-1
$\begingroup$

How to find Inverse Laplace transform of $\tan^{-1}\left(\frac{2}{s^2}\right)$?


I don't understand how to rewrite it and use the table of Laplace transforms.

$\endgroup$

closed as off-topic by Namaste, JonMark Perry, user223391, Simply Beautiful Art, user99914 Oct 6 '17 at 2:08

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Namaste, JonMark Perry, Community, Simply Beautiful Art, Community
If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ Maple gives the result: $2\sinh(t)\sin(t)/t.$ I don't think that this Laplace transform is trivial. You should look this up in a table. $\endgroup$ – MrYouMath Sep 10 '17 at 10:56
  • 1
    $\begingroup$ There is no excuse to be asking such poorly stated questions. You have supposedly "earned" more than 750 in rep, so you're no "newbie". I'd advise you to familiarize yourself with the help section (upmost tab to the right), and pay particular attention to "How to ask a good question." $\endgroup$ – Namaste Oct 6 '17 at 0:32
2
$\begingroup$

You can write $$\tan ^{-1}\left(\frac{2}{s^2}\right)=\tan ^{-1}\left(\frac{1}{s-1}\right)-\tan ^{-1}\left(\frac{1}{s+1}\right)$$ $$\mathcal{L}_s^{-1}\left[\tan ^{-1}\left(\frac{2}{s^2}\right)\right](t)=\mathcal{L}_s^{-1}\left[\tan ^{-1}\left(\frac{1}{s-1}\right)\right](t)-\mathcal{L}_s^{-1}\left[\tan ^{-1}\left(\frac{1}{s+1}\right)\right](t)$$

Now remember the property $\mathcal{L}_s^{-1}[f(s)](t)=-\dfrac{\mathcal{L}_s^{-1}\left[f'(s)\right](t)}{t}$

Derivative of $\tan ^{-1}\left(\dfrac{1}{s-1}\right)$ is

$-\dfrac{1}{(s-1)^2+1}$ and any Laplace transform table can tell that

$\mathcal{L}_s^{-1}\left[-\dfrac{1}{(s-1)^2+1}\right](t)=-e^t \sin t$

Therefore $\mathcal{L}_s^{-1}\left[\tan ^{-1}\left(\dfrac{1}{s-1}\right)\right](t)=-\dfrac{-e^t \sin t}{t}=\dfrac{e^t\,\sin t}{t}$

and in a similar way

$\mathcal{L}_s^{-1}\left[\tan ^{-1}\left(\frac{1}{s+1}\right)\right](t)=\dfrac{e^{-t}\, \sin (t)}{t}$

so that $$\mathcal{L}_s^{-1}\left[\tan ^{-1}\left(\frac{2}{s^2}\right)\right](t)=\dfrac{e^t\,\sin t}{t}-\dfrac{e^{-t}\, \sin (t)}{t}=\dfrac{\left(e^t-e^{-t}\right)\sin t}{t}=\frac{2\sinh t \sin t}{t}$$

Hope this helps

$\endgroup$
1
$\begingroup$

Using Inverse Laplace transform of derivatives theorem : $$\mathcal{L}_s^{-1}\left[\frac{\partial ^nf(s)}{\partial s^n}\right](t)=(-1)^n t^n F(t)$$

$$F(t)=\frac{\mathcal{L}_s^{-1}\left[\frac{\partial ^nf(s)}{\partial s^n}\right](t)}{(-1)^n t^n}$$ in my case : $n=1$

$$-\frac{\mathcal{L}_s^{-1}\left[\frac{\partial }{\partial s}\tan ^{-1}\left(\frac{2}{s^2}\right)\right](t)}{t}=-\frac{\mathcal{L}_s^{-1}\left[-\frac{4 s}{4+s^4}\right](t)}{t}=-\frac{\mathcal{L}_s^{-1}\left[-\frac{4 s}{\left(2-2 s+s^2\right) \left(2+2 s+s^2\right)}\right](t)}{t}=-\frac{\mathcal{L}_s^{-1}\left[-\frac{1}{2-2 s+s^2}+\frac{1}{2+2 s+s^2}\right](t)}{t}=-\frac{\mathcal{L}_s^{-1}\left[-\frac{1}{((1-i)-s) ((1+i)-s)}+\frac{1}{((1-i)+s) ((1+i)+s)}\right](t)}{t}=-\frac{\mathcal{L}_s^{-1}\left[\frac{i}{2 ((-1-i)+s)}-\frac{i}{2 ((-1+i)+s)}-\frac{i}{2 ((1-i)+s)}+\frac{i}{2 ((1+i)+s)}\right](t)}{t}=-\frac{i e^{(-1-i) t} \left(-1+e^{2 i t}\right) \left(-1+e^{2 t}\right)}{2 t}=\frac{2 \sin (t) \sinh (t)}{t}$$

$\endgroup$

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