# I Evaluating improper integrals with singularities

1. Mar 15, 2017

### Mr Davis 97

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

2. Mar 15, 2017

### alan2

What level of class and textbook are you referring to?

3. Mar 15, 2017

### Mr Davis 97

It's a book on integration techniques, called "Inside Interesting Integrals."

4. Mar 16, 2017

### MAGNIBORO

5. Mar 20, 2017

### Mr Davis 97

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. Mar 20, 2017

### MAGNIBORO

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)

7. Mar 22, 2017