I know how to do a basic laplace transform, but how does one deal with transforming complex combination of functions?

For example, how would we handle:

$$\mathcal{L}\left( \ \sqrt{\frac{t}{\pi}}cos(5t) \right) = ... $$

From a table of laplace transforms it is known that: $$\mathcal{L}\left( \ \frac{cos(5t)}{\sqrt{\pi t}} \right) = \frac{e^{-5/s}}{\sqrt{s}}$$

This table value must be of some use to solve this problem, but how?

EDIT: Can we use $\mathcal{L}\left( f(t) *g(t) \right) = \mathcal{L}\left( f(t)\right) * \mathcal{L}\left( g(t)\right) $?

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    $\begingroup$ There's a relation between $\dfrac{d}{ds} \mathcal{L}[f](s)$ and $t\cdot f(t)$. Which? $\endgroup$ – Daniel Fischer Nov 24 '13 at 23:13
  • $\begingroup$ There should be an identity for the Laplace transform of $t.f(t)$ if $\mathcal{L}(f(t))$ is known. $\endgroup$ – Sudarsan Nov 24 '13 at 23:14
  • $\begingroup$ Hrm -- there's a known identity for taking the transform of $g(t)f(t)$ where the transform of $f(t)$ is known? $\endgroup$ – Bob Shannon Nov 24 '13 at 23:15

Hint: Use the property

$$ L(t f(t)) = -F'(s). $$

Added Note that,

$$\frac{\sqrt{t}}{\sqrt{\pi}}{\cos(5t)} = t \frac{\cos(5t)}{\sqrt{\pi t}}.$$

Now, take Laplace transform of both sides of the above equality

$$ \mathcal{L}\left\{ \frac{\sqrt{t}}{\sqrt{\pi}}{\cos(5t)} \right\}= \mathcal{L}\left\{ t \frac{\cos(5t)}{\sqrt{\pi t}} \right\}=-\frac{d}{ds}\frac{e^{-5/s}}{\sqrt{s}}=\dot\,. $$

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  • $\begingroup$ Is $F'$ the transform of $f(t)$ or the derivative of the transform of $f(t)$? $\endgroup$ – Bob Shannon Nov 24 '13 at 23:16
  • $\begingroup$ It's the derivative of the transform. $\endgroup$ – Sudarsan Nov 24 '13 at 23:17
  • $\begingroup$ @Bob: It is the derivative of the Laplace transform of the function which is in your case $f(t)=\frac{\cos(5t)}{\sqrt{\pi t}}$. $\endgroup$ – Mhenni Benghorbal Nov 24 '13 at 23:21
  • $\begingroup$ So what would $t$ be in this case? $\endgroup$ – Bob Shannon Nov 24 '13 at 23:25
  • $\begingroup$ @Bob: I'll add more material. $\endgroup$ – Mhenni Benghorbal Nov 24 '13 at 23:30

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