微積分的improper integrals的問題3？

Which of the following integrals are improper ?why ?

1.∫1/(2x-1)dx 此為定積分上 2下1

2.∫1/(2x-1)dx 此為定積分上 1下0

3.∫sin x/(1+x^2) dx 此為定積分上∞下 -∞

4.∫ln(x-1)dx 此為定積分上 2下1

Update:

convergent和divergent怎麼分？

Update 2:

improper integrals是暇積分？

Update 3:

Update 4:

Update 5:

∫1/x^2 dx 定積分上∞下1-----converge

∫1/x dx 定積分上∞下1-----diverge

Update 6:

Rating

By the definition of improper integra l:

An improper integral is a definite integral that has either or both limits infinite or an integrand that approaches infinity at one or more points in the range of integration. Improper integrals cannot be computed using a normal Riemann integral.

Hence 1 is general integral, 2,3,4 is improper integral.

Noting that 2 is improper at x=1/2, 3is improper at x = ∞, -∞ and 4 is improper at x=1.

2009-03-18 14:59:31 補充：

∫1/x^2 dx 定積分上∞下1-----converge

∫1/x dx 定積分上∞下1-----diverge

為何呢？

∫[1~∞] 1/(x^p) dx

p>1時收斂,p≤1發散! (微積分教科書就有)

第2題和第4題

都沒有∞的定積分

為何是暇積分？

請您注意第2題因為被積分函數在1/2處不連續(因為現在積分的上下限為[0~1] , 1/2∈[0~1]所以為瑕積分(improper integral)

至於第4題 ln(x-1) 因為真數x-1>所以x>1

顯然x=1這個點也是瑕點喔

• linch
Lv 7

improper integrals是暇積分？ 是的!!

2,3,4 都是 improper integral

其中 2,4 發散 3 收斂