# I Telling whether an improper integral converges

1. Mar 23, 2017

### Mr Davis 97

Say we have the following result: $\displaystyle \int_0^{\infty} \frac{\log (x)}{1 - bx + x^2} = 0$. We see that the denominator is 0 for some positive real number when $b \ge 2$. Thus, we obtain a two singularities under that condition. Here's my question. Can we go ahead and say that the integral does not exist in the case that $b \ge 2$? Could it ever be possible that for a specific $b \ge 2$, when we take the limits around the singularities we get an integral that still converges to 0?

2. Mar 23, 2017

### zwierz

you can consider the integral in sense of p.v. when b>2

3. Mar 23, 2017

### Mr Davis 97

I know that I could do that, but what I am asking is whether if b > 2 the integral could still converge to 0 in the normal sense, and if not why not.

4. Mar 24, 2017

### zwierz

in the "normal sense" (not in p.v. sense) this integral does not exist. In p.v. sense it is a priori strange why such a function of b must be identical zero for all b. Extract the singularity and try a numerical experiment first