What is the Pointwise Limit of This Sequence of Integral Functions?

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Here is this week's POTW:

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Find the pointwise limit of the sequence of integral functions

$$ f_n(x) = \int_0^{2\pi} \frac{\cos n\phi}{1 - 2x\cos \phi + x^2}\, d\phi, \quad 0 < x < 1$$-----Remember to read the http://www.mathhelpboards.com/showthread.php?772-Problem-of-the-Week-%28POTW%29-Procedure-and-Guidelines to find out how to http://www.mathhelpboards.com/forms.php?do=form&fid=2!

 

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Congratulations to Opalg and Ackbach for solving this week's POTW!

Here is Opalg's solution:

Spoiler

Substitute $z=e^{i\phi}$. Then $\cos n\phi = \operatorname{re\,}(z^n)$, $2\cos\phi = z + z^{-1}$, $dz = ie^{i\phi}d\phi$ so that $d\phi = \frac{dz}{iz}$, and the $z$-integral is taken round the unit circle. Therefore $$f_n(x) = \operatorname{re\,} \oint \frac{z^n}{(z-x)(z^{-1}-x)} \,\frac{dz}{iz} = \operatorname{re\,}\frac1i \oint \frac{z^n}{(z-x)(1-xz)}\,dz. $$ There is only one pole inside the contour, at $z=x$, where the residue is $\dfrac{x^n}{1-x^2}$. Therefore $f_n(x) = \dfrac{2\pi x^n}{1-x^2}$, which converges pointwise to zero as $n\to\infty$, for $0<x<1$.

You can read Ackbach's solution below.

Spoiler

Suppose $0<x<1$. Let
$$g(\phi):=\frac{1}{1-2x\cos(\phi)+x^2}.$$
Then $g(\phi)$ is continuous on $[0,2\pi]$, since its denominator is continuous on $[0,2\pi]$ and nowhere zero, hence is Riemann integrable. By the Riemann-Lebesgue Lemma,
$$\lim_{n\to\infty}f_n(x)=0.$$