Lim (1-x) [arctanx/(1-x 3)] = pi/12 (where x→ 1-)
That is, p= 1, λ=pi/ 12, so this deficient integral diverges.
(6) Because x=0 is a defect, and
limx^(m-2)[( 1-cosx)/x^m]=lim( 1-cosx)/x^2= 1/2
That is λ= 1/2.
Therefore, when m
(7) Because x=0 is a defect, and
|f(x)|=||≤ 1/x^α
So when 0
1≤α& lt; Quadratic conditional convergence
Divergence when α≥2
(8) Firstly, the integration region is divided into two parts (0, +∞)=[0, e]∞(e, +∞).
∫exp(-x)lnxdx has a defect of x=0 at (0, e) because.
Limx (1/2) exp (-x) lnx = 0, so ∫exp(-x)lnxdx converges to (0, e).
The infinite integral ∫exp(-x)lnxdx also converges to (e, +∞).
Because limx 2lnx/exp (x) = 0 (where x→+∞)
So this loss integral converges.