I have come across the following ODE, for $x \in \mathbb{R}$: $$\dfrac{d^2 p}{dx^2}(x) + sgn(x) \dfrac{dp}{dx}(x) = 0$$ Its solution is of the form $k.e^{-|x|}$ (a Laplace distribution if it integrates to 1). I want to use this simple example to practice using Fourier Transforms, which I don't know much about but might allow me to solve harder DEs later. It seems that the FT of that equation is $$ (2\pi i \xi)^2 \widehat{p}(\xi) + \int_{-\infty}^{\infty} \dfrac{-1}{i\pi(\xi-\xi')} [(2\pi i \xi')\widehat{p}(\xi')]d\xi'=0$$ which simplifies into $$ (2\pi \xi)^2 \widehat{p}(\xi) + \int_{-\infty}^{\infty} \dfrac{ 2\xi' \widehat{p}(\xi')}{\xi-\xi'}d\xi'=0$$ The FT of $k.e^{-|x|}$, i.e. $\dfrac{2}{1+4\pi^2\xi^2}$ should satisfy the above equation for all $\xi$, which leads to: $$ \dfrac{2(2\pi \xi)^2}{1+4\pi^2\xi^2} + \int_{-\infty}^{\infty} \dfrac{ 4\xi'}{(\xi-\xi')(1+4\pi^2\xi'^2)}d\xi'=0$$ How can one solve such an integral (not in $L^1$)? Did I do something wrong so far? Or perhaps it is some sort of generalized integral?
Thank you very much for your help
Martin
Given that if $p=e^{-|x|}$, we have that, in the distributional sense, $p'=-\mbox{sgn}(x)e^{-|x|}$, and $p''=e^{-|x|}-2\delta(x)$. Thus I think what is wrong is that $p=e^{-|x|}$ just does not solve the differential equation. A solution to the equation is $p=2H(-x)+\mbox{sgn}(x)e^{-|x|}$, where $H$ is the Heaviside function. Indeed, in that case, $p'=-e^{-|x|}$ and $p''=\mbox{sgn}(x)e^{-|x|}$. The other solution to the differential equation is the constant solution.