Can anyone tell me the name of this laplace transform? I need to apply this rule to solve a Laplace transform:
$\mathcal{L(\frac{f(t)}{t})}=\int_s^\infty F(u) du$
I've been given a table on laplace transform "rules" but I don't know how to use this one. My textbook doesn't even mention it. What's its name? How do I use it? And if possible, how do I prove it?
Thanks.
Edit: I have to solve the laplace transform of $\frac{-cos(5t)+cos(8t)}{t}$ I guess we can simplify my problem by just saying I have to solve the laplace transform of $\frac{-cos(5t)}{t}$
 A: http://en.wikipedia.org/wiki/Laplace_transform#Properties_and_theorems
Rule "Frequency integration", 7th row in the table. To answer the "how do I use it?", it would be helpful if you'd post your actual problem.

Applying the rule to your problem yields
$$f(t) = -\cos(5t) \ \ \ \Rightarrow \ \ \ F(\tilde{s}) = -\frac{\tilde{s}}{\tilde{s}^2+5^2}$$
Now the Laplace transform in question is
$$\mathcal{L}\left\{\frac{-cos(5t)}{t}\right\} = -\int_s^\infty \frac{\tilde{s}}{\tilde{s}^2+5^2} d\tilde{s}$$
and further
$$-\frac{1}{2}\int_s^\infty \frac{2\tilde{s}}{\tilde{s}^2+5^2} d\tilde{s} 
= -\frac{1}{2}\left.\log(\tilde{s}^2+5^2)\right|_{\tilde{s}=s}^{\tilde{s}=\infty} 
= \left.\log\frac{1}{\sqrt{\tilde{s}^2+5^2}}\right|_{\tilde{s}=s}^{\tilde{s}=\infty}$$
$$= \log\frac{1}{\sqrt{\infty^2+5^2}} + \log\sqrt{s^2+5^2} = \log(0) + \log\sqrt{s^2+5^2}$$
and the solution is
$$\mathcal{L}\left\{\frac{-\cos(5t)}{t}\right\} = \log(0) + \log\sqrt{s^2+5^2}$$
The $\log(0) = -\infty$ is the result of only considering half of the problem. For the full problem see below.

Applied on the full question, this is
$$\mathcal{L}\left\{\frac{-\cos(5t)+\cos(8t)}{t}\right\} = \log\sqrt{s^2+5^2} - \log\sqrt{s^2+8^2} = \frac{1}{2}\log\left(\frac{s^2+5^2}{s^2+8^2}\right)$$
but I think one has to perform the limit (won't write it out)
$$\lim_{u\rightarrow\infty}\int_s^u F(\tilde{s}) d\tilde{s}$$
to reliably argue
$$\lim_{u\rightarrow\infty} \frac{1}{2} \log\left(\frac{u^2+8^2}{u^2+5^2}\right) = \log(1) = 0$$
