# Show existence of $\int_1^{\infty}\frac{\ln(x)}{1+x^2} \, dx$

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?

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 $$1.5 > 1.$$