I'm computing the Inverse Fourier Integral of

             ((a^2 + omega ^2) c^2)
    /((b^2 + omega^2) ((r^2 + omega^2 - omegaInt^2)^2 + (2 omegaInt r)^2))

where all parameters are positive real numbers. Mathematica does a great job of solving it in around 10 seconds. However, if one makes the substitution


So that the new expression is

            ((a^2 + omega ^2) c^2)
    /((b^2 + omega^2) ((r^2 + omega^2 + omegaInt^2)^2 - (2 omegaInt r)^2))

(the difference being the sign in front of the two terms containing omegaInt) the integral is no longer solvable, at least in any reasonable amount of time. I think it is because an additional pole appears, but even then, the over all 1/omega^2 scaling makes me think it should still be integrable. Any ideas anyone? Am going to have to do some contour integral on paper? ugh. please no. Any comments are appreciated.

kind regards.

ps. this expression is the power spectral density of a damped, noisy-driven oscillator (fed through another highpass filter) and I'm after the system's autocorrelation function.

  • 1
    $\begingroup$ On my system with version 8 and 9, both examples above yield a result when I wrap them in InverseFourierTransform[...,omega, t]. So I can't reproduce the problem. $\endgroup$
    – Jens
    May 11, 2013 at 18:54
  • $\begingroup$ Working here too, with version 8. What version are you using @LiaChica? $\endgroup$ May 11, 2013 at 19:01

1 Answer 1


In the second case the denominator vanishes at some points on the real line if r=omegaInt. Excluding this case explicitly gives the result:

InverseFourierTransform[((a^2 + omega^2) c^2)/((b^2 + omega^2) ((r^2 + omega^2 + 
         omegaInt^2)^2 - (2 omegaInt r)^2)), omega, x, 
 Assumptions -> {omegaInt > 0, a > 0, b > 0, c>0, r>0, omegaInt != r}] 

$$\small \frac{1}{4} \sqrt{\frac{\pi }{2}} c^2 \left(\frac{4 \left(a^2-b^2\right) e^{b x} \theta (-x)}{b (b-\text{omegaInt}-r) (b+\text{omegaInt}-r) (b-\text{omegaInt}+r) (b+\text{omegaInt}+r)}+\frac{4 \left(a^2-b^2\right) e^{-b x} \theta (x)}{b (b-\text{omegaInt}-r) (b+\text{omegaInt}-r) (b-\text{omegaInt}+r) (b+\text{omegaInt}+r)}+\frac{\left((\text{omegaInt}-r)^2-a^2\right) \text{sgn}(\text{omegaInt}-r) e^{x (\text{omegaInt}-r)} \theta (-x \text{sgn}(\text{omegaInt}-r))}{\text{omegaInt} r (\text{omegaInt}-r) (-b+\text{omegaInt}-r) (b+\text{omegaInt}-r)}+\frac{\left((\text{omegaInt}-r)^2-a^2\right) \text{sgn}(\text{omegaInt}-r) e^{x (r-\text{omegaInt})} \theta (x \text{sgn}(\text{omegaInt}-r))}{\text{omegaInt} r (\text{omegaInt}-r) (-b+\text{omegaInt}-r) (b+\text{omegaInt}-r)}+\frac{\theta (-x) \left(a^2-(\text{omegaInt}+r)^2\right) e^{x (\text{omegaInt}+r)}}{\text{omegaInt} r (\text{omegaInt}+r) (-b+\text{omegaInt}+r) (b+\text{omegaInt}+r)}+\frac{\theta (x) \left(a^2-(\text{omegaInt}+r)^2\right) e^{x (-(\text{omegaInt}+r))}}{\text{omegaInt} r (\text{omegaInt}+r) (-b+\text{omegaInt}+r) (b+\text{omegaInt}+r)}\right) $$

  • $\begingroup$ sweet! Thanks so much, Andrew. I knew that it diverged, but didn't think to exclude that value in the assumptions. Thanks! $\endgroup$
    – LiaChica
    May 11, 2013 at 19:14

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