Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $L \{ \frac{1}{k}f(\frac{t}{k}) \}= F(ks)$


share|cite|improve this question
I think it might be useful… – 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
@P.M.O.: I think there is a typo in the body of your question. – Babak S. Oct 17 '12 at 20:21
up vote 3 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 $$\mathcal L^{-1}\{\sqrt{3}F(3s)\}=\frac{1}{\sqrt{3}}\frac{\cos(2\sqrt{3t})}{\sqrt{\pi t}}\bigg|_{t\to t/3}=\frac{\cos(2\sqrt{t})}{\sqrt{\pi t}}$$

share|cite|improve this answer

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 $k=2$.

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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.