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

1 Answer
Jun 5, 2018

#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"#