Valid application of integration by parts for piecewise functions

Question

Consider the integral $$ \int_{0}^{n} \left[ f(x)\cdot \left(\color{blue} {x-\left\lfloor x \right\rfloor-\frac{1}{2}} \right) \right] \mathrm{d}x $$ where $f$ is $k\ge 1$ times differentiable. I should reduce this integral to some expression which involves higher derivatives of $f$ (which can still be under an integral sign) meaning I have to find the antiderivative for the blue expression. I know how I would compute a definite integral of the form $$ \int_{0}^{t}\left( {x-\left\lfloor x \right\rfloor-\frac{1}{2}} \right) \mathrm{d}x $$ which yields the term $$ \frac{1}{2}\left( t-\left\lfloor t \right\rfloor \right)^2 - \frac{1}{2}(t-\left\lfloor t \right\rfloor) $$ but can I really apply the integration by parts formula on this to obtain $$ \int_{0}^{n} \left[ f(x)\cdot \left( {x-\left\lfloor x \right\rfloor-\frac{1}{2}} \right) \right] \mathrm{d}x = \left[ \frac{1}{2}\left( x-\left\lfloor x\right\rfloor \right)^2 - \frac{1}{2}(x-\left\lfloor x\right\rfloor)f(x) \right]_0^n +\\[15pt] \int_{0}^{n} \left[ \left( \frac{1}{2}\left( x-\left\lfloor x\right\rfloor \right)^2 - \frac{1}{2}(x-\left\lfloor x\right\rfloor) \right) f'(x) \right]\mathrm{d}x $$ or is this not allowed?

Answer 1

What you did is backed up by a generalized version of integration by parts theorem when one of the functions is absolutely continuous and the other one Lebesgue integrable.

$f$ is absolutely continuous as it is continuously differentiable. And the other one is Lebesgue integrable as it is piecewise continuous.