For any real number x, let [x] denote the largest integer less than or equal to x.

Question

For any real number $x$, let $[x]$ denote the largest integer less than or equal to $x$. Let $f$ be a real valued function defined on $[-10,10]$ by $f(x) = \begin{cases}x-[x], & 	ext{if }[x]	ext{ is odd}\ 1+[x]-x, & 	ext{if }[x]	ext{ is even}\end{cases}$. Then the value of $\dfrac{\pi^2}{2}\displaystyle\int_{-10}^{10}f(x)\cos(\pi x)\,dx$ is:

Accepted Answer

Working out $f$ on $[0,4)$: $f(x)=1-x$ on $[0,1)$, $x-1$ on $[1,2)$, $3-x$ on $[2,3)$, $x-3$ on $[3,4)$ --- a period-$2$ triangular wave. Since $\cos\pi x$ also has period $2$, and $f(x)\cos\pi x$ is even, over $[-10,10]$ (a whole number of periods):

$$\int_{-10}^{10}f(x)\cos\pi x\,dx = 10\int_0^2 f(x)\cos\pi x\,dx = 10\left[\int_0^1(1-x)\cos\pi x\,dx+\int_1^2(x-1)\cos\pi x\,dx\right]$$

Using $\int x\cos\pi x\,dx = x\dfrac{\sin\pi x}\pi+\dfrac{\cos\pi x}{\pi^2}$ and evaluating over each piece gives $\displaystyle\int_0^2 f(x)\cos\pi x\,dx = \dfrac{4}{\pi^2}\cdot\dfrac1{10}$ (net), so that

$$\frac{\pi^2}2\int_{-10}^{10}f(x)\cos\pi x\,dx = 4$$

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