Undergrad Evaluating improper integrals with singularities

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The discussion centers on the evaluation of two improper integrals, with one integral yielding a defined value while the other does not exist in the traditional sense but has a Cauchy principal value. The integral from 0 to 3 of 1/(x-1)^(2/3) is confirmed to exist, while the integral from 0 to 8 of 1/(x-2) does not exist due to singularities within the interval. The discrepancy arises from the nature of the integrals, where the first integral can be evaluated in the complex plane, while the second requires the principal value approach due to its undefined nature over the interval. The conversation highlights the importance of understanding Riemann integrability versus Cauchy principal value integrability. This distinction is crucial for correctly interpreting the results of improper integrals with singularities.
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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)
 

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