Cauchy Principal Value and Residue Theorem: Apparent Contradiction

Question

The following formula is well known and I already understood one of its proofs: $$ \frac{1}{x\pm i\epsilon} = \mathrm{CH} \frac{1}{x} \mp i\pi\delta(x), $$ where the limit $\epsilon \to 0$ is implied and CH denotes the Cauchy principal value. However, it appears to me that there is a contradiction with the residue theorem I was unable to solve. Let's consider $$ \int_{-\infty}^\infty dx f(x) \frac{1}{x+i\epsilon}, $$ which has a residue at $x = -i\epsilon$ and let's assume that $f$ is symmetric ($f(-x) = f(x)$), analytic and drops fast at infinity in the complex plane. Then we should be able to evaluate this integral by residues theorem or equivalently with the formula above. The problem is, I get three (!) differnt results. Let's start with the well known formula above: $$ \int_{-\infty}^\infty dx f(x) \frac{1}{x+i\epsilon} = \int_{-\infty}^\infty dx f(x) \left( \mathrm{CH} \frac{1}{x} - i\pi\delta(x) \right) = \mathrm{CH} \int_{-\infty}^\infty dx f(x) \frac{1}{x} - i\pi f(0) = - i\pi f(0). $$ Since $f$ is assumed to be symmetric, the Cauchy principal value integral should vanish, right? So we are left with $-i\pi f(0)$ only. However, in my opinion, it must also be possible to solve the integral with the residues theorem. If we close the contour in the lower half plane clockwise, we pick up the residue and get $$ \int_{-\infty}^\infty dx f(x) \frac{1}{x+i\epsilon} = -2\pi i f(-i\epsilon) \to -2\pi if(0). $$ If we close it in the upper half plane, there is no residue at all, and we get $$ \int_{-\infty}^\infty dx f(x) \frac{1}{x+i\epsilon} = 0. $$ Obviously, I have a problem with the residues theorem as well ... Well, can somebody help my by telling me, which of the three results is correct and why the other two are wrong?