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I Telling whether an improper integral converges

  1. Mar 23, 2017 #1
    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. jcsd
  3. Mar 23, 2017 #2
    you can consider the integral in sense of p.v. when b>2
     
  4. Mar 23, 2017 #3
    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.
     
  5. Mar 24, 2017 #4
    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
     
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