# Need help figuring out error on Improper Integral Resolution

my friends. I upfront apologize for using any unusual terms below, I'm not used to writing math in english. It's a rather simple integral that, when put on wolframalpha, we find out its sum does not converge.

$$\int_0^3 {{1 \over {x - 1}}dx}$$

The thing is I came across one resolution of it that shows ln(2) as final answer, and I can't find in which step he got something wrong:

$$\int_0^3 {{1 \over {x - 1}}dx} = \mathop {\lim }\limits_{a \to {1^ - }} \int_0^a {{1 \over {x - 1}}dx} + \mathop {\lim }\limits_{b \to {1^ + }} \int_b^3 {{1 \over {x - 1}}dx}$$

$$\mathop {\lim }\limits_{a \to {1^ - }} \left( {\ln \left| {a - 1} \right| - \ln \left| { - 1} \right|} \right) + \mathop {\lim }\limits_{b \to {1^ + }} \left( {\ln 2 - \ln \left| {b - 1} \right|} \right)$$

$$\mathop {\lim }\limits_{a \to {1^ - } \atop b \to {1^ + }} {\rm{ }}\left( {\ln \left( {1 - a} \right) - \ln \left( {b - 1} \right)} \right) + \ln 2$$

$$\mathop {\lim }\limits_{a \to {1^ - } \atop b \to {1^ + } } \ln \left( {{{1 - a} \over {b - 1}}} \right) + \ln 2 = \mathop {\lim }\limits_{x \to 1} \ln \left| {{{1 - x} \over {x - 1}}} \right| + \ln 2$$

The limit on the left was solved using L'Hospital rule which returned an Ln(1)=0, so, final answer would be Ln(2).

Can anyone see where mistakes were made?

• Each of the limits must exist for its own , otherwise the integral does not converge. Google for "principle value" for further details – Peter Apr 11 '18 at 22:43
• thanks for this, mate, helped me understand it! Cheers. – Pedro Alonso Apr 12 '18 at 3:49

$$\mathop {\lim }\limits_{a \to {1^ - } \atop b \to {1^ + }} {\rm{ }}\left( {\ln \left( {1 - a} \right) - \ln \left( {b - 1} \right)} \right) + \ln 2$$

I not aware of that notation being considered acceptable. You can take the limit as the input approaches a point, but that is different from taking two different limits of two different variables. In the comments, @MikeEarnest mentions http://mathonline.wikidot.com/limits-of-functions-of-two-variables Note that here, the definition of "limit" translates easily, with one $\delta$ and one $\epsilon$. Now, we could interpret your notation as being the limit as (a,b) approaches (1,1), but there are several problems with that. First, there is no guarantee that this limit exists (and, indeed, in this case it does not), and second, you have a approaching 1 from below and b approaching 1 from above. It would be quite difficult to translate this into the limit as we approach (1,1).

$$\mathop {\lim }\limits_{a \to {1^ - } \atop b \to {1^ + } } \ln \left( {{{1 - a} \over {b - 1}}} \right) + \ln 2 = \mathop {\lim }\limits_{x \to 1} \ln \left| {{{1 - x} \over {x - 1}}} \right| + \ln 2$$

Here, any pretense that this can be interpreted as the limit as we approach (1,1) has to be abandoned. For the limit as we approach (1,1) to exist, it must be the same regardless of what path we use. You have limited us entirely to the a = b = x path, which is not valid.

Basically, what you're doing is taking this:

$$\int_0^3 {{1 \over {x - 1}}dx} = \mathop {\lim }\limits_{a \to {1^ - }} \int_0^a {{1 \over {x - 1}}dx} + \mathop {\lim }\limits_{b \to {1^ + }} \int_b^3 {{1 \over {x - 1}}dx}$$

and then you're replacing both a and b with the same variable. If you look at the graph of $\frac{1}{x-1}$, there's a singularity at 1. If you approach this singularity at the same rate from both sides, the values cancels out and the integral is finite. On the other hand, if you approach the left side faster than the right side, then you get the positive amounts faster, and you get positive infinity. If you approach the right side faster, you get negative infinity. It's not valid to pick one of these and call it the limit; if these limits aren't equal, then the limit doesn't exist. It's the same principle that says that if the sum of the limit of two series is not the limit of their sums, if their limits are not finite