Calculus Improper Integral Convergence; Which is right: Limits or Areas? Could someone please explain to me the following doubt I have on improper integral: 
$$\int_{-\infty}^{\infty} \frac{1}{x} \ \mathrm{ dx}$$
I still think that since integrals signify areas that this evaluates to 0 instead of DNE (Does not Exist).
I understand that limit notation solution for this improper integral returns DNE, but the areas from $-\infty$ to $\infty$ cancel.
 A: The intuition is a common one.  It may be intuitive to say $$\int_{-\infty}^{\infty} \frac{1}{x} dx=\int_{0}^{\infty} \frac{1}{x} dx+\int_{-\infty}^{0} \frac{1}{x} dx=$$ $$\lim_{a \rightarrow 0^+} \lim_{b \rightarrow \infty} \left(\int_{a}^b \frac{1}{x} dx +\int_{-b}^{-a} \frac{1}{x} dx\right)=\lim_{a \rightarrow 0^+} \lim_{b \rightarrow \infty} \left(\int_{a}^b \frac{1}{x} dx -\int_{a}^b \frac{1}{x} dx\right)=0$$  However, one could just as easily say for any real $c$ 
$$\int_{-\infty}^{\infty} \frac{1}{x} dx=\lim_{a \rightarrow 0^+} \lim_{b \rightarrow \infty} \left(\int_{a\cdot e^{-c}}^b \frac{1}{x} dx -\int_{a}^b \frac{1}{x} dx\right)$$ $$=c$$ or 
$$\int_{-\infty}^{\infty} \frac{1}{x} dx=\lim_{a \rightarrow 0^+} \lim_{b \rightarrow \infty} \left(\int_{a/b}^b \frac{1}{x} dx -\int_{a}^b \frac{1}{x} dx\right)$$ $$=+\infty$$ or 
$$\int_{-\infty}^{\infty} \frac{1}{x} dx=\lim_{a \rightarrow 0^+} \lim_{b \rightarrow \infty} \left(\int_{a}^b \frac{1}{x} dx -\int_{a}^{b/a} \frac{1}{x} dx\right)$$ $$=-\infty$$
A: The limiting argument for the improper integral is clearly the way to go.
Note that $x=0$ is the only point of infinite discontinuity of the integrand $\frac{1}{x}$ in $(-\infty,\infty)$.
Thus, $\displaystyle\int_{-\infty}^{\infty}\frac{1}{x}\,dx$
$=\displaystyle\lim_{\substack{\epsilon \to 0+ \\A \to -\infty }}\displaystyle\int_{A}^{0-\epsilon}\frac{1}{x}\,dx
+\displaystyle\lim_{\substack{\delta \to 0+ \\ B\to \infty}}\displaystyle\int_{0+\delta}^{B}\frac{1}{x}\,dx$ $\quad$ [provided each of the limit in the R.H.S. exists.]
$=\displaystyle\lim_{\substack{\epsilon \to 0+ \\A \to -\infty }}(\ln {\epsilon}-\ln {|A|})+\displaystyle\lim_{\substack{\delta \to 0+ \\B \to \infty }}(\ln {|B|}-\ln {\delta})$ $\quad$ 
Each of those limits do not exist. Hence the improper integral diverges.
If we put $\epsilon=\delta$ and take $\epsilon \to 0+$, then we see that
$\displaystyle\int_{-\infty}^{\infty}\frac{1}{x}\,dx=\displaystyle\lim_{\substack{\epsilon \to 0+ \\ A\to -\infty \\B\to \infty}}(\ln {\epsilon}-\ln {|A|}+\ln {|B|}-\ln {\epsilon})=\displaystyle\lim_{\substack{ A\to -\infty \\B\to \infty}}\left(\ln {\left|\frac{B}{A}\right|}\right)$, which still doesn't exist.
However, if you take the simpler integral $\displaystyle\int_{-1}^{1}\frac{1}{x}\,dx$ (which also diverges), you can see that in the special case when $\epsilon=\delta, \epsilon \to 0+$, the limit evaluates to $0$. 
This is called the Cauchy principal value of the integral, which matches with your intuition of symmetry of areas of the odd function.
A: Since you mean to ask why $\int_{-\infty}^{0}\frac{1}{x} + \int_{0}^{\infty}\frac{1}{x}$ doesn't equal zero; it simply follows since these integrals do not converge.
Consider $\int_{0}^{\infty}\sin(x)$; as we approach infinity the areas always cancel out after the fixed period and the value of the integral at any point larger than zero never exceeds $\int_{0}^{\pi} sin(x) = 2$ but since the $lim_{x \to \infty} \int_{0}^{x}sin(t)dt = \int_{0}^{\infty}sin(t)dt$ doesn't converge that integral isn't defined.
So simply put, if you were to integrate with a newly invented operation like this:
$lim_{x \to 0^{-},0^{+} \gets x} \int_{-x_{0},x_{0}}^{x}\frac{1}{t} dt$ you will get zero, but by definition, these one sided limits do not exist.
A: I think what you said make sense. It is a time to get more understanding of improper integration. By definition improper integral is the limit of proper integral. In this case, you have four integral.
$$\int_{1}^{\infty} \frac{1}{x} \ \mathrm{ dx}$$
$$\int_{-\infty}^{-1} \frac{1}{x} \ \mathrm{ dx}$$
$$\int_{0}^{1} \frac{1}{x} \ \mathrm{ dx}$$
$$\int_{-1}^{0} \frac{1}{x} \ \mathrm{ dx}$$
None of them are DNE. Thus the sum is DNE by current textbook definition unless you make your own definition. Taking the following as example $$\int_{-1}^{1} \frac{1}{x} \ \mathrm{ dx}$$
If you change definition to the following, the integral is zero. 
$$\int_{-1}^{1} \frac{1}{x} \ \mathrm{ dx} = \lim_{t \rightarrow 0} (\int_{-1}^{t} \frac{1}{x}\mathrm{ dx} +\int_{t}^{1} \frac{1}{x}\mathrm{ dx})$$
So intuitively, the areas cancels. By we somehow need to abide to definition. 
