- #1

Antarres

- 170

- 81

$$\int_{-\infty}^\infty dx \frac{e^{iax}}{x^2}$$

using contour integration, with ##a>0## above. So I would calculate a contour integral with contour being a semicircle that goes along the real axis, closing it in positive direction in the upper half-plane, which would ensure exponential decay on that arc. Hence, the contour integral would be equal to the integral along the real axis.

Now to remedy the singularity at zero, I can change the integrand to a function: ##f(z) = \frac{e^{ix}}{(x\pm i\epsilon)^2}##. Here the plus sign would shift the pole from the real axis to the lower half-plane, and the minus sign would shift it to the upper half-plane. Then I would take the limit as ##\epsilon \rightarrow 0##. However, if I shift the pole to the lower part of the plane, it would go outside the contour, hence the contour integral would be equal to zero by Cauchy residue theorem. But if I shift it into the upper part of the plane, it would go inside the contour, and the residue is calculated to be ##iae^{-a\epsilon}##, which isn't zero.

I don't understand what I'm doing wrong here, as it seems that my choice of regularization changes the value of the integral, which shouldn't happen.