Question

关于判断积分的敛散性,我计算到两个积分都是收敛的,第一个是等于arctan(x+1)|-∞到+∞=pai,第二个是等于a

Answer 1

1、∫[-∞--->+∞] 1/(x²+2x+2) dx

=∫[-∞--->+∞] 1/[(x+1)²+1] dx

=arctan(x+1) |[-∞--->+∞]

=π/2-(-π/2)

=π

2、∫ [1--->e] 1/[x√(1-ln²x)] dx

=∫ [1--->e] 1/√(1-ln²x) d(lnx)

=arcsin(lnx) |[1--->e]

=arcsin(lne)-arcsin(ln1)

=arcsin1-0

=π/2