# Laplace transform of two functions

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 $\mathcal{L}\left( f(t) *g(t) \right) = \mathcal{L}\left( f(t)\right) * \mathcal{L}\left( g(t)\right)$?

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

## 1 Answer

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

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

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