Find the inverse Laplace's transform of $\frac{2}{s(1-e^{-2 \pi s})} $.

Question

If $ \large \ \ F(s)=\dfrac{2}{s(1-e^{-2 \pi s})} \ $ , then find the Inverse Laplace's transform.

Hint

If $ f(t) $ be the Inverse Laplace transform of $ F(s) $ , then I think the function $ f(t) $ is periodic function with period $ 2 \pi \ $.

But I can not find the Inverse Laplace's transform .

Is there any help ?

Answer 1

Well, in general:

$$\mathscr{L}_\text{s}^{-1}\left[\frac{1}{\text{s}}\cdot\text{F}\left(\text{s}\right)\right]_{\left(t\right)}=\int_0^t\text{f}\left(\tau\right)\space\text{d}\tau\tag1$$

So, we get:

$$\text{f}\left(t\right):=\mathscr{L}_\text{s}^{-1}\left[\frac{2}{\text{s}\cdot\left(1-e^{-2\pi\text{s}}\right)}\right]_{\left(t\right)}=\mathscr{L}_\text{s}^{-1}\left[\frac{1}{\text{s}}\cdot\frac{2}{1-e^{-2\pi\text{s}}}\right]_{\left(t\right)}=$$ $$\int_0^t\mathscr{L}_\text{s}^{-1}\left[\frac{2}{1-e^{-2\pi\text{s}}}\right]_{\left(\tau\right)}\space\text{d}\tau=2\int_0^t\mathscr{L}_\text{s}^{-1}\left[\frac{1}{1-e^{-2\pi\text{s}}}\right]_{\left(\tau\right)}\space\text{d}\tau\tag2$$

Now, we can use:

$$\frac{1}{1-x}=\sum_{\text{n}=0}^\infty x^\text{n}\tag3$$

When $\left|x\right|<1$

So:

$$\text{f}\left(t\right)=2\int_0^t\mathscr{L}_\text{s}^{-1}\left[\sum_{\text{n}=0}^\infty\left(e^{-2\pi\text{s}}\right)^\text{n}\right]_{\left(\tau\right)}\space\text{d}\tau=2\int_0^t\mathscr{L}_\text{s}^{-1}\left[\sum_{\text{n}=0}^\infty e^{-2\pi\text{s}\text{n}}\right]_{\left(\tau\right)}\space\text{d}\tau=$$ $$2\int_0^t\sum_{\text{n}=0}^\infty\mathscr{L}_\text{s}^{-1}\left[e^{-2\pi\text{s}\text{n}}\right]_{\left(\tau\right)}\space\text{d}\tau=2\sum_{\text{n}=0}^\infty\int_0^t\mathscr{L}_\text{s}^{-1}\left[e^{-2\pi\text{s}\text{n}}\right]_{\left(\tau\right)}\space\text{d}\tau\tag4$$

Now, the inverse laplace transform of $e^{-2\pi\text{s}\text{n}}$ is:

$$\mathscr{L}_\text{s}^{-1}\left[e^{-2\pi\text{s}\text{n}}\right]_{\left(\tau\right)}=\delta\left(\tau-2\pi\text{n}\right)\tag5$$

So, we get:

$$\text{f}\left(t\right)=2\sum_{\text{n}=0}^\infty\int_0^t\delta\left(\tau-2\pi\text{n}\right)\space\text{d}\tau=2\sum_{\text{n}=0}^\infty\left[\theta\left(\tau-2\pi\text{n}\right)\right]_0^t=$$ $$2\sum_{\text{n}=0}^\infty\left(\theta\left(t-2\pi\text{n}\right)-\theta\left(0-2\pi\text{n}\right)\right)=2\sum_{\text{n}=0}^\infty\theta\left(t-2\pi\text{n}\right)\tag6$$

Answer 2

HINT by user Sangchul Lee.

$$\mathcal{L}_s^{-1}\left[\frac{2}{\left(1-e^{-2 \pi s}\right) s}\right](t)=\mathcal{L}_s^{-1}\left[\sum _{n=0}^{\infty } \frac{2 \exp ^n(-2 \pi s)}{s}\right](t)=\sum _{n=0}^{\infty } \mathcal{L}_s^{-1}\left[\frac{2 e^{-2 n \pi s}}{s}\right](t)=\sum _{n=0}^{\infty } 2 \theta (-2 n \pi +t)=2 \left(1+\left\lfloor \frac{t}{2 \pi }\right\rfloor \right)$$

$\lfloor t\rfloor$ where is a Floor and $\theta (t)$ is Heaviside.