How do you determine if the improper integral converges or diverges #int dx/(2x-1)# from 0 to infinity?

2 Answers
Mar 10, 2016

Use the fact that #int dx/(2x-1) =1/2 ln abs(2x-1) +C#. As #xrarroo#, so does #ln(2x-1)#, so the integral diverges.

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 #1/2#

#int_0^oo dx/(2x-1) = int_0^(1/2) dx/(2x-1)+int_(1/2)^oo dx/(2x-1)# provided that both integrals converge (exist).

#int_0^(1/2) dx/(2x-1) = lim_(brarr(1/2)^-) int_0^b dx/(2x-1)#

# = 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 #lim_(brarr(1/2)^-)ln(1-2b) = -oo#.

Therefore the integral diverges, so the big integral diverges.

Mar 10, 2016

Show #int_0^oo 1/(2x-1) dx# diverges by elementary means.

Explanation:

Let #f(x) = 1/(2x-1)#

Let #a_i = 2^(i-2)+1/2# for #i = 1, 2, 3,...#

Then #a_i < a_(i+1)# for all #i# and #a_i -> oo# as #i->oo#

Then:

#f(a_i) = 1/(2^(i-1))#

#f(a_(i+1)) = 1/(2^i)#

Hence #f(x) >= 1/(2^i)# for any #x in [a_i, a_(i+1)]#

We find:

#a_(i+1) - a_i = (2^(i-1)+1/2) - (2^(i-2)+1/2) = 2^(i-2)#

So #int_(a_i)^(a_(i+1)) f(x) dx >= 2^(i-2) * 1/(2^i) = 1/4#

Hence #int_(a_1)^(a_n) f(x) dx >= (n-1)/4 -> oo# as #n->oo#

graph{(y-1/(2x-1)) = 0 [-4.33, 7.984, -2.81, 3.344]}