Question

What is the improper integral 1/(2x-1)dx from 0 to 1/2 ? . .

Answer 1

`I=int_0^(1/2)1/(2x-1)*dx=lim_(brarr1/2)int_0^b1/(2x-1)*dx`

Explanation:

show below:

`I=int_0^(1/2)1/(2x-1)*dx`

Since `1/(2x-1)` not continues at `x=1/2 in [0,1/2]` we must use improper integral method to integrate it :

`I=int_0^(1/2)1/(2x-1)*dx=lim_(brarr1/2)int_0^b1/(2x-1)*dx`

`lim_(brarr1/2)1/2int_0^b2/(2x-1)*dx`

`1/2*lim_(brarr1/2)[ln(2x-1)]_0^b`

`1/2*lim_(brarr1/2)[ln(2b-1)]`

Note: The integral is divergent. The result shown is the Cauchy principal value

Note: It was assumed that `b>0,2b-1>0`

`1/2*lim_(brarr1/2)[ln(2b-1)]="Does not exist"`