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If, $$\mathcal L \left\{ \frac{\cos(2\sqrt{3t})}{\sqrt{\pi t}} \right\}= \frac{\exp\big(\frac{-3}{s}\big)}{\sqrt{s}}$$, $$\mathcal L^{-1} \left\{ \frac{\exp\big(\frac{-1}{s}\big)}{\sqrt{s}}\right\}=?$$

could help

Laplace transform, proof that $L \{ \frac{1}{k}f(\frac{t}{k}) \}= F(ks)$

Thanks!

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What have you tried so far? –  Argon Oct 17 '12 at 19:47
    
I think it might be useful math.stackexchange.com/questions/215831/… –  P. M. O. Oct 17 '12 at 19:49
    
Do you know the answer? –  vesszabo Oct 17 '12 at 19:50
    
@vesszabo No... –  P. M. O. Oct 17 '12 at 19:51
    
Why do you think Laplace transform, proof that ... could help? –  vesszabo Oct 17 '12 at 19:53

2 Answers 2

up vote 2 down vote accepted

You noted that: $$\mathcal L \{ \frac{\cos(2\sqrt{3t})}{\sqrt{\pi t}} \}= \frac{\exp(\frac{-3}{s})}{\sqrt{s}}$$ and $$\mathcal L \{ \frac{1}{k}f\bigg(\frac{t}{k}\bigg) \}= F(ks)$$ and $F(s)=\frac{\exp(\frac{-3}{s})}{\sqrt{s}}$. Set $s$ to $3s$ in $F(s)$, so you have $$\sqrt{3}F(3s)=\frac{\exp(\frac{-1}{s})}{\sqrt{s}}$$ So $$\mathcal L^{-1}\{\sqrt{3}F(3s)\}=\mathcal L^{-1} \bigg(\frac{\exp(\frac{-1}{s})}{\sqrt{s}}\bigg)$$ But $$\mathcal L^{-1}\{\sqrt{3}F(3s)\}=\frac{1}{\sqrt{3}}\frac{\cos(2\sqrt{3t})}{\sqrt{\pi t}}\bigg|_{t\to t/3}=\frac{\cos(2\sqrt{t})}{\sqrt{\pi t}}$$

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A slight modification of the method you used to compute the first transform will give you:

$$\mathcal{L}\left(\frac{\cos\left(k\sqrt{t}\right)}{\sqrt{\pi t}}\right) = \frac{e^{-k^2/4s}}{\sqrt{s}} \, . $$

For your inverse transform you need $k=2$.

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how i can use $\mathcal{L} \{ \frac{1}{k}f(\frac{t}{k}) \}= F(ks)$ –  P. M. O. Oct 17 '12 at 20:44

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