How do you determine if the improper integral converges or diverges #int dx/(2x-1)# from 0 to infinity?
2 Answers
Use the fact that
Explanation:
The answer above is how I determine quickly that the integral diverges.
If I'm going to do a detailed analysis, I start by observing that the integral must be split into two improper integral to avoid integrating across a discontinuity at
# = lim_(brarr(1/2)^-) 1/2 ln abs(2x-1)]_0^b#
# =1/2 lim_(brarr(1/2)^-)ln(1-2x)]_0^b #
# =1/2 lim_(brarr(1/2)^-)ln(1-2b) - ln1 #
But
Therefore the integral diverges, so the big integral diverges.
Show
Explanation:
Let
Let
Then
Then:
#f(a_i) = 1/(2^(i-1))#
#f(a_(i+1)) = 1/(2^i)#
Hence
We find:
#a_(i+1) - a_i = (2^(i-1)+1/2) - (2^(i-2)+1/2) = 2^(i-2)#
So
Hence
graph{(y-1/(2x-1)) = 0 [-4.33, 7.984, -2.81, 3.344]}