How to compute this inverse Laplace transform ?

$$\mathcal{L}}^{\mathcal{-}\mathcal{1}}\left\{\frac{1}{s(\mathrm{exp}(s)+1)}\right\$$

$$\mathcal{L}}^{\mathcal{-}\mathcal{1}}\left\{\frac{1}{s(\mathrm{exp}(s)+1)}\right\$$

daniko883y
2022-09-06
Answered

How to compute this inverse Laplace transform ?

$$\mathcal{L}}^{\mathcal{-}\mathcal{1}}\left\{\frac{1}{s(\mathrm{exp}(s)+1)}\right\$$

$$\mathcal{L}}^{\mathcal{-}\mathcal{1}}\left\{\frac{1}{s(\mathrm{exp}(s)+1)}\right\$$

You can still ask an expert for help

Zayne Wagner

Answered 2022-09-07
Author has **5** answers

Find the Square wave function $\mathbf{s}\mathbf{q}\mathbf{w}(x)=(-1{)}^{\mathbf{f}\mathbf{l}\mathbf{o}\mathbf{o}\mathbf{r}(x)}$ to find out it is a periodic function which is obviously piecewise and by using the L.T. rules we can see that

$$\mathcal{L}\{\mathbf{s}\mathbf{q}\mathbf{w}(x)\}=\frac{1}{s}\mathrm{tanh}(s/2)=\frac{{e}^{s}-1}{s({e}^{s}+1)}$$

So since

$$\mathcal{L}\{1\}=\frac{1}{s}$$

$$\mathcal{L}\{\mathbf{s}\mathbf{q}\mathbf{w}(x)\}=\frac{1}{s}\mathrm{tanh}(s/2)=\frac{{e}^{s}-1}{s({e}^{s}+1)}$$

So since

$$\mathcal{L}\{1\}=\frac{1}{s}$$

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