Laplace inverse of the question on unit step function

Question

$$L^{-1} \left\{\dfrac{e^{-1/s}}{s}\right\}=?$$ First it use the integral property of Laplace and unit step function but I don't how to do it.

Answer 1

Well, we are trying to find the following inverse Laplace transform:

$$f(x):=\mathcal{L}_\text{s}^{-1}\left[\frac{\exp\left(-\frac{1}{\text{s}}\right)}{\text{s}}\right]_{\left(x\right)}\tag1$$

Using the Properties and theorems of the Laplace transform, we get:

$$f(x)=\int_0^x\mathcal{L}_\text{s}^{-1}\left[\exp\left(-\frac{1}{\text{s}}\right)\right]_{\left(\tau\right)}\space\text{d}\tau\tag2$$

Using the fact that:

$$\exp\left(x\right):=\sum_{\text{n}=0}^\infty\frac{x^\text{n}}{\text{n}!}\tag3$$

We get, using the linearity property of the Laplace transform and the integral:

$$f(x)=\int_0^x\mathcal{L}_\text{s}^{-1}\left[\sum_{\text{n}=0}^\infty\left(-\frac{1}{\text{s}}\right)^\text{n}\cdot\frac{1}{\text{n}!}\right]_{\left(\tau\right)}\space\text{d}\tau=\sum_{\text{n}=0}^\infty\frac{\left(-1\right)^\text{n}}{\text{n}!}\int_0^x\mathcal{L}_\text{s}^{-1}\left[\frac{1}{\text{s}^\text{n}}\right]_{\left(\tau\right)}\space\text{d}\tau\tag4$$

Using the table of selected Laplace transforms, we get:

$$f(x)=\sum_{\text{n}=0}^\infty\frac{\left(-1\right)^\text{n}}{\text{n}!}\int_0^x\frac{\tau^{\text{n}-1}}{\Gamma\left(\text{n}\right)}\space\text{d}\tau=\sum_{\text{n}=0}^\infty\frac{\left(-1\right)^\text{n}}{\text{n}!}\cdot\frac{1}{\Gamma\left(\text{n}\right)}\int_0^x\tau^{\text{n}-1}\space\text{d}\tau=$$ $$\sum_{\text{n}=0}^\infty\frac{\left(-1\right)^\text{n}}{\text{n}!}\cdot\frac{1}{\Gamma\left(\text{n}\right)}\cdot\left[\frac{\tau^\text{n}}{\text{n}}\right]_0^x=\sum_{\text{n}=0}^\infty\frac{\left(-1\right)^\text{n}}{\text{n}!}\cdot\frac{1}{\Gamma\left(\text{n}\right)}\cdot\left(\frac{x^\text{n}}{\text{n}}-\frac{0^\text{n}}{\text{n}}\right)=$$ $$\sum_{\text{n}=0}^\infty\frac{\left(-1\right)^\text{n}}{\text{n}!}\cdot\frac{1}{\Gamma\left(\text{n}\right)}\cdot\frac{x^\text{n}}{\text{n}}\tag5$$

We can rewrite that using $\text{n}\left(\text{n}!\right)\Gamma\left(\text{n}\right)=\left(\text{n}!\right)^2$:

$$f(x)=\sum_{\text{n}=0}^\infty\frac{x^\text{n}\left(-1\right)^\text{n}}{\left(\text{n}!\right)^2}\tag6$$