反常积分收敛问题详细证明当1

当a>1时,I=∫ln(1+x^3)/x^adx
=1/(1-a)∫ln(1+x^3)/dx^(1-a)
=1/(1-a){[ln(1+x^3)x^(1-a)]-3∫[x^(3-a)/(1+x^3)]dx}
=A-3J.
其中 A=[ln(1+x^3)x^(1-a)]
=limln(1+x^3)/x^(a-1)-ln2
=lim3x^2/[(a-1)x^(a-2)(1+x^3)]-ln2
=lim3x^2/{(a-1)[x^(a-2)+x^(a+1)]}-ln2
=lim3/{(a-1)[x^(a-4)+x^(a-1)]}-ln2=-ln2,
J=∫[x^(3-a)/(1+x^3)]dx
=∫1/x^adx-∫1/[(1+x^3)x^a]dx
=K1-K2.
式中 K1=1/[(1-a)x^(a-1)]=1/(a-1),
令x=1/t,得 K2=∫[-t^(1+a)/[(t^3+1)]dt
=∫[t^(1+a)/[(t^3+1)]dt