2

I was trying to figure out

$\int_{1}^{3}\frac{1}{x-2}dx$

to solve another integral. But, I ran into a problem. When I split this up into $$ \begin{split} \int_{1}^{3}\frac{1}{x-2}dx&=\lim_{A \to 2^-}\int_{1}^{A}\frac{1}{x-2}dx + \lim_{B \to 2^+}\int_{B}^{3}\frac{1}{x-2}dx\\ \\ &=\lim_{A \to 2^-}[\ln|x-2|]_{1}^{A} + \lim_{B \to 2^+}[\ln|x-2|]_{B}^{3} \end{split} $$

I end up with $\infty - \infty$. My professor said that if I have an indeterminate form, then I can't conclude that it's divergent. I would have to rework it. But, I have no idea how to do that in this case. I don't think that it's okay to just combine the limits. So, how would I deal with this?

Of course, I may have done something wrong.

Daniele Tampieri
  • 9,727
Fennly Xerxich
  • 21
  • You could try by a change of variable $x-2=y$. In this way, you get $\int_{-1}^1\frac{1}{y}dy$. Do you know how to take it from here? –  Mar 06 '19 at 16:19

1 Answers1

2

Actually, since the integrals on both sides of $2$ diverge, the integral is said to diverge.

However, there is the Cauchy Principal Value, which, in this case, would be $$ \begin{align} \operatorname{PV}\int_1^3\frac1{x-2}\,\mathrm{d}x &=\lim_{\epsilon\to0^+}\left(\int_1^{2-\epsilon}\frac1{x-2}\,\mathrm{d}x+\int_{2+\epsilon}^3\frac1{x-2}\,\mathrm{d}x\right)\\ &=\lim_{\epsilon\to0^+}(\log(\epsilon)-\log(\epsilon))\\[3pt] &=0 \end{align} $$

robjohn
  • 345,667