we are discussing improper integrals in Calc II, and I am failing to understand why the integral from $-a$ to $a$ of $f(x)=1/x$ is not zero.
Since the function is odd and thus symmetric about the origin and the line $y=x$ (also $y=-x$), it looks like the area from $-a$ to $0$ is just the negative of the area from $0$ to $a$. Anything plus its opposite is zero. I get that it can get weird when dealing with infinity (the areas are infinite in size), but they seem to be the "same" infinity.
I understand how to find out that the integral diverges, but am wondering why that is so.
I am also interested in understanding why the integral from negative infinity to infinity of $f(x)=1/x$ isn't $0$, but I figure that I should understand the first question more thoroughly before adding more infinities into the mix!
Thanks in advance for any and all help!
The improper integral $\int_{-a}^a\frac{dx}{x}$ is by definition equal to $$ \lim_{r_1\to 0^{+}}\int_{r_1}^a\frac{dx}{x}+\lim_{r_2\to 0^{-}}\int_{-a}^{r_2}\frac{dx}{x}$$ and since both integrals diverge, the integral is undefined.
On the other hand, the Cauchy principal value is: $$ \lim_{r\to 0^+}\Big[\int_{r}^a\frac{dx}{x}+\int_{-a}^{-r}\frac{dx}{x}\Big]=0 $$ since the integrals cancel as you have observed.