A peculiar integral using Leibnitz rule of integration?

Question

Mathworld while explaining (rather very briefly) the Leibnitz rule of integration, aka derivative under an integral sign, mentions that this method may be used to evaluate peculiar integrals such as $$\phi(\alpha)=\int_0^{\pi}\ln(1-2\alpha \cos x+\alpha^2)~\mathrm dx=2\pi \ln|\alpha| \\ \text{for}~~ |\alpha|>1$$

This seems a tricky (or peculiar one) indeed!

How would one go about proving it?

Answer 1

I was able to find the full text for the source that Wolfram Mathworld mentions. See page 144 of the text, or page 160 of the electronic document. It reads:

The principle of differentiating under the integral sign may sometimes be used to evaluate a definite integral. For example, take $$\phi(\alpha) =\int_0^\pi \log(1-2\alpha\cos(x)+\alpha^2)\mathrm dx\tag{11}$$ $$ \frac{\mathrm d\phi}{\mathrm d\alpha}=\int_0^\pi \frac{-2\cos(x)+2\alpha}{1-2\alpha\cos(x)+\alpha^2}\mathrm dx \\ =\frac{1}{\alpha}\int_0^\pi \left[1-\frac{1-\alpha^2}{1-2\alpha\cos(x)+\alpha^2}\right]\mathrm dx \\ =\frac{\pi}{\alpha}-\frac{2}{\alpha}\left[\tan^{-1}\left(\frac{1+\alpha}{1-\alpha}\tan\frac{x}{2}\right)\right]\bigg|^{x=\pi}_{x=0}$$

As $x$ varies for (from?) $0$ to $\pi$, $\frac{1+\alpha}{1-\alpha}\tan\frac{x}{2}$ varies through positive values for (from?) $0$ to $\infty$ when $-1<\alpha<1$, and $\frac{1+\alpha}{1-\alpha}\tan\frac{x}{2}$ varies through negative values from $0$ to $-\infty$ when $\alpha<-1$ or $\alpha>1$. Hence

$$\left[\tan^{-1}\left(\frac{1+\alpha}{1-\alpha}\tan\frac{x}{2}\right)\right]\bigg|^{x=\pi}_{x=0}=\frac{\pi}{2}~~~\text{when}~~-1<\alpha<1 \\ \text{and}~~\left[\tan^{-1}\left(\frac{1+\alpha}{1-\alpha}\tan\frac{x}{2}\right)\right]\bigg|^{x=\pi}_{x=0}=-\frac{\pi}{2}~~~\text{when}~~\alpha<-1~\text{or}~\alpha>1$$ Therefore $$\frac{\mathrm d\phi}{\mathrm d\alpha}=0~~\text{when}~~-1<\alpha<1, \\ \text{and}~~\frac{\mathrm d\phi}{\mathrm d\alpha}=\frac{2\pi}{\alpha}~~\text{when}~~\alpha<-1~\text{or}~\alpha>1;$$ whence $\phi=C_1$ when $-1<\alpha<1$, and $\phi=\pi\log(\alpha^2)+C_2$ when $\alpha<-1$ or $\alpha>1$.

We may determine $C_1$ by placing $\alpha=0$ in $(11)$. Then, $C_1=0$. Hence $$\phi=0~ \text{when}~ -1<\alpha<1\tag{12}$$ To determine $C_2$ in the same manner we should need to substitute in $(11)$ a value of $\alpha$ greater numerically than $1$. This is not convenient. Instead we will place $\alpha=\frac{1}{\beta}$, where $-1<\beta<1$. Then $$\phi(\alpha)=\int_0^\pi \big[\log(1-2\beta\cos x+\beta^2)-\log\beta^2\big]\mathrm dx \\ =-\log \beta^2~\int_0^\pi\mathrm dx~~~~~\text{[by (12)]} \\ =-\pi\log\beta^2 \\ =\pi \log\alpha^2~~~\text{when}~\alpha<-1~\text{or}~\alpha>1$$ Therefore $C_2=0$. The definition of $\phi(\alpha)$ is now complete.

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And it goes on. Note that by the properties of $\log$, $\pi\log(\alpha^2)=2\pi\log |\alpha|$, so there are no problems.

Answer 2

You may want to refer to this paper "arXiv 2308.09619v1 On the Leibnitz Rule for Differentiating Under the Integral Sign", which provides a very short solution using the complex representation of $\cos(x)$ (Example 2, page 3).