I want to know the Laplace transform of the function: $\mathcal{L} \int_t^{\infty} \frac{f(u)}{u}du$.
I tried the following method
$\int_0^{\infty} e^{-ts}\int_t^{\infty} \frac{f(u)}{u}du dt=\int_0^{\infty} (\frac{e^{-ts}}{-s})'\int_t^{\infty} \frac{f(u)}{u}du dt =-\frac{1}{s}\int_0^{\infty}e^{-ts}\frac{f(t)}{t} dt$.
Using the relationship
$\int_0^{\infty}e^{-ts}\frac{f(t)}{t} dt = \int_s^{\infty}\mathcal{L}f(t)dt$
I obtained the following final expression:
$-\frac{1}{s}\int_s^{\infty}\mathcal{L}f(t)dt$
But it may be wrong because it is different from the expression in the comment of Find the inverse Laplace transform of $f(t) = \int_t^\infty \frac{e^{ - u}}{u}du$, which is
$\mathcal{L}\int_t^{\infty} \frac{f(u)}{u}du = \frac{1}{s}\int_0^{s}\mathcal{L}f(t)dt$
I'm not sure if the following holds
$-\frac{1}{s}\int_s^{\infty}\mathcal{L}f(t)dt = \frac{1}{s}\int_0^{s}\mathcal{L}f(t)dt$
I'm very grateful if you could show me the derivation.