Posted by **sarah** on Saturday, February 9, 2008 at 6:34pm.

how would you do this improper integral

1/(x-1)

from 0 to 2

this is improper at one, so I split it up into two integrals

ln(x-1) from 0-1 and

ln(x-1) from 1-2

I then did for the first one the (lim t->1(-) of ln(t-1))-(ln(0-1))

and then the same thing for the second part

I didn't know if this was right though, or what the answer would be

- calc -
**Count Iblis**, Saturday, February 9, 2008 at 6:55pm
You need to include absolute value signs in the argument of the logarithm:

ln|x-1| from 0-1 and

ln|x-1| from 1-2

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.

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