<s>[INST] How to compute the following definite integral Studying some integral table, I came across the following definite integral
$$\int_0^{\pi} \log [ a^2 + b^2 -2 a b \cos \phi ]\,d\phi$$ for $a,b \in \mathbb{R}$. Does somebody know a nice way to get the results?
 [/INST] Here is a different way, and is strangely the first thing that came to my mind.  It uses generating series, and power series as well as other various techniques:
(Warning!: It is significantly more complicated)
Assume that $a\geq b\geq0$ without loss of generality.  Notice that our integral is $$\pi \log(a^2+b^2)+\int_{0}^{\pi}\log\left( 1-\frac{2ab\cos\phi}{a^2+b^2}\right). $$ Since $\log(1-x)=-\sum_{i=1}^{\infty}\frac{x^{i}}{i}$ we have 
$$\int_{0}^{\pi}\log\left( 1-\frac{2ab\cos\phi}{a^2+b^2}\right)d\phi=-\int_{0}^{\pi}\sum_{i=1}^{\infty} \frac{1}{i}\left(\frac{2ab\cos\phi}{a^2+b^2}\right)^i d\phi$$
Switch the order or integration and summation to find:  (More on why this is legal at the end)
$$-\int_{0}^{\pi}\sum_{i=1}^{\infty} \frac{1}{i}\left(\frac{2ab\cos\phi}{a^2+b^2}\right)^i d\phi=-\sum_{i=1}^{\infty} \frac{1}{i}\left(\frac{2ab}{a^2+b^2}\right)^i \int_{0}^{\pi} \cos^i\phi d\phi$$
For $n$ odd we have $\int_0^\pi \cos^n xdx=0$.  For even $n=2r$ we see by integration by parts that $\int_{0}^{\pi}\cos(x)^{2r}dx=\frac{2r-1}{2r}\int_{0}^{\pi}\cos(x)^{2r-2}dx$, so that induction yields $$\int_{0}^{\pi}\cos(x)^{2r}dx=\pi\prod_{i=1}^{r}\frac{2i-1}{2i}=\pi\frac{\left({2r\atop r}\right)}{4^{r}}$$
Thus our sum becomes 
$$-\pi \sum_{r=1}^{\infty} \frac{1}{2r}\left(\frac{ab}{a^2+b^2}\right)^{2r} \left({2r\atop r}\right ) $$
Now let $$f(z)=\sum_{r=1}^{\infty} \frac{1}{r} \left({2r\atop r}\right) \frac{z^r}{4^r}$$
Then we need to find $-\frac{\pi}{2} f\left(\left( \frac{2ab}{a^2+b^2} \right)^2\right)$.
Let $Y=\left( \frac{2ab}{a^2+b^2} \right)^2$.
Recall the generating series $$\sum_{n=0}^{\infty}\left({2n\atop n}\right)x^{n}=\frac{1}{\sqrt{1-4x}}. $$  Differentiating yields $1+xf^{'}(x)=\frac{1}{\sqrt{1-x}}$ and hence $f(Y)=\int_{0}^{Y}\frac{1-\sqrt{1-x}}{x\sqrt{1-x}}dx$.
Make the substitution $x=\sin^{2}(u)$, and we get $$f(Y)=\int_{0}^{\theta}\frac{1-\cos(u)}{\sin^{2}(u)\cos(u)}2\sin(u)\cos(u)du=2\int_{0}^{\theta}\csc(u)-\cot(u)du$$ where $\theta$ is in the first quadrant and satisfies $\sin (\theta)=\frac{2ab}{a^2+b^2}.$  
Then since $\int\csc x\, dx=-\log\left|\csc x+\cot x\right|$, and $\int\cot x dx=\log\left|\sin x\right|$, we see that $\int\csc(u)-\cot(u)du=-\log|\sin(u)\csc(u)+\sin(u)\cot u|=-\log|1+\cos(u)|$. Thus $$f(Y)=-2\log|1+\cos(u)|\biggr|_{u=0}^{u=\theta}$$
Drawing the triangle with sides $2ab$, $a^2-b^2$ and $a^2+b^2$ tells us that $\cos(\theta)=\frac{a^2-b^2}{a^2+b^2}$, and hence $$f(Y)=2\log 2- 2\log \left(1+\frac{a^2-b^2}{a^2+b^2}\right)$$
Thus the final answer is  $$\pi \log(a^2+b^2)-\frac{\pi}{2}f(Y)=\pi\log(a^2+b^2)-\pi \log 2 +\pi\log\left(1+\frac{a^2-b^2}{a^2+b^2}\right)=2\pi\log(a)$$
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