明显是0,下面是无穷大,而上面一定是个有限值:2>=∫[0->x] sintdt >= -2sint是绝对值sint，答案不是0，是派/2|sint|是周期为π的函数∫[0->π] |sint|dt = 2则有∫[0->nπ] |sint|dt = 2n,∫[0->(n+1)π] |sint|dt = 2(n+1)由于f(x)=∫[0->x] |sint|dt是个增函数，所以当x∈[nπ, (n+1)π]时,f(nπ)<=f(x)<=f((n+1)π),易得:f(nπ)/((n+1)π)<=f(x)/x<=f((n+1)π)/nπ同时取极限:x->∞时，n->∞，lim[n->∞] f(nπ)/((n+1)π) = lim[n->∞] 2n/((n+1)π) = 2/πlim[n->∞] f((n+1)π)/nπ 亦等于2/π，由夹逼定理得lim[x->∞] ∫[0->x] |sint|dt / x =2/π