# Generelized Integral Convergence

Here what i have to solve:

$$\int_{0}^2 \frac{1}{x-1} \mathrm dx$$

I have to say if that integral convergse.

What I did (which is false) :

$$\int_{0}^2 \frac{1}{x-1} \mathrm dx=\int_{0}^1 \frac{1}{x-1} \mathrm dx+\int_{1}^2 \frac{1}{x-1} \mathrm dx$$

$$= \lim_{n \to 1} \int_{0}^n \frac{1}{x-1} \mathrm dx + \lim_{n \to 2} \int_{0}^n \frac{1}{x-1} \mathrm dx$$

$$= \lim_{n \to 1^-} \ln(|n-1|) - \lim_{n \to 1^+} \ln(|n-1|) = 0$$

Then it converges. But I don't undestand why the real solution is that it doesn't converge.

-

If $\int_{0}^2 \frac{1}{x-1} \mathrm dx$ would converge, also $\int_{0}^1 \frac{1}{x-1} \mathrm dx$ had to be convegent.

-
Why ? if a term of the limit is not convergent then the limit is not convergent? –  The Answer Jun 16 '13 at 14:48
For $a<c<b$ it holds $\int_{a}^b f = \int_{a}^c f + \int_{c}^b f$ which follows directly from the definition of the riemann integral (via riemann sums) –  Hans Meierwurst Jun 16 '13 at 14:54

$$\lim_{n \to 1^-} \int_0^n \frac1{x-1}\ dx + \lim_{n \to 1^+}\int_n^2 \frac1{x-1}\ dx,$$
where the limits have to be taken separately. And remember that $\int \frac1{x}\ dx = \ln |x| + C$. This is important because $\ln(n-1)$ is undefined when $n < 1$.
@TheAnswer: It does solve your problem, you just didn't compute the limits correctly. Both are $-\infty$, therefore the limit doesn't exist and the integral doesn't converge. –  Javier Jun 16 '13 at 14:55