Convergence of $\int_{-\infty}^{\infty}f(x)dx$ I posed a question to my calculus professor, asking how to evaluate the Riemann integral
$$\int_{-\infty}^\infty f(x) \, dx$$
I can simplify the above integral as 
$$\int_{-\infty}^{\infty}f(x)\,dx = \int_{-\infty}^a f(x)\,dx + \int_a^\infty f(x)\,dx $$
$$\lim_{t\rightarrow -\infty} \int_t^c f(x) \, dx + \lim_{w\rightarrow \infty} \int_c^w f(x)\,dx$$
What happens if I get $ -\infty + \infty$ ? 
My professor answered that we cre unable to know where the integral convergences. Is that really the case or not?  

With the integral
$$
\int_{-\infty}^\infty \frac{x}{x^2+1}\,dx
$$
you actually do get $-\infty+\infty$.  And
$$
\lim_{a\to\infty} \int_{-a}^a \frac{x}{x^2+1}\,dx =0. \tag 1
$$
And
$$
\lim_{a\to\infty} \int_{-a}^{2a} \frac{x}{x^2+1}\,dx = \log_e 2. \tag 2
$$
The value in $(1)$ is the "Cauchy principal value" of the improper integral.
It is only when the positive and negative parts are both infinite that rearranging like this can change the value of the integral.  Convergence like that in $(1)$, where the limit changes with a rearrangement because both the positive and negative parts are infinite, is called "conditional convergence".  When the positive and negative parts are both finite, then one has "absolute convergnce" and rearrangement will not change the value.
