I was inspired by BlackPenRedPen's video about why $$\int_{-1}^1\frac1xdx$$ doesn't exist. However, I was able to prove the following claim which gives the above integral as $0$. I want to know what is wrong with my proof:
Claim: Let $f(x)$ be continuous in the interval $[a,b]$ except for an asymptote $a<c<b$. Additionally, $f(x+c)=-f(-x+c)$. Then, $$\int_a^b f(x)dx=\int_{2c-a}^b f(x)dx$$Proof: Using the fact that $f(x+c)=-f(-x+c)$, it follows that $$\int_a^{c-\epsilon}f(x)dx=-\int_{c+\epsilon}^{2c-a}f(x)dx\tag{1}$$Where $\epsilon$ is chosen to be small. We also have that $$\int_a^bf(x)dx=\lim_{\epsilon\rightarrow0}\left(\int_a^{c-\epsilon}f(x)dx+\int_{c+\epsilon}^bf(x)dx\right)\tag{2}$$The second integral on the RHS can be written as $$\int_{c+\epsilon}^bf(x)dx=\int_{c+\epsilon}^{2c-a}f(x)dx+\int_{2c-a}^bf(x)dx$$Substituting this into $(2)$, we have that $$\int_a^bf(x)dx=\lim_{\epsilon\rightarrow0}\left(\int_a^{c-\epsilon}f(x)dx+\int_{c+\epsilon}^{2c-a}f(x)dx+\int_{2c-a}^bf(x)dx\right)=\int_{2c-a}^bf(x)dx$$The last equality follows from $(1)$.
Substituting $c=0$ and $f(x)=\frac1x$ into the claim, we get that the very first integral is zero.
An important thing to note is that BlackPenRedPen turned the integral into two limits, while I had one limit in my proof. I think this is what is wrong, but why?
You've had some comments already but I'd like to point out what I consider to be the main "error":
This is false. In fact, the right hand side describes the principal value integral which does exist (and equal zero) for $f: x\mapsto1/x$.
Instead, in the standard treatments of improper Riemann integration, we demand the following limit to exist: $$\lim_{\epsilon\to0^+\\\delta\to0^+}\left(\int_a^{c-\delta}f(x)\,\mathrm{d}x+\int_{c+\epsilon}^bf(x)\,\mathrm{d}x\right)\tag3$$And then we call this the integral. This double limit does not exist for $f:x\mapsto1/x$.