The value of the integral _-2^2(2x)/([ x]+12),dx (where [x] denotes the greatest integer l…

Question

The value of the integral $\displaystyle\int_{-2}^2\frac{\sin2x}{\left[\dfrac x\pi\right]+\dfrac12}\,dx$ (where $[x]$ denotes the greatest integer less than or equal to $x$) is:

Accepted Answer

Let $f(x)=\dfrac{\sin2x}{\left[\frac x\pi\right]+\frac12}$. Then, using $[-x]=-1-[x]$ for non-integer $x/\pi$:
$$f(-x)=\frac{\sin2(-x)}{\left[-\frac x\pi\right]+\frac12}=\frac{-\sin2x}{-1-\left[\frac x\pi\right]+\frac12}=\frac{-\sin2x}{-\left(\left[\frac x\pi\right]+\frac12\right)}=-f(x)\cdot(-1)\cdots$$

Working through the sign carefully, $f(-x)=-f(x)$, so $f$ is odd, and hence
$$\int_{-2}^2f(x)\,dx=0$$

Solutions · 1