Convergence of improper integral with parameter

Question

I'm studying improper integrals for an exam and solving some exercises from previous exams and I came across this. It says: "Give two definitions of the following integral and analyse the convergence for each case. Explain the result." $$\int_0^3 1/(x-a) dx$$ $$ 0<a<3 $$ What does he mean by two definitions? (Maybe I'm not translating correctly). But all I know is I have to split the integral because it's discontinuous at x=a. So I have $\lim_{k\to a} \int_0^k 1/(x-a)dx + \lim_{k\to a}\int_k^3 1/(x-a) dx $
The result for the first one is $-\infty$ so the integral is divergent.Based on what I'm asked to do I'm sure this is not the solution.

Answer 1

I think what they might mean (and sorry if I'm wrong here) is that you should take two different limits that could define the improper integral and compare them. (By definition, the integral exists only if all possible defining limits are the same.)

The Cauchy principal value mentioned in the comments is one: $$ \lim_{\epsilon\rightarrow 0^+}\left(\int_0^{a-\epsilon}\frac{1}{x-a}dx + \int_{a+\epsilon}^3\frac{1}{x-a}dx\right) = \lim_{\epsilon\rightarrow 0^+}\left(\ln\left(\frac{\epsilon}{a}\right) + \ln\left(\frac{3-a}{\epsilon}\right)\right) =\ln\left(\frac{3-a}{a}\right) $$

Another might be $$ \lim_{\epsilon\rightarrow 0^+}\left(\int_0^{a-\epsilon}\frac{1}{x-a}dx + \int_{a+2\epsilon}^3\frac{1}{x-a}dx\right)= \lim_{\epsilon\rightarrow 0^+}\left(\ln\left(\frac{\epsilon}{a}\right) + \ln\left(\frac{3-a}{2\epsilon}\right)\right) =\ln\left(\frac{3-a}{2a}\right). $$ The fact that these are not equal shows that the improper integral does not exist.