How to calculate the integral $\int\limits_0^{\pi} \ln(1-2\alpha\cos{x}+\alpha^2) dx$ by Fourier Series? I have $\dfrac{q\sin{x}}{1-2q\cos{x}+q^2}=\sum\limits_{n=0}^{\infty} q^n\sin{nx} (q&lt;1)$ now, I tried term wise integration but I cannot figure it out.   The answer is $0$ if $|\alpha|\leqslant 1$, $2\pi\ln|\alpha|$ if $|\alpha|&lt;1$.
For $a\ge 0$ let's define $$I(a)=\int_{0}^{\pi}\ln\left(1-2a\cos x+a^2\right)dx.$$ Find explicit formula for $I(a)$.   My attempt: Let  $$\begin{align*} f_n(x) &amp;= \frac{\ln\left(1-2 \left(a+\frac{1}{n}\right)\cos x+\left(a+\frac{1}{n}\right)^2\right)-\ln\left(1-2a\cos x+a^2\right)}{\frac{1}{n}}\\ &amp;=\frac{\ln\left(\displaystyle\frac{1-2 \left(a+\frac{1}{n}\right)\cos x+\left(a+\frac{1}{n}\right)^2}{1-2a\cos x+a^2}\right)}{\frac{1}{n}}\\ &amp;=\frac{\ln\left(1+\dfrac{1}{n}\left(\displaystyle\frac{2a-2\cos x+\frac{1}{n}}{1-2a\cos x+a^2}\right)\right)}{\frac{1}{n}}. \end{align*}$$ Now it is easy to see that $f_n(x) \to \frac{2a-2\cos x}{1-2a\cos x+a^2}$ as $n \to \infty$. $|f_n(x)|\le \frac{2a+2}{(1-a)^2}$ RHS is integrable so $\lim_{n\to\infty}\int_0^\pi f_n(x)dx = \int_0^\pi \frac{2a-2\cos x}{1-2a\cos x+a^2} dx=I'(a)$. But  $$\int_0^\pi \frac{2a-2\cos x}{1-2a\cos x+a^2}=\int_0^\pi\left(1-\frac{(1-a)^2}{1-2a\cos x+a^2}\right)dx.$$ Consider $$\int_0^\pi\frac{dx}{1-2a\cos x+a^2}=\int_0^\infty\frac{\frac{dy}{1+t^2}}{1-2a\frac{1-t^2}{1+t^2}+a^2}=\int_0^\infty\frac{dt}{1+t^2-2a(1-t^2)+a^2(1+t^2)}=\int_0^\infty\frac{dt}{(1-a)^2+\left((1+a)t\right)^2}\stackrel{(*)}{=}\frac{1}{(1-a)^2}\int_0^\infty\frac{dt}{1+\left(\frac{1+a}{1-a}t\right)^2}=\frac{1}{(1-a)(1+a)}\int_0^\infty\frac{du}{1+u^2}=\frac{1}{(1-a)(1+a)}\frac{\pi}{2}.$$ So $$I'(a)=\frac{\pi}{2}\left(2-\frac{1-a}{1+a}\right)\Rightarrow I(a)=\frac{\pi}{2}\left(3a-2\ln\left(a+1\right)\right).$$  It looks too easy, is there any crucial lack?  $(*)$ — we have to check $a=1$ here by hand and actually consider $[0,1), (1,\infty)$ but result on these two intervals may differ only by constant - it may be important but in my opinion not crucial for this proof.
Set a global policy for floats positioning I have a document with many floats (figures, specifically).  I want to be able to change the global floats positioning policy from htbp to just h (or even H with the floats package).  For my-own reasons, I don't want to do this by appending the [h] argument to each float env.  One option that I can think of is using a macro:  \newcommand{\mypos}{h} % or htbp \begin{figure}[\mypos]...   I also prefer not to do that - It is cumbersome.  Is there a way to set the global float positioning argument globally? Does Auto White-Balance Really Work? How? SEDE should pick up on changes to community metadata It appears that site meta data isn't updated in SEDE. As an example, Moderators.SE was  to  in December 2014. It is still listed in SEDE as Moderators.SE, as of July 2015.  To search for Community Building data, we need to search the "Moderators" community:    It is not listed as Community Building: Why doesn't my systemd user unit start at boot? 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So how would I express this?