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.