calc
posted by sarah on .
how would you do this improper integral
1/(x1)
from 0 to 2
this is improper at one, so I split it up into two integrals
ln(x1) from 01 and
ln(x1) from 12
I then did for the first one the (lim t>1() of ln(t1))(ln(01))
and then the same thing for the second part
I didn't know if this was right though, or what the answer would be

You need to include absolute value signs in the argument of the logarithm:
lnx1 from 01 and
lnx1 from 12
What you find is that both the limits diverge logarithmically, so the integral doesn't exist. However, if you add both the terms together and take the limit in one go, then the divergent terms cancel. This is called the Cauchy principal value of the integral.