Answer:
Step-by-step explanation:
a)
this can be done by substitute lnx = u
dx/x = du
When x =1, u =0 and when x =2, u = ln 2
So integral =
We find that this integral value is not definid for p =1
Hence for values of p other than 1, this converges.
When we substitute limits
and converges for p ≠1
b)
So not converging for p =1
But ln x is defined only for x >0
So integral 0 to 1 makes this integral not valid and hence not convergent.
