# integrate $\int_0^{2\pi} e^{\cos \theta} \cos( \sin \theta) d\theta$ [closed]

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

• Sir, yes, sir! How about you show your progress so it doesn't sound so imperative? – Lazar Ljubenović Apr 11 '13 at 17:42
• @LazarLjubenović stuck at evaluating $\int_0^{2\pi} e^{e^{i \theta }} d\theta$ – Mula Ko Saag Apr 11 '13 at 17:43

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: $\displaystyle \int _\gamma \frac{e^z}{z}\mathrm dz=2\pi i$.

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

• comparing real and imaginary part right!! – Mula Ko Saag Apr 11 '13 at 17:51
• @MulaKoSaag ^ ^ – Git Gud Apr 11 '13 at 17:52
• You can choose between "Fatality", "Friendship", "Babality" or "Brutality" ;-) – Matemáticos Chibchas Apr 11 '13 at 19:50

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

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

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

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

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

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

Consider $I+iJ$

• $\int_0^{2\pi} e^{e^{i \theta }} d\theta$ then? – Mula Ko Saag Apr 11 '13 at 17:42
• @MulaKoSaag Yeah, I am trying to figure out what that is, maybe the hint is not so useful.... – Lost1 Apr 11 '13 at 17:42
• @MulaKoSaag acutally now i think it works – Lost1 Apr 11 '13 at 17:47
• what did you apply? – Mula Ko Saag Apr 11 '13 at 17:49

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.
The last integral can be transformed as follows $$\oint\limits_{\gamma}{\frac{e^{\bar{z}}dz}{iz}}=\int\limits_{0}^{2\pi}{{e^{\rho{e^{-i\theta}}}d{\theta}}}=\left|\matrix{\varphi=2\pi-\theta \\ d{\theta}=-d{\varphi} }\right|= \\ =-\int\limits_{2\pi}^{0}{{e^{\rho{e^{i(\varphi-2\pi)}}}d{\varphi}}}=\int\limits_{0}^{2\pi}{{e^{\rho{e^{i\varphi}}}d{\varphi}}}=\oint\limits_{\gamma}{\frac{e^{w}dw}{iw}}.$$
• how do I integrate the last part? $\oint_\gamma e^{\bar z}/z dz$? – Mula Ko Saag Apr 11 '13 at 17:56