 11 '13 at 17:43
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Let $\gamma$ be the unitary circumference positively parametrized going around just once.

Consider $\displaystyle \int _\gamma \frac{e^z}{z}\,dz$.

On the one hand $$\begin{align} \int _\gamma \frac{e^z}{z}\mathrm dz&=\int \limits_0^{2\pi}\frac{e^{e^{i\theta}}}{e^{i\theta}}ie^{i\theta}\mathrm d\theta\\ &=i\int _0^{2\pi}e^{\cos (\theta)+i\sin (\theta )}\mathrm d\theta\\ &=i\int _0^{2\pi}e^{\cos (\theta )}[\cos (\sin (\theta))+i\sin (\sin (\theta))\textbf{]}\mathrm d\theta. \end{align}$$

On the other hand Cauchy's integral formula gives you: $\displaystyle \int _\gamma \frac{e^z}{z}\mathrm dz=2\pi i$.

$\large \color{red}{\text{FINISH HIM!}}$

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  • $\begingroup$ comparing real and imaginary part right!! $\endgroup$ – Mula Ko Saag Apr 11 '13 at 17:51
  • $\begingroup$ @MulaKoSaag ^ ^ $\endgroup$ – Git Gud Apr 11 '13 at 17:52
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    $\begingroup$ You can choose between "Fatality", "Friendship", "Babality" or "Brutality" ;-) $\endgroup$ – Matemáticos Chibchas Apr 11 '13 at 19:50
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$$\int_0^{2\pi} e^{\cos\theta}\cos(\sin\theta)d\theta=\Re\left(\int_0^{2\pi} e^{e^{i\theta}}d\theta\right)$$

$$I(\lambda)=\int_0^{2\pi} e^{\lambda e^{i\theta}}d\theta$$

$$I'(\lambda)=\frac{1}{i\lambda}\int_0^{2\pi} i\lambda e^{i\theta}e^{\lambda e^{i\theta}}d\theta =\frac{1}{i\lambda}\bigg[e^{\lambda e^{i\theta}}\bigg]_0^{2\pi}=0$$

Hence:

$$I(\lambda)=\mathcal{C}$$

Taking $\lambda =0$ we have $\mathcal{C}=2\pi$ so:

$$\int_0^{2\pi} e^{\cos\theta}\cos(\sin\theta)d\theta=2\pi$$

Similarly, since $\Im\, (2\pi)=0:$

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

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Hint: If the first integral is called $I$ and the second is called $J$

Consider $I+iJ$

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  • $\begingroup$ $\int_0^{2\pi} e^{e^{i \theta }} d\theta$ then? $\endgroup$ – Mula Ko Saag Apr 11 '13 at 17:42
  • $\begingroup$ @MulaKoSaag Yeah, I am trying to figure out what that is, maybe the hint is not so useful.... $\endgroup$ – Lost1 Apr 11 '13 at 17:42
  • $\begingroup$ @MulaKoSaag acutally now i think it works $\endgroup$ – Lost1 Apr 11 '13 at 17:47
  • $\begingroup$ what did you apply? $\endgroup$ – Mula Ko Saag Apr 11 '13 at 17:49
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Hint:
$$\cos(\sin{\theta})=\frac{e^{i\sin{\theta}}+e^{-i\sin{\theta}}}{2}$$ so $$e^{\cos{\theta}}\cos(\sin{\theta})=\frac{1}{2}\left(e^{\cos{\theta}+i\sin{\theta}}+e^{\cos{\theta}-i\sin{\theta}}\right)=\frac{1}{2}\left(e^z+e^{\bar{z}} \right), \\ d{\theta}=\frac{dz}{iz},$$ thus
$$ \int\limits_0^{2\pi} e^{\cos{\theta}}\cos(\sin{\theta})d{\theta}=\frac{1}{2}\left(\oint\limits_{\gamma}{\frac{e^{z}dz}{iz}}+\oint\limits_{\gamma}{\frac{e^{\bar{z}}dz}{iz}} \right), $$ where $\gamma$ denotes the unit circle.
Added:
The last integral can be transformed as follows $$\oint\limits_{\gamma}{\frac{e^{\bar{z}}dz}{iz}}=\int\limits_{0}^{2\pi}{{e^{\rho{e^{-i\theta}}}d{\theta}}}=\left|\matrix{\varphi=2\pi-\theta \\ d{\theta}=-d{\varphi} }\right|= \\ =-\int\limits_{2\pi}^{0}{{e^{\rho{e^{i(\varphi-2\pi)}}}d{\varphi}}}=\int\limits_{0}^{2\pi}{{e^{\rho{e^{i\varphi}}}d{\varphi}}}=\oint\limits_{\gamma}{\frac{e^{w}dw}{iw}}.$$

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  • $\begingroup$ how do I integrate the last part? $\oint_\gamma e^{\bar z}/z dz $? $\endgroup$ – Mula Ko Saag Apr 11 '13 at 17:56
  • $\begingroup$ I edit my answer with respect to your question $\endgroup$ – M. Strochyk Apr 11 '13 at 19:52

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