# calculus – Integral transform reduced?

I discovered that the following integrals are equal:

$$int_0^1sx^{s-1}expbigg(frac{t}{log(x)}bigg)~dx=int_0^1expbigg(frac{st}{log(x)}bigg)~dx$$

Let $$f^s(x)=x^s,$$ then the LHS can be written as $$int_0^1frac{d}{dx}bigg(x^sbigg)expbigg(frac{t}{log(x)}bigg)~dx$$

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:

$$int_0^1 expbigg(frac{r}{log(x)}bigg)~dx$$

So for this specific kernel, I think it’s Mellin transform can be written only using the kernel itself.

Is that correct?