To check the convergence of an integral (2) I tried to check if this integral is convergent:
$\int _{-\infty }^{\infty }\left(\frac{\sin\left(x\right)\ln\left|x\right|}{1+x^2}\right)dx\:$
so, there are 3 points to check: $\pm\infty$ and $0$. We have
$$\int _{-\infty }^{\infty }\left(\frac{\sin x\ln\left|x\right|}{1+x^2}\right)dx = \int _{-\infty }^{0}\left(\frac{\sin x\ln\left|x\right|}{1+x^2}\right)dx + \int _{0 }^{\infty }\left(\frac{\sin x\ln\left|x\right|}{1+x^2}\right)dx\:$$
My question is, if it's right to check only the second part, and because the function is odd both parts will substantially be the same?
And if it's possible to do that:
$$\int _{0 }^{\infty }\left(\frac{\sin\left(x\right)\ln\left|x\right|}{1+x^2}\right)dx\: \le \int _{0 }^{\infty }\left|\frac{\sin\left(x\right)\ln\left|x\right|}{1+x^2}\right|dx\: = \int _{0 }^{\infty }\left(\frac{|\sin\left(x\right)||\ln\left|x\right||}{1+x^2}\right)dx\: \le \int _{0 }^{\infty }\left(\frac{|\ln\left|x\right||}{1+x^2}\right)dx\: \le \int _{0 }^{\infty }\left(\frac{1}{x^2}\right)dx\:$$
so the integral is convergence?
And how can I check the behavior near $x=0$ ?
 A: Checking the limit you obtain that $\frac{\sin x \log |x|}{1+x^2} $ is continuous on the compact $I=[-2\pi, 2\pi]$ ant then assume minimum (call $m$) and maximum (call $M$) on $I$.
Far from this compact holds $\left|\frac{\sin x \log |x|}{1+x^2}\right|\leq \left|\frac{\log |x|}{1+x^2}\right|$.
Now, your integral is equal to $$\int_I\frac{\sin x \log |x|}{1+x^2}dx +2\int^\infty_{2\pi}\frac{\sin x \log |x|}{1+x^2}dx = \int_I\frac{\sin x \log |x|}{1+x^2}dx +2 \sum _{j=2}^\infty \int_{j\pi}^{(j+1)\pi}\frac{\sin x \log |x|}{1+x^2}dx \leq 4\pi (M-m) + 2 \sum _{j=2}^\infty \int_{j\pi}^{(j+1)\pi}\frac{\sin x \log |x|}{1+x^2} dx $$
We have just to prove the convergence of $\sum _{j=2}^\infty \int_{j\pi}^{(j+1)\pi}\frac{\sin x \log |x|}{1+x^2} dx $.
Now the function $\frac{\log |x|}{1+x^2}$ is decreasing on $[2\pi, +\infty]$ and $$\int_{j\pi}^{(j+1)\pi}\frac{\sin x \log |x|}{1+x^2}dx \leq (-1)^{j+1} \pi\frac{\log(j\pi)}{1+(j\pi)^2} $$
then the sum  $$\sum _{j=2}^\infty \int_{j\pi}^{(j+1)\pi}\frac{\sin x \log |x|}{1+x^2} dx \leq \sum _{j=2}^\infty (-1)^{j+1} \pi\frac{\log(j\pi)}{1+(j\pi)^2} $$
is convergent by Leibnitz Criterion.
A: We have $$\int_{0}^{\infty}\frac{\left|\log\left|x\right|\right|}{x^{2}+1}dx=\int_{0}^{\infty}\frac{\left|\log(x)\right|}{x^{2}+1}dx=-\int_{0}^{1}\frac{\log\left(x\right)}{x^{2}+1}dx+\int_{1}^{\infty}\frac{\log\left(x\right)}{x^{2}+1}dx=2C
 $$ where $C
 $ is the Catalan's constant. 
A: I "guess" the problem is to show that $\int _{0 }^{\infty }\left(\frac{|\ln\left|x\right||}{1+x^2}\right)dx$ is bounded. If this guess is correct then 
\begin{align}
I&=\int _{0 }^{\infty }\left(\frac{|\ln\left|x\right||}{1+x^2}\right)dx\\
&=\int _{0 }^{1 }\left(\frac{|\ln\left|x\right||}{1+x^2}\right)dx+\int _{1 }^{\infty }\left(\frac{|\ln\left|x\right||}{1+x^2}\right)dx\\
&=2\int _{0 }^{1 }\left(\frac{|\ln\left|x\right||}{1+x^2}\right)dx\\
&=-2\int _{0 }^{1 }\frac{\ln x}{1+x^2}dx\\
&=-2\sum _{0 }^{\infty }\frac{(-1)^n}{(2n+1)^2}\\
&\leq 2\sum _{1 }^{\infty }\frac{1}{n^2}=\frac{\pi^2}{3}
\end{align}
A: It is easy to observe that $$\int_{-100}^{100} \frac{\sin x \ln |x| }{1+x^2 } dx <\infty $$ since the function integrant is continuous.
So it is enough to show that $$\int_{100}^{\infty} \frac{\sin x \ln |x| }{1+x^2 } dx <\infty$$ since the integrant function is odd. 
But if $x>10$ then $\ln x <\sqrt{x} $ hence $$\int_{100}^{\infty} \frac{\sin x \ln |x| }{1+x^2 } dx \leq \int_{100}^{\infty}\frac{\sqrt{x} }{1+x^2 } dx \leq \int_{100}^{\infty} \frac{dx}{x^{\frac{3}{2}}} =\frac{1}{5}< \infty $$
so the integral is convergent.
