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Let us say I want Mathematica to compute the integral

$$\int dx \, \text{sign} (x) \, \frac{1}{1+x^2} = 0 \, .$$

Indeed

Integrate[Sign[x] 1/(1+x^2), {x, -Infinity, Infinity}]

gives the answer 0. I am however interested in computing the integral in the Fourier space, i.e. in the (equivalent) form

$$\int dt \, \mathcal{F} \big[ \text{sign} (x) \big] (t) \, \mathcal{F} \bigg[ \frac{1}{1+x^2} \bigg] (t) \, ,$$

where $\mathcal{F}[f(x)](t)$ is the Fourier transform of $f(x)$. In our case this is

$$\int dt \, \frac{i}{t} \, e^{- |t|} \, .$$

But now

Integrate[Exp[-Abs[t]] I/t, {t, -Infinity, Infinity}]

gives Integrate::idiv: Integral of (I E^-Abs[t])/t does not converge on {-\[Infinity],\[Infinity]}. Why? Is there something I can tell Integrate to give me the right answer in this kind of situations?

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  • $\begingroup$ @yarchik: Can you ground your claim giving us an accessible reference? TIA. $\endgroup$ Commented Feb 23, 2022 at 12:05
  • $\begingroup$ @yarchik: Wiki says "The Fourier transform translates between convolution and multiplication of functions. If f(x) and g(x) are integrable functions with Fourier transforms f̂(ξ) and ĝ(ξ) respectively, then the Fourier transform of the convolution is given by the product of the Fourier transforms f̂(ξ) and ĝ(ξ) (under other conventions for the definition of the Fourier transform a constant factor may appear).". $\endgroup$ Commented Feb 23, 2022 at 12:13
  • $\begingroup$ @yarchik: Sorry, don't understand you. Ungrounded statements do not make a good impression. Can you elaborate your previous comment, giving us a possible answer to the question? TIA. $\endgroup$ Commented Feb 23, 2022 at 12:47
  • $\begingroup$ @yarchik: Sorry, again ungrounded $$\int dx f(x) g(x) = \int dt \mathcal{F}[f](t) \mathcal{F}[g](-t).$$ It should be noticed that InverseFourierTransform[Convolve[Exp[-Abs[t]], I/t, t, y], y, t] returns the input. $\endgroup$ Commented Feb 23, 2022 at 12:51
  • $\begingroup$ @yarchik: If I state a non-trivial claim, I try to give an accessible reference to it. $\endgroup$ Commented Feb 23, 2022 at 15:20

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You wrote "Why? Is there something I can tell Integrate to give me the right answer in this kind of situations?". Mathematica is right: the improper integral Integrate[Exp[-Abs[t]] I/t, {t, -Infinity, Infinity}] diverges because of the singularity at the origin. One can calculate its principal value by

Limit[Integrate[Exp[-RealAbs[t]] I/t, {t, -Infinity, -\[Epsilon]}, Assumptions -> \[Epsilon] > 0] + Integrate[Exp[-RealAbs[t]] I/t, {t, \[Epsilon], Infinity}, Assumptions -> \[Epsilon] > 0], \[Epsilon] -> 0]

0

It should be noticed that

Integrate[Exp[-RealAbs[t]] I/t, {t, -Infinity, 0, Infinity}, PrincipalValue -> True]

returns the input.

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  • $\begingroup$ And why adding PrincipalValue -> True does not return 0? $\endgroup$ Commented Feb 23, 2022 at 12:45
  • $\begingroup$ @MBolin: Don't know it. $\endgroup$ Commented Feb 23, 2022 at 12:52

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