# Laplace transform of a signal over t

I tried to identify a formula that is appropriate for computing the Laplace transform of $$f(t)=\displaystyle\frac{\cos 2t-\cos 3t}{t}$$ but I couldn't find one. Give me a suggestion please. Thanks, Alex.

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Can you prove that if the Laplace transform of $f(t)$ is $F(s)$, then the Laplace transform of $f(t)/t$ is $\int_s^\infty F(r)\ dr$? – B R Apr 25 '12 at 6:22
As @BR suggested, take laplace of the numerator and use the laplace transform for f(t)/t to get the final answer. – TenaliRaman Apr 25 '12 at 7:10
@B R and @ TenaliRaman Thank you very much! – Alex Apr 25 '12 at 7:36

## 1 Answer

\begin{align*} \int\limits_s^\infty G(u)\cdot\text du &=\int\limits_s^\infty\int\limits_0^\infty g(t)e^{-ut}\cdot\text dt\cdot\text du \\&=\int\limits_0^\infty g(t)\int\limits_s^\infty e^{-ut}\cdot\text du\cdot\text dt \\&=\int\limits_0^\infty g(t)\left.\frac{e^{-ut}}{-t}\right|_s^\infty \cdot\text dt \\&=\int\limits_0^\infty \frac{g(t)}te^{-st}\cdot\text dt \\&=\mathcal L\left\{\frac {g(t) }t \right\} \end{align*}

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$$f(t)=\frac{g(t)}t$$ – Elements in Space Jan 1 '13 at 9:01