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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 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.

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    Are you familiar with the [Cauchy principal value](https://en.wikipedia.org/wiki/Cauchy_principal_value) of an improper integral? – 2017-01-22
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    Your comment mostly cleared it up for me. I'm aware of it but our teacher didn't explain it very well. So the other definition is with Cauchy's principal value which gives a finite number and the integral then converges.Now I have to look up how this is possible. Thanks! – 2017-01-22
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    What class are you in? Usually CPV is a concept in classes past BC. – 2017-01-22
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    Electrical and computer engineering , first year, Differential Calculus and Mathematical analysis if I translate it correctly. It's the first math class. – 2017-01-22

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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.