Proving convergence of $\int^{\infty}_{0}\frac{\ln{x}}{1+x^{2}}\,dx$ I was trying to prove the convergence of the following integral:
$$\int^{\infty}_{0}\frac{\ln{x}}{1+x^{2}}\,\mathrm{d}x$$
The only way (and indeed quite a convenient one) that came to mind was using the fact that if it is convergent, then the series 
$$\sum^{\infty}_{n=1}\frac{\ln{n}}{1+n^{2}}$$
should also converge. Indeed, it is quite easy to show with Cauchy condensation test that it does. However, it seems to me that we cannot use it: the theorem which states that the convergence of $\int^{\infty}_{a}f(x)dx$ and $\sum^{\infty}_{n=[a+1]}f(n)$ for $a\geq{0}$ relies on the assumption that $f$ is decreasing. This is not the case for our function, since we have 
$$\lim_{x\to{0^{+}}}\frac{\ln{x}}{1+x^{2}}=-\infty$$
and our function in increasing up to some $x_{0}\in(1,3)$. I though of breaking it up into $\int^{y}_{0}f(x)dx$ and $\int^{\infty}_{y}f(x)dx$ for some $y\geq{3}$ and then applying the theorem I mentioned. Unfortunately, it seems like the first intregral doesn't converge...
My question is therefore as following: can we violate the assumption that $f$ is decreasing? If yes, than how can we show that it is still legitimate to use that theorem? If no, in what other way can we prove the convergence of the mentioned integral?
 A: Some cancellation can occur if we calculate
$$\int_{1/n}^n \frac{\log x}{1+x^2} dx$$
Namely, substitute $y = 1/x$, then we get:
$$\int_{1/n}^n \frac{\log x}{1+x^2} dx = \int_n^{1/n} \frac{\log y^{-1}}{1+y^{-2}} \cdot \frac{-1}{y^2} dy = \int_n^{1/n} \frac{\log y}{y^2+1}dy$$
Swapping the integration bounds on the integral over $y$ leads to the conclusion that the integral from $1/n$ to $n$ equals zero, for all $n \in \Bbb N$ (or $n \in \Bbb R$, for that matter). Taking the limit for $n \to \infty$ concludes that the integral is zero (with this special limiting behaviour of the bounds). The unrestricted integral (with independent convergence to both bounds) does not exist by the earlier argument.
EDIT: My point is that:
$$\lim_{n \to \infty} \int_{1/n}^n \frac{\log x}{1+x^2} dx = 0$$
EDIT2: As pointed out in the comments, my assertion about $\log x$ was rubbish. If we can show that:
$$\lim_{(m,n) \to (0,\infty)} \int_m^n \frac{\log x}{1+x^2} dx$$
exists, it necessarily equals $0$ by the above observation.
Now since for $x \in (0,1)$ we have:
$$\left\vert\frac{\log x}{1+x^2}\right\vert \le |\log x|$$
Lebesgue dominated convergence establishes the first integral converges on $(0,1)$ since $\displaystyle \int_0^1 |\log x| dx = 1$. For $(1,\infty)$, we observe:
$$\left\vert\frac{\log x}{1+x^2}\right\vert \le \frac {\log x}{x^2}$$
and the convergence of the latter integral over $(1,\infty)$ follows since upon substituting $y = x^{-1}$ again we obtain:
$$\int_1^\infty \frac{\log x}{x^2} dx = -\int_0^1 \frac{\log y^{-1}}{y^{-2}} \frac{-1}{y^2} dy = - \int_0^1 \log y \ dy = 1$$
Thus, again by dominated convergence $\displaystyle \int_1^\infty \frac{\log x}{1+x^2} dx$ converges; we conclude the integral is $0$.
My sincere apologies for the errors that were present earlier.
A: We don't really need it to be decreasing at all times. We simply need it to be eventually decreasing. We can always just consider the tail of the series during which it is decreasing, and apply the condensation test to that.
However, that's not really a useful approach, since (as you pointed out) the first integral doesn't converge. Try a limiting approach of $$\int_{1/n}^nf(x)dx$$ as $n\to\infty$, with a substitution $u=1/x$.
A: I think : $\displaystyle I=\int\limits_{0}^{1}\dfrac{\ln x}{1+x^2}dx+\int\limits_{1}^{+\infty}\dfrac{\ln x}{1+x^2}dx.$
+) We have : $\lim_{x\to 0^+}\dfrac{x^{\frac{1}{2}}\ln x}{1+x^2}=0$ and $\displaystyle\int\limits_{0}^{1}\dfrac{dx}{x^{\frac{1}{2}}}$ is convergent.
+) $\lim_{x\to +\infty}\dfrac{x^{\frac{3}{2}}\ln x}{1+x^2}=0$ and $\displaystyle\int\limits_{0}^{1}\dfrac{dx}{x^{\frac{3}{2}}}$ is convergent.
$\Rightarrow I $ is convergent.
