∫(0->2) (e^2x + 1/x) dx
= (1/2)e^2x + lnx：(0->2)
= (1/2)e^4 + ln2 - (1/2*1 + ln0)
= (1/2)e^4 + ln2 - 1/2 + ln0,由于ln0趋向负无穷大
∴这个积分发散
∫(0->π/2) sin²(x/2) dx
= (1/2)∫ (1-cosx) dx
= (1/2)(x - sinx)：(0->π/2)
= (1/2)[π/2 - sin(π/2)] - (1/2)(0 - sin0)
= (1/2)(π/2 - 1)
= π/4 - 1/2