To test the convergence of $\int _0^1 \frac{\ln x}{(x^2 +1)}\, \mathrm dx $

Question

The question is to test the convergence of $$\int _0^1 \frac{\ln x}{(x^2 +1)}\, \mathrm dx $$ I tried comparison test or comparison it with a form of the Gamma Function or something, but was led to nowhere.

Answer 1

First, use the substitution $x=\dfrac1t$, $\;\mathrm d x=-\dfrac{\mathrm d t}{t^2}$. The convergence of the given improper integral is equivalent to the convergence of $$-\int_1^\infty\frac{\ln t\,\mathrm d t}{t^2+1}.$$ This one converges because $$\frac{\ln t}{t^2+1}=o\biggl(\frac1{t^{3/2}}\biggr)\quad\text{near }\infty.$$

Answer 2

In $[0,1]$, $\frac1{x^2+1}\in\left[\frac12,1\right]$ and therefore the convergence of your integral is equivalent to the convergence of $\int_0^1\log(x)\,\mathrm dx$. Now, doing $x=e^{-t}$ and $\mathrm dx=-e^{-t}\,\mathrm dt$, your integral becomes$$\int_0^{+\infty}te^{-t}\,\mathrm dt.$$ But, if $t\gg0$, $te^{-t}\leqslant e^{-\frac t2}$ and the integral $\displaystyle\int_0^{+\infty}e^{-\frac t2}\,\mathrm dt$ converges. Therefore, your integral converges.