Evaluating improper integrals with singularities

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Mr Davis 97
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For two improper integrals, my textbook claims that ##\displaystyle \int_0^3 \frac{dx}{(x-1)^{2/3}} = 3(1+2^{\frac{1}{3}})## and that ##\displaystyle \int_0^8 \frac{dx}{x-2} = \log 3##. However, when I put these through Wolfram Alpha, the former exists but the latter does not, and it says that the "principle value" is ##\log 3##. I am not sure why there is this discrepancy, but it would be nice if someone could explain
 
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What level of class and textbook are you referring to?
 
alan2 said:
What level of class and textbook are you referring to?
It's a book on integration techniques, called "Inside Interesting Integrals."
 
MAGNIBORO said:
look https://en.wikipedia.org/wiki/Cauchy_principal_value
the P.V is a way to give a value for the integral, because is not define if in the interval of integration the function takes values ##\pm \infty##
But what is the difference between the two integrals such that Wolfram Alpha would say that the former exists and has that value, while the latter does not exist but has a principal value of log3?
 
the first integral give complex results from ##x<1##, and the second take real values over the all the interval of integration.
I'm not completely sure if that's the main reason But I remember that in complex integration the P.V Is very useful for finding the value of real integrals by integrating in the complex plane Separating the path of integration

edit: in the first integral, wolfram give a complex value.
https://www.wolframalpha.com/input/?i=integral+0+to+3+1/(x-1)^(2/3)