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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 $\frac{-cos(5t)}{t}$

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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

$$\lim_{u\rightarrow\infty} \frac{1}{2} \log\left(\frac{u^2+8^2}{u^2+5^2}\right) = \log(1) = 0$$

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  • $\begingroup$ Sorry, but all I see there is the same formula I have. I have no idea what substitution to make, or why it works. If I click on the link on "frequency integration" i'm directed to a very basic article on what frequency is. $\endgroup$ – DLV Oct 12 '14 at 16:48
  • $\begingroup$ Well, you asked for the name and I delivered ;) For the question about the usage, you best post your actual problem. $\endgroup$ – GDumphart Oct 12 '14 at 16:50
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    $\begingroup$ I added what I got, but it's no solution, sorry. $\endgroup$ – GDumphart Oct 12 '14 at 17:07
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    $\begingroup$ I solved it, I hope everything is correct. $\endgroup$ – GDumphart Oct 12 '14 at 17:19
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    $\begingroup$ This is wrong. You can not transform one cosine term at a time, since each of them diverges. Only both term combined gives a convergent answer. Therefore set $f(t)=\cos(8t)-\cos(5t)$ then apply the formula. $\endgroup$ – Hans Oct 12 '14 at 17:26

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