When trying to solve an ODE using Fourier methods, I met the following solution:
$$\omega \left(e^{-\omega}\operatorname{Ei}(\omega) - e^{ax}\operatorname{Ei}(-\omega)\right)$$
Which I have to Fourier transform back. I attempted using my 'dirty physics' approach:
Use the definition of the Exponential Integral to write, $$ \operatorname{Ei}(\omega) = -\int_{-\omega}^\infty \text{d}t~ t^{-1} e^{-t} = e^\omega \int_0^\infty \, \text{d}u\frac{e^{-u}}{\omega-u}$$ And thus we can write the combination in the brackets as $$e^{-\omega}\operatorname{Ei}(\omega)-e^{\omega}\operatorname{Ei}(-\omega) = 2\omega\int_0^\infty \text{d}u \, \frac{e^{-u}}{\omega^2-u^2} $$
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} \operatorname{Ei}(\omega)-e^\omega \operatorname{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\operatorname{sgn}(x)\int_{0}^{\infty} \text{d}u~ e^{-u} \cos(xu)\\ &=i~\frac{\operatorname{sgn}(x)}{1+x^2} \end{align}
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}\operatorname{Ei}(\omega)-e^\omega \operatorname{Ei}(-\omega)\right] = \partial_x \left(\frac{\operatorname{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?