化为二重积分来讨论:
∫[0->1](x^(p-1)-x^(q-1))dx/lnx
=∫[0->1]dx∫[q->p] x^(y-1) dy
=∫[q->p]dy∫[0->1] x^(y-1) dx
=∫[q->p] (1/y) dy
=ln|p|-ln|q|
故当p≠0或q≠0,或者p,q同时为0时,积分收敛
化为二重积分来讨论:
∫[0->1](x^(p-1)-x^(q-1))dx/lnx
=∫[0->1]dx∫[q->p] x^(y-1) dy
=∫[q->p]dy∫[0->1] x^(y-1) dx
=∫[q->p] (1/y) dy
=ln|p|-ln|q|
故当p≠0或q≠0,或者p,q同时为0时,积分收敛