Get integral of function containing integer part of x

Question

I have an integral: $$ \int [x]\cdot |\sin(\pi x)| dx;\ \ x > 0 $$ I‘ m trying to calculate it by integrating by parts, but a don’t know how to define [x]’. This is what I got: $$ [x]’ = \lim_{\Delta x \to 0} \left(\frac{[x + \Delta x] - [x]}{\Delta x}\right) \not\exists $$ I can’t define f(x) = [x]

Answer 1

The function $f : x \mapsto \lfloor x \rfloor \sin(\pi x)$ is continuous on $\mathbb{R} \setminus \mathbb{Z}$ and also on $\mathbb{Z}$, since $\sin(\pi x) \to 0$ and $\lfloor x \rfloor$ remains bounded when $x \to k \in \mathbb{Z}$. Hence it admits a primitive. Call $F$ the primitive such that $F(0)=0$.

For $k \in \mathbb{Z}$ and $x \in [k,k+1]$, $$F(x)-F(k\pi) = \int_{k\pi}^x k(-1)^k \sin(\pi t) \mathrm{d}t = k(-1)^k \frac{(-1)^k -\cos(\pi x)}{\pi}) = \frac{k}{\pi} ( 1-(-1)^k\cos(\pi x))$$ In particular $$F((k+1)\pi)-F(k\pi) = \frac{2k}{\pi}.$$ Two recursions (one for the positive integers, one for the negative integers) shows that for every $k \in \mathbb{Z}$, $F(k\pi) = k(k-1)/\pi$. As a result, for $k \in \mathbb{Z}$ and $x \in [k,k+1]$, $$F(x) = \frac{k(k-1)}{\pi} + \frac{k}{\pi} ( 1-(-1)^k\cos(\pi x)) = \frac{k^2}{\pi} - (-1)^k\frac{k}{\pi} \cos(\pi x).$$