When trying to solve an ODE using Fourier methods, I met the following solution:

$$\omega \left(e^{-\omega}\text{Ei}(\omega) - e^{ax}\text{Ei}(-\omega)\right)$$

Which I have to Fourier transform back. I attempted using my 'dirty physics' approach:

  1. Use the definition of the Exponential Integral to write, $$ \text{Ei}(\omega) = -\int_{-\omega}^{\infty} \text{d}t~ t^{-1} e^{-t} = e^{\omega}\int_{0}^{\infty} \text{d}u\frac{e^{-u}}{\omega-u}$$ And thus we can write the combination in the brackets as $$e^{-\omega}\text{Ei}(\omega)-e^{\omega}\text{Ei}(-\omega) = 2\omega\int_{0}^{\infty} \text{d}u \frac{e^{-u}}{\omega^2-u^2} $$
  2. Do the Fourier transform for the above by changing the order of integration, \begin{align} \int_{-\infty}^{\infty}\frac{\text{d}\omega}{2\pi} ~e^{i\omega x}\left[ e^{-\omega}\text{Ei}(\omega)-e^{\omega}\text{Ei}(-\omega)\right] &= \int_{-\infty}^{\infty}\frac{\text{d}\omega}{2\pi}~e^{i\omega x} 2\omega\int_{0}^{\infty} \text{d}u \frac{e^{-u}}{\omega^2-u^2}\\ &= \int_{0}^{\infty} \text{d}u~ e^{-u} \int_{-\infty}^{\infty}\frac{\text{d}\omega}{2\pi}~\frac{2\omega}{\omega^2-u^2}e^{i\omega x}\\ &=i~\text{sgn}(x)\int_{0}^{\infty} \text{d}u~ e^{-u} \cos(xu)\\ &=i~\frac{\text{sgn}(x)}{1+x^2} \end{align}

    1. Use the usual property of the Fourier Transform that $\mathcal{F}[i\omega f(\omega)] = \partial_x f(x)$, \begin{align} \int_{-\infty}^{\infty}\frac{\text{d}\omega}{2\pi} ~e^{i\omega x}\omega\left[ e^{-\omega}\text{Ei}(\omega)-e^{\omega}\text{Ei}(-\omega)\right] = \partial_x \left(\frac{\text{sgn}(x)}{1+x^2}\right) = \frac{2\delta(x)}{1+x^2}-\frac{2|x|}{(1+x^2)^2} \end{align}

However if you evaluate the above in Mathematica, I get simply $2\delta(x)$ as the result, which is the first piece of the above.

Does anyone know where I am messing it up and would be able to help with a rigorous treatment?

  • $\begingroup$ The solution cannot be simply $2\delta(x)$ because its Fourier transform is constant, which $w (e^{-w}\text{Ei}(w) - e^w \text{Ei}(-w)$ isn't. Did you double check the Mathematica input? Anyway I went over the solution and it looks good to me. When in doubt you can of course simply verify your result with the original ODE. $\endgroup$ – Hyperplane Nov 12 '17 at 11:44
  • $\begingroup$ @Hyperplane, thank you for pointing out. It is true that it cannot be simply $2\delta(x)$. I more or less have pinned down the problem with Mathematica. First I noticed that asking for the FT of $\omega(\dots+\dots)$ returns the $2\delta(x)$ while asking for $(\omega\times\dots+\omega\times\dots)$ returns the result I quote above. Further, I noticed that doing this only happens in my lab machine, and not on my personal laptop. So they must have corrected some mistake in the FT function. $\endgroup$ – kurtachovo Nov 14 '17 at 12:21
  • $\begingroup$ Btw I am happy to give the bounty for someone that can justify rigorously this result above - the existence of the first FT, why Fubini theorem applies and why the lemma $i\omega \to \partial_x$ can be applied to the resulting function. $\endgroup$ – kurtachovo Nov 14 '17 at 12:23

It might be simpler to work out the direct transform in terms of distributions. If $1/x$ is defined by $$\left( \frac 1 x, \phi \right) = \operatorname{v.\!p.} \int_{-\infty}^\infty \frac {\phi(x)} x dx,$$ then $$\mathcal F[\operatorname{sgn} x] = -\frac {2 i} w, \\ \mathcal F\left[ \frac 1 {x^2+1} \right] = \pi e^{-\left| w \right|}, \\ \mathcal F\left[ \frac {i \operatorname{sgn} x} {x^2+1} \right] = \frac 1 w * e^{-\left| w \right|} = \operatorname{v.\!p.} \int_{-\infty}^\infty \frac {e^{-\left| w - \tau \right|}} \tau d\tau = \\ e^{-w} \operatorname{v.\!p.} \int_{-\infty}^w \frac {e^\tau} \tau d\tau + e^w \operatorname{v.\!p.} \int_w^\infty \frac {e^{-\tau}} \tau d\tau = \\ e^{-w} \operatorname{Ei}(w) - e^w \operatorname{Ei}(-w).$$ Taking the distributional derivative, we indeed obtain $$\mathcal F^{-1}\left[ w e^{-w} \operatorname{Ei}(w) - w e^w \operatorname{Ei}(-w) \right] = -\frac {2 \left| x \right|} {(x^2+1)^2} + 2 \delta(x),$$ where the first term is an ordinary function.


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