Is this Complex Integration correct? I want to integrate $\displaystyle \int_{-\infty}^\infty dx \, e^{iax}\frac{1-e^{-bx^2}}{x^2}$ for a>0. I am going to try and do this using the method of contour integration. I will choose a semicircular contour that runs from minus to plus infinity along the real line followed by a semicircular arc of infinite radius that lies entirely in the upper half plane. Now there are no poles enclosed by this contour as the integrand doesn't have poles, so the contour integral is zero. Also the integral along the semicircular arc goes to zero as on the upper half plane the term $e^{iax}$ goes to zero. Therefore the integration along the real line should go to zero. $$\displaystyle \int_{-\infty}^\infty dx \, e^{iax}\frac{1-e^{-bx^2}}{x^2} = 0$$ Is this correct or am I making a mistake somewhere? Thanks for your help.

You are mistaken. The integral along the circular arc emphatically does not vanish. You can see this by writing it out:

$$i R \int_0^{\pi} d\theta \, e^{i \theta} \, e^{i a R e^{i \theta}} \frac{1-e^{-b R^2 e^{i 2 \theta}}}{R^2 e^{i 2 \theta}} $$

The magnitude of this integral is bounded by

$$\frac1{R} \int_0^{\pi} d\theta \, e^{-a R \sin{\theta}} \left |1-e^{-b R^2 \cos{2 \theta}} \right |$$

Note that, while $\sin{\theta} \gt 0$ over the region of integration, $\cos{2 \theta}$ is not. Rather, when $b \gt 0$, $\cos{2 \theta} \lt 0$ when $\theta \in [\pi/4,3 \pi/4]$. Within this region, the gaussian term rapidly overwhelms the exponential term and the integral blows up. We thus learn nothing from bounding the integral around the circular arc except that you cannot make a statement about the Fourier transform integral as you have done.