Evaluating improper integrals with singularities

In summary, the conversation discusses the discrepancy between two improper integrals: ##\displaystyle \int_0^3 \frac{dx}{(x-1)^{2/3}} = 3(1+2^{\frac{1}{3}})## and ##\displaystyle \int_0^8 \frac{dx}{x-2} = \log 3##. While the former has a defined value, the latter does not and is given a "principle value" of ##\log 3##. This is due to the fact that the first integral gives complex results for ##x<1##, while the second takes real values over the entire interval of integration. The discrepancy is also
  • #1
Mr Davis 97
1,462
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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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  • #2
What level of class and textbook are you referring to?
 
  • #3
alan2 said:
What level of class and textbook are you referring to?
It's a book on integration techniques, called "Inside Interesting Integrals."
 
  • #5
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?
 
  • #6
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)
 

Related to Evaluating improper integrals with singularities

What is an improper integral?

An improper integral is an integral where one or both of the bounds of integration are infinite or where the integrand has a singularity (a point where the function is undefined or infinite) within the interval of integration.

How do I know if an integral is improper?

An integral is improper if one or both of the bounds of integration are infinite or if the integrand has a singularity within the interval of integration. This can also be determined by evaluating the integral and seeing if it diverges (goes to infinity) or converges (has a finite value).

What are singularities?

Singularities are points where a function is undefined or infinite. In the context of improper integrals, singularities can cause the integral to diverge or have a finite value.

How do I evaluate an improper integral with a singularity?

To evaluate an improper integral with a singularity, the singularity must be isolated and the integral must be split into two separate integrals. The first integral will have the singularity as one of its bounds, and the second integral will have the other bound. Both integrals must then be evaluated separately, and the sum of the two will be the value of the original integral.

What is the purpose of evaluating improper integrals with singularities?

Evaluating improper integrals with singularities allows us to find the value of integrals that would otherwise be impossible to evaluate. It also helps us understand the behavior of functions near their singularities and how they affect the overall value of the integral.

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