I discovered that the following integrals are equal:
$$ \int_0^1sx^{s-1}\exp\bigg(\frac{t}{\log(x)}\bigg)~dx=\int_0^1\exp\bigg(\frac{st}{\log(x)}\bigg)~dx $$
Let $f^s(x)=x^s,$ then the LHS can be written as $$ \int_0^1\frac{d}{dx}\bigg(x^s\bigg)\exp\bigg(\frac{t}{\log(x)}\bigg)~dx$$
It reminds me of the Mellin transform on a bounded support.
Is the RHS a sort of Mellin transform in disguise?
It seems to be. This means that for a constant $r>0$ the following integral represents a Mellin transform where the only object in the integrand is the kernel:
$$ \int_0^1 \exp\bigg(\frac{r}{\log(x)}\bigg)~dx $$
So for this specific kernel, I think it's Mellin transform can be written only using the kernel itself.
Is that correct?
For $s > 0$, assuming it converges absolutely on some interval $(a,b)$
$$\int_0^\infty x^{s-1}f(x)dx=\int_0^\infty x^{s-1}f(e^{\log x})dx=\frac1s \int_{0}^\infty f(e^{\frac1s \log t})dt\tag{1}$$
So yes the RHS is a Mellin and Laplace transform in disguise.
More fun when asking (by analytic continuation) for which $f$ and in which sense (analytic functional) the equality stays true for $\Re(s)\in (a,b)$ complex.