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.

  • 1
    $\begingroup$ if u really want to solve this integral, look at $\partial_b I(a,b)$. $\endgroup$
    – tired
    Jan 11, 2016 at 11:41
  • $\begingroup$ @tired : If I take that derivative I am left with a Gaussian integral, performing which I have $Ce^{\frac{-a^2}{4b}}/\sqrt{b}$ which I now need to integrate w.r.t b which doesn't seem doable. Is this the track you had in mind or have I messed up? $\endgroup$ Jan 11, 2016 at 12:38
  • 1
    $\begingroup$ u may find this integral to be equal to $\sqrt{\pi } \left(\sqrt{\pi } k \text{erf}\left(\frac{k}{2 \sqrt{b}}\right)+2 \sqrt{b} e^{-\frac{k^2}{4 b}}\right)$ $\endgroup$
    – tired
    Jan 11, 2016 at 13:10
  • 1
    $\begingroup$ this can be shown using a transformation $b\rightarrow 1/y^2$ followed by an integration by parts together with the defintion of the errorfunction. $\endgroup$
    – tired
    Jan 11, 2016 at 13:28
  • $\begingroup$ @tired That's right, don't know why mathematica failed to integrate it the first time around. Thanks for the help! $\endgroup$ Jan 11, 2016 at 13:28

1 Answer 1


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.

  • $\begingroup$ Are you claiming that $i R \int_0^{\pi} d\theta \, e^{i \theta} \, e^{i a R e^{i \theta}} \frac{1-e^{-b R^2 e^{i 2 \theta}}}{R^2 e^{i 2 \theta}}$ blows up as $R \to \infty$? I think the limit has to be finite since $\int_{-\infty}^\infty dx \, e^{iax}\frac{1-e^{-bx^2}}{x^2} $ converges. $\endgroup$ Jan 11, 2016 at 20:12
  • $\begingroup$ @RandomVariable: I'm claiming that because the magnitude blows up, the fact that the integral about the arc is bounded by something that blows up tells us nothing useful whatsoever. (It says nothing whatsoever about the value of the integral about the real line, which clearly converges.) $\endgroup$
    – Ron Gordon
    Jan 11, 2016 at 20:27

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