Why does my derivation of $\mathcal{L(\frac{f(t)}{t})}$ lead to a wrong answer? I'm trying to prove that
$$\mathcal{L(\frac{f(t)}{t})(s)} = \int_s^{\infty}\mathcal{L(f(t))}(u)du$$
Here's my attempt:
$$\mathcal{L(\frac{f(t)}{t})}(s)=\int_{0}^{\infty} \frac{f(t)}{t}e^{-st}dt$$
By using Leibnitz' differentiation under the integral sign, we get:
$$ \dfrac{d\mathcal{L(\frac{f(t)}{t})}(s)}{ds}=-\int_{0}^{\infty} f(t)e^{-st}dt=-\mathcal{L(f(t))(s)}$$
$$\mathcal{L(\dfrac{f(t)}{t})}(s)= - \int \mathcal{L(f(t))(s)} ds$$
It can be rewritten as:
$$\mathcal{L(\dfrac{f(t)}{t})}(s)= - \int_{0}^{s} \mathcal{L(f(t))(u)} du$$
But this is apparently wrong. I can't figure out what I'm doing wrong on my own.
 A: Don't you insert your integration limits in a wrong way? Why are you taking the one limit as $0$ instead of $\infty$? Take your intermediate result
$$ \frac{d}{du}\mathcal{L(f(t)/t)}(u)=-\mathcal{L(f(t))(u)}$$
then integrate from $k\to\infty$ to $s$ to get
$$
\mathcal{L(f(t)/t)}(s)-\lim_{k\to\infty}\mathcal{L(f(t)/t)}(k) = \int_s^\infty\mathcal{L(f(t))(u)}du
$$
Finally you have to show that
$$
\lim_{k\to\infty}\mathcal{L(f(t)/t)}(k) = \lim_{k\to\infty}\int_{0}^{\infty} \frac{f(t)}{t}e^{-kt}dt \overset{?}{=} 0
$$
Now under certain conditions (which I don't remember, but which probably are fulfilled here) you can interchange the limit and the integration and note that "$e^{-\infty}=0$" to get the result
A: This is frequency integration.
Hint: RHS  = $\int_{s}^{\infty} F(u) du$ = $\int_{s}^{\infty} \int_{0}^{\infty} f(t)e^{-ut} dt du$
If you give up:
http://www.eee.hku.hk/~u3500898/LTProperty.pdf
I think the justification for the double integral swap is the same as the justification for your swapping of derivative and integral in your attempt.
A: The hitch is at the end : 
$\mathcal{L(\dfrac{f(t)}{t})}(s)= - \int \mathcal{L(f(t))(s)} ds$ + constant
It can be rewritten as:
$\mathcal{L(\dfrac{f(t)}{t})}(s)= - \int_{0}^{s} \mathcal{L(f(t))(u)} du +C$
Now, you will have to determine $C$ as 
$C=  \int_{0}^{\infty} \mathcal{L(f(t))(u)} du $
