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?(adsbygoogle = window.adsbygoogle || []).push({});

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# I Telling whether an improper integral converges

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