$$\frac{1}{2\pi} \int_{0}^{2\cdot\pi}\left |1+e^{j\cdot k\cdot d\cdot \cos(\theta )} \right |^{2}d\theta =2(1+ J_{0}\left ( k d \right ))$$ Where: $$j=\sqrt{-1}$$ and $J_{0}$ is the Bessel function of the first kind of order zero. k,d are constants.
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Did you mean $(\frac{1}{2\cdot\pi})\cdot \int_{0}^{2\cdot\pi}\left (1+e^{j\cdot k\cdot d\cdot cos(\theta )} \right )^{2} d\theta$ when writing $(\frac{1}{2\cdot\pi})\cdot \int_{0}^{2\cdot\pi}\left |1+e^{j\cdot k\cdot d\cdot cos(\theta )} \right |^{2}$? – Noureddine Ouertani Nov 01 '20 at 13:13
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Is $j$ the imaginary unit or just a random real variable? What's $J_0$ ? – Noureddine Ouertani Nov 01 '20 at 13:16
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j is the imaginary unit indeed and J represents a Bessel function – Yechiel Asraff Nov 01 '20 at 13:21
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OK. I see. Thank you! – Noureddine Ouertani Nov 01 '20 at 13:22
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Did you check this question already? https://math.stackexchange.com/questions/2468863/what-is-the-integral-of-e-cos-x – Noureddine Ouertani Nov 01 '20 at 13:23
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You're welcome Noureddine Ouertani – Yechiel Asraff Nov 01 '20 at 13:28
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|x| is the absolute value of x. I used absolute in the equation and not parentheses. – Yechiel Asraff Nov 01 '20 at 18:49
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It's the same because |x|²= x² for all x – Noureddine Ouertani Nov 01 '20 at 18:53
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Sorry we are in C .... I think you're right. You mean $ \sqrt{(re(z))² + (Im(z))²}$ ? – Noureddine Ouertani Nov 01 '20 at 18:58
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Here is my idea despite the fact that I didn't finish the calculation (yet)... I also don't promise to do so. I hope this helps though somehow ... Maybe you can continue from where I stopped my reasoning.
Let $ I = (\frac{1}{2\cdot\pi})\cdot \int_{0}^{2\cdot\pi}\left (1+e^{j\cdot k\cdot d\cdot cos(\theta )} \right )^{2}d\theta $
Prove that $ I = 2 (1+ J_{0} ( 2 k d ))$
Let $x=cos \theta $ which means that $d\theta = \frac{-dx}{sin x}$
Therefore $I = -\frac{1}{2 \pi} \int_{0}^{2 \pi} (1 + e^{jkdx})². \frac{1}{sin x}dx$ : (1)
On the other hand (source: https://en.wikipedia.org/wiki/Bessel_function)
$J_0(2kd) = \frac{1}{2 \pi} \int_{- \pi}^{\pi} e^{2jkdsin \alpha}d \alpha = \frac{1}{\pi} \int_{0}^{\pi} cos(-2kdsin(\beta)) d \beta$ :(2)
Now back to (1): $I = -\frac{1}{2 \pi}( \int_{0}^{2 \pi} \frac{1}{sin x} dx + 2 \int_{0}^{2 \pi} e^{jkdx}. \frac{1}{sin x}dx + \int_{0}^{2 \pi}e^{2jkdx}. \frac{1}{sin x} dx)$
Can you continue from here? Or did you do this kind of reasoning already?
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I found a mistake in this equation, on the right side and corrected it. I corrected the question and I managed to find the answer. the equation was taken from an article. I will post the full answer later. – Yechiel Asraff Nov 02 '20 at 12:57
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Hint.\begin{align} \left|1+e^{j\cdot k\cdot d\cdot \cos(\theta )} \right|^{2}&= \big(1+e^{j\cdot k\cdot d\cdot \cos(\theta )}\big) \big(1+e^{-j\cdot k\cdot d\cdot \cos(\theta )}\big) \\ &= e^{-j\cdot k\cdot d\cdot \cos(\theta )} + 2 + e^{j\cdot k\cdot d\cdot \cos(\theta )} \end{align} $$=2+2\cdot \cos(k\cdot d\cdot \cos(\theta))=2\cdot(1+\cos(k\cdot d\cdot \cos(\theta))$$
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$$\frac{1}{2\pi}\int_{0}^{2\cdot\pi}|1+e^{j\cdot k\cdot d\cdot \cos(\theta )}|^{2}d\theta = \frac{1}{2\pi}\int_{0}^{2\cdot\pi}(1+e^{j\cdot k\cdot d\cdot \cos(\theta )})(1+e^{-j\cdot k\cdot d\cdot \cos(\theta )})d\theta $$ $$=\frac{1}{2\pi}\int_{0}^{2\cdot\pi}(2(1+cos(cos(k\cdot d\cdot cos(\theta)))d\theta $$ Using the following identity: $$\frac{1}{2\pi}\int_{0}^{2\cdot\pi}cos(a\cdot cos(\theta)d\theta =J_{0}(a)$$ We get: $$\frac{1}{2\pi}\int_{0}^{2\cdot\pi}(2(1+cos(cos(k\cdot d\cdot cos(\theta)))d\theta=2(1+J_{0}(k\cdot d))$$