I am trying to find the integral of:
$\int \frac{\pi}{x} \sin (\frac{\pi}{x})dx$.
I am supposed to use integration by parts to obtain: $x\cos (\frac{\pi}{x}) - \int \cos (\frac{\pi}{x})dx$ but can't seem to be getting anywhere.
First, I tried choosing $u = \sin (\frac{\pi}{x})$ and $dv = \frac{\pi}{x}$ but obtained the dreaded unhelpful expression $0 = 0$.
I then concluded I should choose $u = \frac{\pi}{x}$ and $dv = \sin (\frac{\pi}{x})dx$, which isn't like anything I've ever done before. Still I proceeded because I've accepted that math gets weird sometimes.
I calculated $\int \sin (\frac{\pi}{x})dx$ and obtained $x\sin (\frac{\pi}{x}) + \int \frac{\pi}{x} \cos (\frac{\pi}{x})dx$ and used that as $v$:
$$\int \frac{\pi}{x} \sin \Big (\frac{\pi}{x}\Big )dx = \pi \sin \Big (\frac{\pi}{x}) + \frac{{\pi}^2}{x} \int \frac{1}{x} \cos \Big (\frac{\pi}{x}\Big )dx + {\pi}^2 \int \Bigg (\frac{\sin \Big (\frac{\pi}{x}\Big )}{x} + \frac{1}{x^2}\int \frac{\pi}{x} \cos \Big (\frac{\pi}{x}\Big )dx \Bigg ) dx$$
I even tried simplifying it further by finding $\int \frac{\pi}{x} \cos (\frac{\pi}{x})dx$, at which point my work became very muddled and I wasn't any closer to getting the desired result.
Does anyone have any hints or suggestions? Thanks in advance. (PS: if you want some context behind this problem, I am exploring measure theory and the connection between Riemann and Lebesgue integration, so I figured trying to solve this integral shouldn't be the focus of my problem.)
Let $I=\int \frac{\pi}{x}\sin(\pi/x)\,dx$.
Integrating by parts with $u=x$ and $dv=(\pi/x^2)\sin(\pi/x)$. Then, $du=dx$ and $v=\cos(\pi/x)$. Proceeding, we have
$$I=x\cos(\pi/x)-\int \cos(\pi/x)\,dx$$
as was to be shown.