Question

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

Answer 1

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.

Answer 2

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]}