Symmetric integral of 1/x I know that $\int_{-\epsilon}^\epsilon\frac{1}{x}dx$ is undefined, except $\frac{1}{x}$ is an odd function and this is a symmetric integral. So shouldn't the integral evaluate to $0$? Can someone rigorously reconcile the idea of symmetric integrals of odd functions being $0$ with this  integral being undefined?
 A: Well, we need a definition of an integral. The typical definition to work with in this context is the improper Riemann integral where we attempt to get around the unboundedness at the origin by taking a limit $$\int_{-\epsilon}^\epsilon \frac{1}{x}dx := \lim_{a\to 0^+,b\to 0^+} \left(\int_{-\epsilon}^{-a} \frac{1}{x} dx + \int_b^\epsilon \frac{1}{x}dx\right).$$
(The proper Riemann integral of any unbounded function diverges due to an infinite upper Darboux sum, so if we're working with that definition we won't get a value either.)
We can do the integrals inside the limit exactly and get $$\int_{-\epsilon}^\epsilon \frac{1}{x}dx = \lim_{a\to 0^+,b\to 0^+} \ln(b/a)$$ which does not exist. To see it doesn't exist just let $a = x$ and $b=x^2$ and send $x$ to zero and then take $a=b=x.$ You'll get two different answers.
There is a more liberal notion of integral called the Cauchy principal value where we restrict the limit be taken along the line $a=b.$ In this case the integral evaluates to zero since for any $a>0,$ $$\int_{-\epsilon}^{-a} \frac{1}{x} dx + \int_a^\epsilon \frac{1}{x}dx = 0$$ by symmetry, so the limit is zero.
You might be wondering why we don't take Cauchy principal value as our definition of improper integral, since it yields the intuitive "by symmetry" answer here, which may seem superior. Well, first, who's to say the symmetric limit is the only one we want. It may seem like the most natural choice, but it seems a odd that we'd define something in such a way that it depends precisely how the limit is approached. Another way of thinking about this issue is to realize that the symmetry that the Cauchy principal value assumes is not something that is invariant under a change of variables. So you can make some naive change of variables in your integral $u=f(x)$ where $f$ isn't symmetric and end up with a new integral. However when you take the Cauchy principal value of the new integral, it may not be the same as your old one. We don't want the existence of the integral to depend on which integration variables you choose.
Perhaps a snappier way of putting it is that $\int_{-\epsilon}^\epsilon \frac{1}{x}dx$ is something of the form $\infty-\infty,$ which we all know is not zero but rather indeterminate. 
A: The function $f(x)=\frac {1}{x}$ is not defined at $x=0$ therefor we have to consider   $$ \int_{-\epsilon}^\epsilon \frac {1}{x}dx  =  \int_{-\epsilon}^0 \frac{1}{x}dx+  \int_{0}^\epsilon \frac{1}{x}dx $$
Note that both integrals on the right side are improper integrals. Evaluating improper integrals results in $\infty - \infty $ which is not a real number. The symmetry argument does not work in this case. 
Now if instead of $f(x)= \frac {1}{x}$ we consider $f(x)= \frac {1}{x^{1/2}}$ , then the symmetry argument works.
Because the improper integrals $$ \int_{-\epsilon}^0 \frac {1}{x^{1/2}}dx, \text {  and   }   \int_{0}^\epsilon \frac {1}{x^{1/2}}dx $$   converge to opposite real numbers resulting in    $$ \int_{-\epsilon}^\epsilon \frac {1}{x^{1/2}}dx  =0$$
