# Single variable improper integral

Say I have an integral of $x/(1+x^2)$ that goes from negative infinity to infinity, and then part it into two integrals $A + B$ (let $A + B = I_\text{tot}$) where $A$ and $B$ have the limits from R to Infinity and negative infinity to R, respectively, and where R is any R-number. Now, if I integrate e.g. A I get 0.5*(natural log of u) after u-substitution and A having the limit from R to Infinity we know that the integral (and even I_tot because infinity+B=infinity) will diverge, and now I want to know if it is correct to say "The integral diverges or The integral diverges to infinity"? Because I compared many improper integrals that diverged but I couldn't find any patterns to confirm any difference.

• The integral diverges: yes. To $\infty$, no. Integral $A$ diverges to $+\infty$. Integral $B$ diverges to $-\infty$. – GEdgar Jun 11 '15 at 20:57

By definition, $$\int_{-\infty}^{\infty}\frac{x}{1+x^2}\,dx= \lim_{t\to-\infty}\int_{t}^{0}\frac{x}{1+x^2}\,dx+ \lim_{t\to\infty}\int_0^{t}\frac{x}{1+x^2}\,dx$$ provided both limits exist and are finite. Instead of $0$ any number can be chosen.
With the substitution $x=-y$, the first integral becomes $$\int_{-t}^0 \frac{u}{1+u^2}\,du=-\int_{0}^{-t}\frac{u}{1+u^2}\,du$$ and so we are reduced to seeing whether $$\lim_{t\to\infty}\int_0^{t}\frac{x}{1+x^2}\,dx$$ exists and is finite. Now $$\int_0^{t}\frac{x}{1+x^2}\,dx=\Bigl[\frac{1}{2}\log(1+x^2)\Bigr]_0^t =\frac{1}{2}\log(1+t^2)$$ and so $$\lim_{t\to\infty}\int_0^{t}\frac{x}{1+x^2}\,dx=\infty$$ Thus your integral doesn't converge.
You can't say that it diverges to $\infty$ or $-\infty$; the integral involving $-\infty$ diverges to $-\infty$ and the other one diverges to $\infty$.
Note that terminology is not standard: I prefer to say that the integral doesn't converge; others consider “divergent” as a synonym for “not convergent”. I prefer to use “divergent” for the case the limit is $\infty$ or $-\infty$; just personal preference.
$$\int_{-\infty}^r \frac x {1+x^2} \,dx = -\infty \text{ and }\int_r^\infty \frac x {1+x^2}\,dx = +\infty.$$ One can say that one of these diverges to $-\infty$ and the other diverges to $+\infty$. However, notice also that $$\lim_{r\to\infty} \int_{-r}^r \frac x {1+x^2}\,dx = 0\text{ and } \lim_{r\to\infty} \int_{-r}^{2r} \frac x {1+x^2}\,dx = \log_e 2>0.$$ The fact that the rearrangement in the last line can yield two different numbers is something that can happen only if the positive and negative parts both diverge to infinity.