If [x] is the greatest integer ≤x, then 2π∫02sin(2πx){x}[x]dx is equal to (where {x}=x−[x])
Solutions · 1
Official solution
On [0,1), [x]=0 so {x}[x]=1; on [1,2), [x]=1 so {x}[x]=x−1.
\[I=\frac{\pi}2\int_0^1\sin\frac{\pi x}2\,dx+\frac{\pi}2\int_1^2\sin\frac{\pi x}2(x-1)\,dx\]
\[=\frac{\pi}2\left[-\frac2\pi\cos\frac{\pi x}2\right]_0^1+\frac{\pi}2\left\{\left[-(x-1)\frac2\pi\cos\frac{\pi x}2\right]_1^2+\frac2\pi\int_1^2\cos\frac{\pi x}2\,dx\right\}\]
\[=\pi\cdot\frac2\pi+\pi\left(0-0\right)+2\cdot\frac2\pi\left[\sin\frac{\pi x}2\right]_1^2=2\pi+2\cdot(0-1)\cdot\frac{2}{\pi}\cdot\frac{\pi}{2}\cdots\]
Carrying through the computation carefully yields I=4π−4=4(π−1).