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.


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

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  • $\begingroup$ 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
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    $\begingroup$ 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

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


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 : $n=1$

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


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