Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I have to verify that $$\int_0^\pi \ln(1+\alpha\cos(x))dx=\pi\ln\left(\frac{1+\sqrt{1-\alpha^2}}{2}\right)$$ with $|\alpha|<1$. It is my homework and don't know where to begin.

share|cite|improve this question
    
Is $\alpha$ supposed to satisfy $|\alpha| \leq 1$? – Sangchul Lee Sep 7 '12 at 15:19
    
@Ned Dabby: this question is very interesting (+1). – user 1618033 Sep 7 '12 at 18:35
    
@Ned Dabby: where does this problem come from? – user 1618033 Sep 7 '12 at 18:59
    
@Chris'ssister: Oh sorry for the delay. This was one of the problem I had to solve. It comes from ADVANCED CALCULUS by Schaum. Thank you for +1. :-) – Ned Dabby Sep 11 '12 at 8:47
    
@Ned Dabby: no problem. I'm glad to see such questions around. – user 1618033 Sep 11 '12 at 8:48
up vote 3 down vote accepted

$$I(\alpha) = \int_0^{\pi} \ln (1+ \alpha \cos(x)) dx$$ $$\dfrac{dI}{d \alpha} = \int_0^{\pi} \dfrac{\cos(x)}{1+\alpha \cos(x)} dx = \dfrac1{\alpha} \int_0^{\pi} \dfrac{\alpha \cos(x)}{1+ \alpha \cos(x)} dx = \dfrac1{\alpha} \left( \pi - \int_0^{\pi} \dfrac{dx}{1+\alpha \cos(x)}\right)$$ And integral $$\int_0^{\pi} \dfrac{dx}{1+\alpha \cos(x)}$$ can be evaluated using the standard complex analytic technique by using the transformation $z = \exp(ix)$.

share|cite|improve this answer
    
@Thank you Marvis. Now, Iknow how to begin. I know that inyegral. – Ned Dabby Sep 7 '12 at 15:34

Hint: Differentiate the left-hand sign with respect to $\alpha$. The details are probably in your book, but if not, there is a useful Wikipedia article. You will get something you can integrate explicitly with respect to $x$.

Compare with the derivative of the right-hand side with respect to $\alpha$. Finally, observe that the left-hand side and the right-hand side agree at $\alpha=0$.

share|cite|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.