悬赏200分：瑕积分求证题

Question

已知积分(0 -> pi/2) tanx dx = +无穷（发散）。
那么积分(0 -> pi/2) (tanx)^(1/2) dx 是否发散？
原题是英文所以翻译过来对于国内的高手看来会有些奇怪（本人在国外读书），所以不能理解题目的话请告诉我。
回答采纳后追加100分。
非常感谢！
英文原题：
Note integral from 0 to pi/2 (tanxdx) = +infinity diverges.
Does I = integral from 0 to pi/2 (tanx)^0.5 dx diverge?

Tip: This trigonometric identity must be used in solving the question: tanx = sinx / cosx.
Tip 2: Consider sinx^0.5 / cosx

Answer 1

令x=arctan t^2
则积分(0 -> pi/2) (tanx)^(1/2) dx
=积分(0 -> 正无穷) 2t^2/(1+t^4) dt
=积分(0 -> 1) 2t^2/(1+t^4) dt + 积分(1 -> 正无穷) 2t^2/(1+t^4) dt
其中，积分(0 -> 1) 2t^2/(1+t^4) dt<积分(0 -> 1) 2t^2/(2t^2) dt=1，是有限数
积分(1 -> 正无穷) 2t^2/(1+t^4) dt
< 积分(1 -> 正无穷) 2t^2/(t^4) dt
=积分(1 -> 正无穷) 2/t^2 dt
= (-2/t) | (1 -> 正无穷)
=2，且积分(1 -> 正无穷) 2t^2/(1+t^4) dt>0，是有限数
因此应该是收敛，即converge

Answer 2

(0 -> pi/2) tanx dx = +无穷（发散)
(0 -> pi/2) (tanx)^(1/2) dx
+无穷开方还是正无穷
所以还是发散

Answer 3

可以发英文的吗？我也是在国外的，中文的不太懂。

追问

(tanx)^0.5 就是根号下tanx 但我不会打根号

追答

明白了。
答案是diverges。
∫(0)(π/2)=lim(t->π/2^( - ))∫(0)(t) (tanx)^0.5 dx
显然这个极限不是finite number，那瑕积分就是diverges的。

∫(0)(t)是 integral from 0 to t.

追问

I'm afraid this answer is not good enough for my prof.. Please read the supplementary info again.

追答

To complement your proof, you can do the integration of [integral from 0 to t (tanx)^0.5 dx] first.
The supplementary will help prove the limit of that integration is +infinity.
Then I=... diverges. ^ ^