How to prove the convergence of $\int_0^\infty \frac{\ln{x}}{1+x^2}dx$? How to prove the convergence of $\int_0^\infty \frac{\ln{x}}{1+x^2}dx$.Since it's unbounded on both sides, we need to prove the convergence of both $\int_0^1 \frac{\ln{x}}{1+x^2}dx$ and $\int_1^\infty \frac{\ln{x}}{1+x^2}dx$.
$\frac{\ln{x}}{1+x^2}\le \frac{x}{1+x^2}$ on $[1,\infty)$ and both of them are nonnegative, and $\int_1^\infty \frac{x}{1+x^2}dx=\frac12[\ln{|1+x^2|}]^\infty_1=\infty$. Apparently basic comparison does not work in this case. 
For $\int_0^1 \frac{\ln{x}}{1+x^2}dx$, the function$\frac{\ln{x}}{1+x^2}$ is not even nonnegative, we can't apply any comparision test. How are we supposed to know its convergence?
 A: Note that $$\lim_{x \to +\infty} \frac{x^{3/2}\ln(x)}{1+x^2}=0$$ hence the function is integrable at infinity.
A: HInt. Concerning the first integral, the denominator does not play a significant role for convergence, thereby we bound by
$$ \int_{0}^{1} \frac{| \log x|}{1+x^2} \, dx
\leq \int_{0}^{1} | \log x| \, dx. $$
Can you check that this bound is finite? (In fact, we can even compute its value.)
For the second integral, notice that $\log x = 2016 \log x^{1/2016} \leq 2016 x^{1/2016}$. (Here we utilized the identity $\log x \leq x$ which is true for all $x > 0$.) So
$$ \int_{1}^{\infty} \frac{\log x}{1+x^2} \, dx
\leq 2016 \int_{1}^{\infty} \frac{x^{1/2016}}{1+x^2} \, dx. $$
Can you check that this bound converges?
A: On the 0 side, $\;\lvert \ln x\rvert=o\biggl(\dfrac1{\sqrt{x}}\biggr)$ which has a convergent improper integral on $[0,1]$. A fortiori, $\dfrac{\lvert \ln x\rvert}{1+x^2} =o\biggl(\dfrac1{\sqrt{x}}\biggr)$, hence $\;\displaystyle\int_0^1\frac{\ln x}{1+x^2}\,\mathrm d\mkern 1mu x\;$ converges.
On the $\infty\;$ side, as $\ln x=o(\sqrt x)$, $\;\dfrac{\lvert \ln x\rvert}{1+x^2} =o\biggl(\dfrac{\sqrt x}{x^2}\biggr)=o\biggl(\dfrac 1 {x^{3/2}}\biggr)$, and the integral of the latter converges.
A: You say that the function is not even non-negative.  Have you studied its sign properly on $[0,1]$?  What is the matter about its sign?
The function is actually negative on $(0,1]$, so minus the function is positive, and now all comparison tests work just fine.
Since $1+x^2$ lies between $1$ and $2$, we have for $\epsilon\gt 0$:
$$0\leq \int_{\epsilon}^1 \frac{-\log x}{1+x^2} dx\leq \int_{\epsilon}^1 (-\log x) dx,$$ 
so the function $|f(x)|$ (where $f$ is your integrand) is integrable, and hence so is $f$.
A: Actually, by sheer luck, this integral can be evaluated. Set $x=1/t$. If the given integral is called $I$, then after u-subbing, the integral becomes its own negative; $I=-I$ from which we find $I=0$ so the integral is convergent and zero. Now if integration does not work (in a lot of cases), then Dubussy's answer is the way to go in my view 
