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?

  • $\begingroup$ can you describe what you mean by $\mathrm{CH} \frac1{x}$? I'm only familiar with Cauchy principal values of integrals... $\endgroup$ – 0x539 Jan 13 at 19:59
  • $\begingroup$ You are completely ignoring the contributions along the semicircular arc above or below. How do you know they vanish? $\endgroup$ – Ron Gordon Jan 13 at 20:36
  • $\begingroup$ Note that the only holomorphic function on $\mathbb{C}$ which falls off to zero in every direction is constant $0$. Therefore Your two last results using the residue theorem are the same. $\endgroup$ – 0x539 Jan 13 at 22:33
  • $\begingroup$ @0x539: Apparently, $\mathrm{CH} 1/x$ can be unterstood as a distribution (like the Dirac-Delta), which is only well defined, when integrated over. $\endgroup$ – BosonRom Jan 14 at 10:17
  • $\begingroup$ I ingored the contributions of the semicirculars, since I assumed, that $f$ drops of fast enough at infinity. But that may be the problem here, since this automatically means the $f = 0$ everywhere, as 0x539 pointed out ... Thank you a lot, already! $\endgroup$ – BosonRom Jan 14 at 10:32

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