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Hello I am solving an integral with a natural logarithm that has a parameter. Let say $I(a)=\int_0^\pi\ln(1-2a\cos(x)+a^2)dx$ Then for differentiation under integral sign and that yields $I'(a)= \int_0^\pi\frac{2a-2\cos(x)dx}{ 1-2a\cos(x)+a^2}$.

I chose take $aI'(a)$ and then I have That $I(a)= \pi\ln(a^2)+a\pi$

My questions:

  1. is there any wrong over result?
  2. where I shoud to take considerations about $a$? magnitude.
  3. what is the integral problem about this exercise?

Thanks, any help, any correction.

Camilo Acevedo.
  • 774

1 Answers1

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Suppose that $0 \le a < 1$. Then we will prove that $$ I(a) = \int_{0}^{\pi}\log(1-2a\cos x+ a^{2})\, dx = 0. $$ If $\log$ is the principal branch then $f(z) = \dfrac{\log(1-z)}{z}$ is an analytic function in $|z| <1$. Consequently $$ 0 = \oint_{|z|=a}f(z)\, dz = \int_{-\pi}^{\pi}\dfrac{\log(1-ae^{it})}{ae^{it}}iae^{it}\, dt = i\int_{-\pi}^{\pi}\log(1-ae^{it})\, dt. $$ The imaginary part of the right hand side is $$ 0 = \int_{-\pi}^{\pi}\ln|1-ae^{it}|\, dt = \int_{-\pi}^{\pi}\ln\left(\sqrt{1-2a\cos t + a^{2}}\right)\, dt = \int_{0}^{\pi}\ln(1-2a\cos t +a^{2})\, dt = I(a). $$ If $a>1$ we observe that $$ \ln(1-2a\cos x + a^{2}) = \ln(a^{2}) + \ln\left(\frac{1}{a^{2}}- 2\frac{1}{a}\cos x + 1\right). $$ Consequently $I(a) = \pi\ln(a^{2})$.

We can handle the case $a = 1$ via continuity.

JanG
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