I'm working with transforms of the kind $$\int_{-\infty}^\infty f(x)e^{it\log(x)} dx \qquad (t\in \mathbb{R}),$$
where $f \in L^1(\mathbb{R})$ and you have fixed a branch of the complex logarithm in $\mathbb{C}$ which is continuous on $\mathbb{R}\setminus \{0\}.$
This type of transformation looks like a kind of Fourier transform, but the kernel of integration has a $\log(x)$ instead of $x$ in the exponent. Of course, this kernel could be also expressed like $x^{it}.$
My question is if this transform coincides or is close to one of the known integral transforms acting on $L^1(\mathbb{R})$. It seems to me that this integral transform is natural enough to have been considered by someone... Please, any kind of references in this direction would be of great help.
Thank you!
Corrected version: note that $e^{i(\pi i)t} = e^{-\pi t}$. If we have set $\log(-x) = \ln x + i\pi$ for $x > 0$, and if $$ f_+(x) \; = \; f(x)\;\;, \hspace{0.5cm} f_-(x) = f(-x) \hspace{2cm} x > 0 $$ then your integral is $$ \int_{\mathbb{R}}f(x)e^{it\log(x)}\,dx \; = \; (\mathcal{M}f_+)(1+it) + e^{-\pi t}(\mathcal{M}f_-)(1+it) $$ in terms of the Mellin transform $\mathcal{M}g$ of $g$. Given the singularity at $0$ in the definition of $\log z$, it is not suprisimg, I guess, that your transform splits into separate transforms of the halves of $f$ for positive and negative $x$ separately.