Absolute convergence of $\int_{0}^{1} \left(x\cos\left(\frac{\pi}{x}\right)\right)' dx$ $\int_{0}^{1} \left(x\cos\left(\frac{\pi}{x}\right)\right)' dx$ converges. Does it absolutely converges?
I found easy to show that it converges (by a direct computation or the subtraction of limits). I'm certain that it doesn't converges absolutely. But to show that it doesn't coverges seems to be a bit tricky. If it wasn't for $x=0$, the integral would absolutely converge. $\int_{0}^{1} \left|\left(x\cos\left(\frac{\pi}{x}\right)\right)'\right| dx$ = $\int_{0}^{1} \left|\cos\left(\frac{\pi}{x} \right)+\frac{\pi}{x}\sin\left(\frac{\pi}{x}\right)\right|dx$. Then if you try to use de triangle inequality it won't work, since the second term diverges. So it would tell you nothing. My second idea was to use the sum of triangles underneath the function. I mean $\sum_{n=1}^{\infty} \left|\int_{n\pi}^{(n+1)\pi} \cos\left(\frac{\pi}{x} \right)+\frac{\pi}{x}\sin\left(\frac{\pi}{x}\right)\right|dx$. I tried also the substitution $u = \frac{\pi}{x}$, which allows you to consider the integral of $\left|\frac{\cos(u)}{u^2} + \frac{\sin(u)}{u}\right|$. Thus you can compare that sum with $\sum \frac{1}{n^2} + \frac{1}{n}$. But I'm not sure if these is the path I should follow. Could someone help me or give further insights?
 A: $$
\int_{0}^{1} \left(x\cos\left(\frac{\pi}{x}\right)\right)' dx=\int_0^1\cos\frac{\pi}{x}\,dx-\pi\int_0^1\sin\left(\frac{\pi}{x}\right)\frac{dx}x.
$$
The first integral on the RHS is absolutely convergent since $|\cos y|\le 1,\, y\in\mathbb{R}$, so the initial integral is absolutely convergent iff the second integral on the RHS is such. 
We shall show that it is not. Changing variables $y(x)=\frac{1}{x}$ we get
$$
\int_0^1\sin\left(\frac{\pi}{x}\right)\frac{dx}x=\int_1^\infty \sin (\pi y)\frac{dy}{y}.
$$
We have
\begin{align*}
\int_1^\infty |\sin (\pi y)|\frac{dy}{y}&\ge \sum_{k=1}^\infty \int_{k+1/4}^{k+3/4}|\sin (\pi y)|\frac{dy}{y}\ge \frac{1}{\sqrt 2} \sum_{k=1}^\infty \int_{k+1/4}^{k+3/4}\frac{dy}{y}\\
&=\frac{1}{\sqrt 2} \sum_{k=1}^\infty \log\left(\frac{k+3/4}{k+1/4}\right)=\frac{1}{\sqrt 2} \sum_{k=1}^\infty \log\left(1+\frac{1/2}{k+1/4}\right)\\
&\simeq \sum_{k=1}^\infty \frac{1}k=+\infty,
\end{align*}
where in the penultimate relation we used 
$$
\lim_{x\rightarrow 0}\frac{\log(1+x)}{x}=1.
$$
