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I want to show the existence of $ \displaystyle \int\limits_1^\infty \frac{\ln(x)}{1+x^2} \, dx$.

I know that there are already numerous proofs about evaluating this improper integral. However, I am not interested in evaluating the integral but rather would like to know if my apporach which only concerns the the existence is correct.


We substitute $e^t=x$ and receive: $$ \int\limits_1^{\beta}\left|\frac{\ln(x)}{1+x^2}\right| \, dx=\int\limits_{\ln(1)}^{\ln(\beta)}\frac{t}{1+e^{2t}} e^t \, dt \leq \int\limits_{0}^{\ln(\beta)}\frac{t}{e^t}dt= \int\limits_0^{\ln(\beta)} t^1e^{-t} \, dt, \text{ for all } \beta >0. $$

We know that $\lim\limits_{\beta\to\infty}\ln(\beta)=\infty $ so we can conclude $$ \lim\limits_{\beta\to\infty} \int\limits_0^{\ln(\beta)} t^1e^{-t} \, dt = \int\limits_0^\infty t^1e^{-t} \, dt = \Gamma(2). $$ So we have found another improper integral which exists and is an upper bound of the original one. Hence, $\int\limits_1^\infty \frac{\ln(x)}{1+x^2} \, dx$ exists.

Is this correct?

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1 Answer 1

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This works but is much more complicated than needed. Just note that $\ln (x) = O(\sqrt{x})$ on $[1, \infty)$ and so the integrand is $O(x^{-1.5})$, which converges when integrated on $[1, \infty)$ since $1.5 > 1.$

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