$I(\lambda)=\int^\pi_0\log(1-2\lambda \cos\theta + \lambda^2)d\theta$ 
Consider the function $I(\lambda)=\int^\pi_0\log(1-2\lambda \cos\theta + \lambda^2)d\theta$. Prove $I$ is even and further that $I(\lambda ^2)=2I(\lambda)$. Hence by considering $I(\lambda ^{-1})$, evaluate $I(\lambda).$

I have solved the first "Prove $I$ is even and further ... " but I can't see how to do the "Hence ...". I think the first part is self-contained, its fine for answers to just quote the result. For the second part I would like a hint(please hold off on a full solution). If you do $I(\lambda^{-1})$ and pull $\lambda^{-1}$ out the logarithm, you can easily find $I(\lambda)-I(\lambda^{-1})=\pi \log\lambda^2$. Now I'm pretty sure we have to use the previous result with $I(\lambda^2)$. I tried $I(\lambda^2)+I(\lambda^{-1})$ and expanding inside the logarithm but it just gives a load of spaghetti without further direction. Thanks in advance. To reiterate, hint only please(at least not a full answer straight away).
 A: I ignore how to use $I(\lambda^2)=2I(\lambda)$ thus far. It could be that it only plays a  role in estimating $I(\pm1)$. The relation between $I(\lambda)$ and $I(\lambda^{-1})$ is indeed helpful. To evaluate $I(\lambda)$ here is a generous hint:
Notice that
$\log(1+2\lambda \cos\theta + \lambda^2)=\log\big(|1+\lambda e^{i\theta}|^2\big)$
so
$$I(\lambda)=I(-\lambda)=\int^{\pi}_0\log\big(|1+\lambda e^{i\theta}|^2\big)\,d\theta=\int^{\pi}_0\log(1+\lambda e^{i\theta}) + \log(1+\lambda e^{-i\theta})\,d\theta\,d\theta
$$
If $|\lambda|<1$, then by dominated convergence
$$I(\lambda)=\int^\pi_0\Big(\sum^\infty_{n=0}\frac{(-1)^{n+1}}{n}\lambda^n\cos(n\theta)\Big)\,d\theta=\sum^{\infty}_{n-1}\frac{(-1)^{n+1}}{n}\lambda^n\int^\pi_0\cos(n\theta)\,d\theta=0$$

Checking all identities in your OP for my own sanity:

*

*$I$ is even: Since $|z|=|\overline{z}|$,
$$\begin{align}
I(\lambda)&=\int^\pi_0\log\big(|1-\lambda e^{i\theta}|^2\big)\,d\theta
\stackrel{u=\pi-\theta}{=}-\int^0_{\pi}\log\big(|1-\lambda e^{ i(\pi-\theta)}|^2\big)\,d\theta\\
&=\int^\pi_0\log\big(|1+\lambda e^{ -i\theta}|^2\big)\,d\theta=I(-\lambda)
\end{align}$$
2: $I(\lambda)-I(\lambda^{-1})=\pi\log(\lambda^2)$: Notice
$$\log\big(|1-\lambda^{-1}e^{i\theta}|^2\big)=\log(|\lambda^{-1}e^{i\theta}|^2)+\log\big(|\lambda e^{-i\theta} -1|^2\big)=\log(\lambda^{-2})+\log\big(|1-\lambda e^{i\theta}|^2\big)$$


*$I(\lambda^2)=2I(\lambda)$:

$$\begin{align}
2I(\lambda)&=\int^\pi_0 \log(1-2\lambda \cos\theta + \lambda^2)+\log(1+2\lambda \cos\theta + \lambda^2)\,d\theta\\
&=\int^\pi_0\log((1+\lambda^2)^2-4\lambda^2\cos^2\theta)\,dt\theta\\
&=\int^\pi_0\log(1+2\lambda^2(1-2\cos^2\theta)+\lambda^4)\,d\theta\\
&=\int^\pi_0\log(1-2\lambda^2\cos(2\theta)+\lambda^4)\,d\theta\\
&=\frac12\int^{2\pi}_0\log(1-2\lambda^2\cos\theta+\lambda^4)\,d\theta\\
&=\frac{1}{2}\Big(I(\lambda^2)+\int^\pi_0\log(1-2\lambda^2\cos(\theta+\pi)+\lambda^4)\,d\theta\Big)\\
&=\frac12\Big(I(\lambda^2)+I(-\lambda^2)\Big)=I(\lambda^2)
\end{align}
$$
