Let $f=f(x)$ a well-behaved function integrable on the whole real axis, whose Fourier transform $g(k)= \int_{\mathbb{R}} d x e^{\mathrm{i} k x} f(x)$ possesses a logarithm. My Fourier conventions are such that $f(x)= \int_{\mathbb{R}} \frac{d k}{2 \pi} e^{-\mathrm{i} k x} g(k)$. I also assume that $f$ is real on the real line, and even in $x$.
The sub-dominant, constant term in the strong form of Szeg\"{o}'s theorem for the associated continuous Fredholm determinant with "symbol" f involves the following integral:
$$ A= \int_0^{+ \infty} \! \! \! dx [x f(x) f(-x)] $$
Notice that the integral just runs along the positive half-axis:$f(x)f(-x)$ is even, and with the $x$-factor the integrand is odd anyhow.
Yet this is annoying to compute in practice, and one would like to have an equivalent expression in terms of an unrestricted integral. It seems that this exists in the Toeplitz/Fredholm determinant research literature, through the limit $B=B(0)$ as $\epsilon \to 0$ of the following expression
$$ B(\epsilon)= \frac{1}{4 \pi} \int_{\Im(k) > \epsilon} dk \Big{|}\frac{g'(k)}{g(k)}\Big{|}^2 $$
$|\dots |$ is the modulus, the integral is a contour one in the complex plane, that reduces here to an ordinary real one under my assumptions ($g$ even real with a well-defined log.), and whose integrand is $g'(k) g'(-k)/[g(k) g(-k)]$
I have found this formula in https://arxiv.org/abs/0908.4049, equations (300) and (301) (for the latter, his function $l(x)$ is my $f(x)$, our $g$ coincide, with the same Fourier conventions). But the "simple calculations" leading to the proof that $A= B$ have eluded me. Apparently, the proof proceeds by viewing $A$ as the regularized limit as $\epsilon \to 0^+$ of $$ A(\epsilon)= \int_0^{+ \infty} \! \! \! dx [x e^{-2 \epsilon x} f(x) f(-x)] $$ and by showing that $A(\epsilon)=B(\epsilon)$.
Any hint welcome.