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.

  • $\begingroup$ Is $\alpha$ supposed to satisfy $|\alpha| \leq 1$? $\endgroup$ – Sangchul Lee Sep 7 '12 at 15:19
  • $\begingroup$ @Ned Dabby: this question is very interesting (+1). $\endgroup$ – user 1357113 Sep 7 '12 at 18:35
  • $\begingroup$ @Ned Dabby: where does this problem come from? $\endgroup$ – user 1357113 Sep 7 '12 at 18:59
  • $\begingroup$ @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. :-) $\endgroup$ – Ned Dabby Sep 11 '12 at 8:47
  • $\begingroup$ @Ned Dabby: no problem. I'm glad to see such questions around. $\endgroup$ – user 1357113 Sep 11 '12 at 8:48

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

  • $\begingroup$ @Thank you Marvis. Now, Iknow how to begin. I know that inyegral. $\endgroup$ – 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 $\alpha=0$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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