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November 24, 2014

November 24, 2014

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

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 6:55pmYou 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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