Definite integral challenge ∫ln(2−2cosx)dx=0

In summary, the purpose of the definite integral challenge ∫ln(2−2cosx)dx=0 is to find the value of x that makes the expression equal to 0 by solving for the antiderivative and evaluating the definite integral. The antiderivative of ln(2−2cosx) can be found using integration techniques and will have a constant of integration attached to it. The limits of integration for this challenge are not specified, so they should be chosen appropriately to make the integral solvable. Two special considerations when solving this challenge are the undefined natural logarithm function at x = 0 and the periodic nature of the given function with a period of 2π. The solution to this challenge has practical applications and
  • #1
lfdahl
Gold Member
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Prove, that the definite integral

$$\int_{0}^{\pi}\ln (2-2\cos x)dx = 0.$$
 
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  • #2
Call the integral $I$. Let $x \mapsto \pi-x$ then we get $\displaystyle \int_0^{\pi}\ln(2+2\cos{x})\,{dx}$.

Then $\displaystyle 2I = \int_0^{\pi}\log(4\sin^2{x})\,{dx} \implies I = \int_0^{\pi}\log(\sin{x}) +\log(2)\,{dx}$

But $\displaystyle\log (\sin x)=-\sum\limits_{j=1}^{\infty }{\frac{\cos (2jx)}{j}}-\log( 2)$. Using this we end up with

$\displaystyle I = -\int_0^{\pi} \sum_{j=1}^{\infty} \frac{1}{j} \cos{2jx} \,{dx}= -\sum_{j=1}^{\infty} \frac{1}{j} \int_0^{\pi} \cos(2jx) \,{dx} = 0$ as required.
 
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  • #3
June29 said:
Call the integral $I$. Let $x \mapsto \pi-x$ then we get $\displaystyle \int_0^{\pi}\ln(2+2\cos{x})\,{dx}$.

Then $\displaystyle 2I = \int_0^{\pi}\log(4\sin^2{x})\,{dx} \implies I = \int_0^{\pi}\log(\sin{x}) +\log(2)\,{dx}$

But $\displaystyle\log (\sin x)=-\sum\limits_{j=1}^{\infty }{\frac{\cos (2jx)}{j}}-\log( 2)$. Using this we end up with

$\displaystyle I = -\int_0^{\pi} \sum_{j=1}^{\infty} \frac{1}{j} \cos{2jx} \,{dx}= -\sum_{j=1}^{\infty} \frac{1}{j} \int_0^{\pi} \cos(2jx) \,{dx} = 0$ as required.

Welcome, June29! - to the challenge forum. What a nice entry you make with this fine solution of yours!
Thankyou for your participation.
 

FAQ: Definite integral challenge ∫ln(2−2cosx)dx=0

1. What is the purpose of the definite integral challenge ∫ln(2−2cosx)dx=0?

The purpose of this challenge is to find the value of x that makes the expression ∫ln(2−2cosx)dx equal to 0. This involves solving for the antiderivative of the given function and then plugging in the limits of integration to evaluate the definite integral.

2. How do you solve for the antiderivative of ln(2−2cosx)?

The antiderivative of ln(2−2cosx) can be found by using integration techniques such as substitution or integration by parts. It is important to note that ln(2−2cosx) is an indefinite integral, so the resulting antiderivative will have a constant of integration (C) attached to it.

3. What are the limits of integration for the definite integral challenge ∫ln(2−2cosx)dx=0?

The limits of integration for this challenge are not specified, so it is up to the solver to choose appropriate limits based on the given function and the desired value of 0. Generally, the limits should be chosen such that the resulting integral is solvable.

4. Are there any special considerations when solving this definite integral challenge?

Yes, there are two main considerations when solving this challenge: 1) the natural logarithm function becomes undefined for x = 0, and 2) the given function is periodic with a period of 2π. These factors should be taken into account when choosing the limits of integration and solving for the antiderivative.

5. What is the significance of the solution to this definite integral challenge?

The solution to this challenge represents the value of x that makes the given function's definite integral equal to 0. This value can have practical applications in various fields such as physics, engineering, and economics. Additionally, solving this challenge can help improve problem-solving skills and deepen understanding of integration techniques.

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