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I recently got to realize I need the Fourier transform of the function $\log(1/e + x e^x)$.

I need this in a context of the following integral

$$ \int \limits_{-\infty}^\infty \mathrm{d} x \, \log (1/e + x e^x) \int \limits_{-\infty}^\infty \mathrm{d} k \, e^{- i k x} \phi (k) = \int \limits_{-\infty}^\infty \mathrm{d} k \, \phi (k) \int \limits_{-\infty}^\infty \mathrm{d} x \, e^{- i k x}\log (1/e + x e^x) $$

As you can see, generalized functions such as $\delta$, its derivatives etc. are allowed, because the function is then integrated over with another function $\phi$, that has any and all of the good properties you'd want for this to work. Also, $\phi$ is analytic everywhere (except some isolated points on the imaginary axis we probably won't touch this way), so results involving complex numbers are fine, too.

Is there an analytic expression for the Fourier transform of $\log(1/e+x e^x)$? Since there's function $x e^x$ involved, using $W$ (Lambert logarithm) in any of its branches is considered "analytic". Since the behavior for $\log (1/e + x e^x)$ is $-1$ for large negative $x$ and linear for large positive $x$, I expect $\theta$ and $\delta^\prime$ to be involved, but I don't know how to proceed.

Darshan P.
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user16320
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  • what happens at $x=-1$? – user619894 Jul 20 '23 at 12:13
  • @user619894 the logarithm diverges at -1. This is not a problem, as we're talking about generalized functions. For example, Yukawa potential diverges at x = 0 (even stronger than my function) and yet we are able to compute it's Fourier transform. Function $x$ diverges on both ends and yet its Fourier transform is $2 \pi i \delta^\prime (k)$.

    EDIT: take Fourier transform of $1/x$, it's equal to $\text{sign}(k)/(2 i)$. That's even more straightforward.

     – user16320 Jul 20 '23 at 12:18
  • How about linking a definition of "Fourier transform" in the sense you want. – GEdgar Jul 20 '23 at 12:23
  • does theFT of the Yukawa indeed converge in 1d? I thought the radial part of the integral regularizes it. – user619894 Jul 20 '23 at 12:27
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    @user619894 let's forget about Yukawa. Check out this post for the Fourier transform of $1/|x|$. Despite diverging at $x = 0$, its Fourier transform is defined. – user16320 Jul 20 '23 at 13:02
  • not sure an analytic form exists, but what I would do is look for a function with the correct asymptotes that is analytically integrable, subtract it from the function to get rid of the singular behavior and then approximate the rest via some expansion in decaying functions. – user619894 Jul 20 '23 at 13:13
  • @GEdgar The definition $\mathcal{F}xf(x)=\int\limits{-\infty }^{\infty } f(x), e^{-i k x} , dx$ seems most relevant to this question. – Steven Clark Jul 20 '23 at 20:17
  • @StevenClark That is a fine defintion, and exsits for all $f \in L^1(\mathbb R)$ ...Next what if the integral does not converge? You use some sort of weak sense for it? – GEdgar Jul 20 '23 at 23:23
  • @GEdgar An earlier comment indicated the Fourier transform of $x$ is $2 \pi i \delta'(k)$ which is consistent with this definition, and it's possible the result for $\log(\frac{1}{e}+x e^x)$ will be a more complicated expression involving generalized functions. – Steven Clark Jul 21 '23 at 00:07
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    @GEdgar here's what we do if the integral itself does not converge. We pretend to put it under another integral with a "test" function, just like the first equation in my question. This test function has all of the beautiful properties, like it converges to 0 at infinity fast enough no matter how ugly the other part is. Now if, together with this test function integrated over with the Fourier transform in question, gives a result as a, let's say, this function evaluated at a certain point, we say, that the result of the original Fourier transform is $\delta$ and so on. That's the "weak sense". – user16320 Jul 21 '23 at 01:26

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