Floor function from India

For every $x \in \mathbb{R}$ let $[x]$ denote the floor function (that is the greater integer that is less or equal to $x$.) We define the real function $f:[-10, 10 ]\rightarrow \mathbb{R}$ as:

$$f(x)= \left\{\begin{matrix}
x-[x] & , & [x]=2k+1\\
 1+[x]-x&,  & [x]=2k
\end{matrix}\right., \quad k \in \mathbb{Z}$$

Evaluate the integral: $\displaystyle \mathcal{J}= \frac{\pi^2}{10}\int_{-10}^{10}f(x)\cos \pi x \, {\rm d}x$.

(IIT JEE 2010, India)

Solution

Noting that in the interval $[0, 1)$ the greatest integer that is less or equal to $x$ is zero this means that $f(x)=1-x, \; x \in [0, 1]$. Same in the interval $[1, 2)$ the greatest integer that is less or equal to $x$ is one , hence the function is of the form $f(x)=x-1, \; x \in [1, 2]$. The graph of the functions $f$ and $g(x)=\cos \pi x$ are seen below:

Obviously the function $g$ is integrable , since it is continuous. Now we note that:

$$\int_{0}^{1}f(x)\cos \pi x \, {\rm d}x = \int_{1}^{2}f(x) \cos \pi x\, {\rm d}x =\cdots$$

etc. The above hold due to symmetry. Hence the integral we are interested in is equal to:

$$\mathcal{J}= 20 \cdot \frac{\pi^2}{10}\cdot \int_{0}^{1}f(x)\cos \pi x \, {\rm d}x = 2\pi^2 \int_{0}^{1}\left ( 1-x \right )\cos \pi x\, {\rm d}x = 4$$