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The integral seems like

$$ \exp\left(\int_{0}^{\pi}\left[\ln(2+\cos\left(x\right)\right)]\,{\rm d}x\right) $$

but the above will be a number bigger than $5$ by appraisement.

This is the original problem.

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Don't forget that when you go from product to sum you have to move the logarithm inside to the individual terms! Your 'final form' is missing that logarithm. –  Steven Stadnicki Sep 25 '13 at 17:17

1 Answer 1

up vote 3 down vote accepted

Taking log, \begin{align*} \log \left( \prod_{i=0}^{n-1} \left( 2 + \cos \left( \frac{i \pi}{ n } \right ) \right )^{\frac \pi n}\right ) &= \sum_{i=0}^{n-1} \frac \pi n \log \left( 2 + \cos \left( \frac{ i \pi}{n}\right ) \right ) \\ &= \pi \cdot \frac {1}{n } \sum_{i=0}^{n-1} \log \left( 2 + \cos \left( \frac{ i \pi}{n}\right ) \right )\\ \end{align*}

Taking limit $$\lim_{n\to\infty } \frac \pi n \sum_{i=0}^{n-1} \log \left( 2 + \cos \left( \frac{ i \pi}{n}\right ) \right ) = \int_0^\pi \log(2 + \cos(x))dx = \pi \log \left(\frac{1}{2} \left(2+\sqrt{3}\right)\right) $$ Taking back to $e$, we get $$\lim_{n \to \infty } \prod_{k=0}^\infty \left( 2 + \cos \left( \frac{i \pi }{n } \right ) \right )^{\frac \pi n } = e^{\pi \log \left(\frac{1}{2} \left(2+\sqrt{3}\right)\right) } = \left( \frac 1 2 ( 2 + \sqrt 3) \right )^\pi $$

Putting

N[Product[(2 + Cos[k Pi/10^3])^(Pi/10^3), {k, 0, 10^3}]]

on Mathematica produces the value of $7.10988$ which is close to our calculated value so I think what you claim on title must be false.

To evaluate the integral, we proceed in the following way. $$\int_0^{\pi} \log \left( 2 + \cos(\theta) \right )d\theta = \frac 1 2 \int_{-\pi}^{\pi} \log \left( 2 + \cos(\theta) \right )d\theta = \frac{1}{2} \int_0^{2\pi} \log(2 + \cos(\theta))d\theta $$

From Gauss MVT we get, $$\int_0^{2\pi } \left(\log ( \sqrt 3 + 2 + e^{i\theta})+\log ( \sqrt 3 + 2 + e^{-i\theta}) \right ) d\theta = 4 \pi \log (\sqrt 3 + 2 )$$

\begin{align*} 4 \pi \log (\sqrt 3 + 2 ) &=\int_0^{2\pi } \left(\log (\sqrt 3 + 2 + e^{i\theta})+\log (\sqrt 3 + 2 + e^{-i\theta}) \right ) d\theta \\ &= \int_0^{2\pi} \log \left( ( \sqrt 3+2 )^2 + 1 + ( \sqrt 3 +2)2 \cos \theta \right )d\theta\\ &= \int_0^{2\pi } \log \left( 4(2 + \sqrt 3) + 2 (2 + \sqrt 3) \cos (\theta) \right ) d\theta \\ &= \int_0^{2\pi }\log(2(2 + \sqrt 3))d\theta + \int_0^{2\pi } \log(2 + \cos\theta)d\theta\\ &= 2 \pi \log (2 (2 + \sqrt 3 )) + \int_0^{2\pi } \log(2 + \cos\theta)d\theta \\ \end{align*}

Which gives $$2\pi \log \left( \frac{(2 + \sqrt 3 )^2}{2 (2 + \sqrt 3 )}\right )= \int_0^{2\pi } \log(2 + \cos\theta)d\theta $$ Taking half of it give the value of our integral as evaluated by Mathematica earlier.

$$\pi \log \left( \frac{2 + \sqrt 3 }{2 }\right )= \int_0^{\pi } \log(2 + \cos\theta)d\theta $$

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1  
A '2' magically became a '1' in your penultimate equality... –  Steven Stadnicki Sep 25 '13 at 17:19
    
@experimentX How do you copmute the integration? –  Jebei Sep 25 '13 at 17:29
    
@frame99 currently I am using Mathematica to evaluate the intgral, and thinking ... I think i'll have to resort to complex analysis to evaluate this integral. I'll update if it find it. –  Santosh Linkha Sep 25 '13 at 17:30
    
@experimentX The $pi$ is redundant. –  Jebei Sep 25 '13 at 17:34

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