How to prove that $\int_{-\infty}^{+\infty}\frac{1+x}{1+x^2}dx$ doesn't exist I'm trying to prove directly that $$\int_{-\infty}^{+\infty}\frac{1+x}{1+x^2}dx$$
doesn't exist:
Integrating,
\begin{align*}\int_u^v\frac{1+x}{1+x^2}dx&=\arctan x\Big|_u^v+\frac{1}{2}\log(1+x^2)\Big|_u^v\\
&= \arctan v-\arctan u+\frac{1}{2}\log\left(\frac{1+v^2}{1+u^2}\right).\end{align*}
Since $$\lim_{\substack{u\to -\infty\\ v\to +\infty}}\arctan v-\arctan u=\pi,$$
we need to show that
$$\lim_{\substack{u\to -\infty\\ v\to +\infty}}\log\left(\frac{1+v^2}{1+u^2}\right)$$
doesn't exist. But I don't know how to prove it. An idea is to use that,
$$\lim_{v\to +\infty}\lim_{u\to -\infty}\log\left(\frac{1+v^2}{1+u^2}\right)\neq\lim_{u\to -\infty}\lim_{v\to +\infty}\log\left(\frac{1+v^2}{1+u^2}\right),$$
but is this sufficient? Thanks.
 A: Let
$$f(u,v)=\ln\left(\frac{1+v^2}{1+u^2}\right)$$
we will prove that
$$\lim_{(-u,v)\to(\infty,\infty)}f(u,v)$$
does not exist.
if we take $ (-u,u) $ we find
$$\lim_{(-u,u)\to(\infty,\infty)}f(u,u)=0$$
but if we take $ (-u,2u) $ we  get
$$\lim_{(-u,2u)\to(\infty,\infty)}f(u,2u)\ne 0$$
So, the limit cannot exist.
A: Alternatively, by definition $$\int_{-\infty}^{+\infty}\frac{1+x}{1+x^{2}}\, {\rm d}x<+\infty \quad \text{ if and only if}\quad \begin{cases} \int_{-\infty}^{\varepsilon}\frac{1+x}{1+x^{2}}\,{\rm d}x<+\infty,\\ \int_{\varepsilon}^{+\infty}\frac{1+x}{1+x^{2}}\, {\rm d}x<+\infty \end{cases}$$ for some constant $\varepsilon >0$ as pointed out in the comment. That is, if at least one integral doesn't converges so our integral doesn't converges.
Now notice that $\displaystyle  x\mapsto \frac{1+x}{1+x^{2}}$ is positive and continuous function over $[\varepsilon, +\infty[$ and the same for the mapping $\displaystyle x\mapsto \frac{1}{x}$.
Since $$\frac{1+x}{1+x^{2}}\underset{+\infty}{\sim}\frac{1}{x}$$ and $\displaystyle \int_{\varepsilon}^{+\infty}\frac{1}{x}\, {\rm d}x$ does not converges so $\displaystyle \int_{\varepsilon}^{+\infty}\frac{1+x}{1+x^{2}}\, {\rm d}x$ doesn't converges.
Therefore $$\int_{-\infty}^{+\infty}\frac{1+x}{1+x^{2}}\, {\rm d}x$$ doesn't converges.
