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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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    @BabakSorouh Correct, i will edit2012-10-17

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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)$2012-10-17