Shehu transform

The Shehu transform of a function ${\displaystyle f(t)}$ is defined over the set of functions

${\displaystyle A=\{f(t):\exists M,p_{1},p_{2}>0,|f(t)|<M\exp(|t|/p_{i}),\,\,\,{\text{if}}\,\,\,t\in (-1)^{i}\times [0,\,\infty )\}}$

as

${\displaystyle \mathbb {S} [f(t)]=F(s,u)=\int _{0}^{\infty }\exp \left(-{\frac {st}{u}}\right)f(t)\,dt=\lim _{\alpha \rightarrow \infty }\int _{0}^{\alpha }\exp \left(-{\frac {st}{u}}\right)f(t)\,dt,\,s>0,\,u>0,\,\,\,\,(1)}$

where ${\displaystyle s}$ and ${\displaystyle u}$ are the Shehu transform variables.^[1] The Shehu transform converges to Laplace transform when the variable ${\displaystyle u=1}$.

The inverse Shehu transform of the function ${\displaystyle f(t)}$ is defined as

${\displaystyle f(t)=\mathbb {S} ^{-1}[F(s,u)]=\lim _{\beta \rightarrow \infty }{\frac {1}{2\pi i}}\int _{\alpha -i\beta }^{\alpha +i\beta }{\frac {1}{u}}\exp \left({\frac {st}{u}}\right)F(s,u)ds,\,\,\,\,(2)}$

where ${\displaystyle s}$ is a complex number and ${\displaystyle \alpha }$ is a real number.^[1]

Properties of the Shehu transform^[1][3]

Property	Explanation
Linearity	Let the functions ${\displaystyle \alpha f(t)}$ and ${\displaystyle \beta w(t)}$ be in set A. Then ${\displaystyle {\mathbb {S} }\left[\alpha f(t)+\beta w(t)\right]=\alpha {\mathbb {S} }\left[f(t)\right]+\beta {\mathbb {S} }\left[w(t)\right].}$
Change of scale	Let the function ${\displaystyle f(\beta t)}$ be in set A, where ${\displaystyle \beta }$ in an arbitrary constant. Then ${\displaystyle {\mathbb {S} }\left[f(\beta t)\right]={\frac {1}{\beta }}F\left({\frac {s}{\beta }},u\right).}$
Exponential shifting	Let the function ${\displaystyle \exp \left(\alpha t\right)f(t)}$ be in set A and ${\displaystyle \alpha }$ is an arbitrary constant. Then ${\displaystyle {\mathbb {S} }\left[\exp \left(\alpha t\right)f(t)\right]=F(s-\alpha u,u).}$
Multiple shift	Let ${\displaystyle {\mathbb {S} }\left[f(t)\right]=F(s,u)}$ and ${\displaystyle f(t)\in A}$. Then ${\displaystyle {\mathbb {S} }\left[t^{n}f(t)\right]=(-u)^{n}{\frac {d^{n}}{ds^{n}}}F(s,u).}$

${\displaystyle {\mathbb {S} }\left[\int _{0}^{t}f(\zeta )d\zeta \right]={\frac {u}{s}}F(s,u),}$

where ${\displaystyle {\mathbb {S} }\left[f(\zeta )\right]=F(s,u)}$ and ${\displaystyle f(\zeta )\in A.}$^[1][3]

If the function ${\displaystyle f^{(n)}(t)}$ is the nth derivative of the function ${\displaystyle f(t)\in A}$ with respect to ${\displaystyle t}$, then ${\displaystyle {\mathbb {S} }\left[f^{(n)}(t)\right]=\left({\frac {s}{u}}\right)^{n}F(s,u)-\sum _{k=0}^{n-1}\left({\frac {s}{u}}\right)^{n-(k+1)}f^{(k)}(0).}$^[1][3]

Let the functions ${\displaystyle f(t)}$ and ${\displaystyle g(t)}$ be in set A. If ${\displaystyle F(s,u)}$ and ${\displaystyle G(s,u)}$ are the Shehu transforms of the functions ${\displaystyle f(t)}$ and ${\displaystyle g(t)}$ respectively. Then

${\displaystyle {\mathbb {S} }\left[(f*g)(t)\right]=F(s,u)G(s,u).}$

Where ${\displaystyle f*g}$ is the convolution of two functions ${\displaystyle f(t)}$ and ${\displaystyle g(t)}$ which is defined as

${\displaystyle \int _{0}^{t}f(\tau )g(t-\tau )d\tau =\int _{0}^{t}f(t-\tau )g(\tau )d\tau .}$^[1][3]