令√x=u,则 ∫e^(-√x)dx = 2∫ue^(-u)du = -2∫ude^(-u)
= -2ue^(-u)+2∫e^(-u)du = -2(u+1)e^(-u)+C
=-2(√x+1)e^(-√x)+C.
∫
-2[1+√(n+1)]e^[-√(n+1)]+2(1+√n)e^(-√n)
∑
= 4/e - 2(1+√2)e^(-√2) + 2(1+√2)e^(-√2) - 2(1+√3)e^(-√3)
+2(1+√3)e^(-√3)-2(1+√4)e^(-√4)+......
+2(1+√n)e^(-√n)-2[1+√(n+1)]e^[-√(n+1)]+......
= 4/e - 2lim
因 lim
= lim
故 ∑