If [x] is the greatest integer $\le$ x, then

$${\pi ^2}\int\limits_0^2 {\left( {\sin {{\pi x} \over 2}} \right)(x - [x]} {)^{[x]}}dx$$ is equal to :

<p>$$I = {\pi ^2}\int_0^2 {\sin \left( {{{\pi x} \over 2}} \right){{(x - [x])}^{[x]}}dx} $$</p> <p>$$ = {\pi ^2}\int_0^1 {\sin \left( {{{\pi x} \over 2}} \right){x^0}dx + {\pi ^2}\int_1^2 {\sin \left( {{{\pi x} \over 2}} \right)(x - 1)dx} } $$</p> <p>$$ = {\pi ^2}\left[ {{{ - 2} \over \pi }\cos {{\pi x} \over 2}} \right]_0^1 + {\pi ^2}\left[ {(x - 1){2 \over \pi }\left( { - \cos {{\pi x} \over 2}} \right)} \right]_1^2 + {\pi ^2}\int_1^2 {{2 \over \pi }\cos {{\pi x} \over 2}dx} $$</p> <p>$$ = \left. {{\pi ^2}\left( {{2 \over \pi }} \right) + {{2{\pi ^2}} \over \pi }(1 - 0) + 2\pi \,.\,{2 \over \pi }\left( {\sin {{\pi x} \over 2}} \right)} \right|_1^2$$</p> <p>$= 2\pi + 2\pi + 4(0 - 1) = 4\pi - 4 = 4(\pi - 1)$</p>