# For f(t)=e^{-t}-1 , find the Laplace transform of \{\frac{f(t)}{t}\}.For g(t)=e^{-t}-2 ,examine if the Laplace transform of \{\frac{g(t)}{t}\} exists

For $f\left(t\right)={e}^{-t}-1$ , find the Laplace transform of $\left\{\frac{f\left(t\right)}{t}\right\}$. Then , for $g\left(t\right)={e}^{-t}-2$ , examine if the Laplace transform of $\left\{\frac{g\left(t\right)}{t}\right\}$ exists.

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Step 1
Given that, $f\left(t\right)={e}^{-t}-1$
Now , we know laplace transformation of $f\left(t\right)$ is $L\left(f\left(t\right)\right)$ is given by $L\left\{f\left(t\right)\right\}={\int }_{0}^{\mathrm{\infty }}f\left(t\right){e}^{-st}dt$
Also we know , property of unilafenal Linear transformation is $L\left[\frac{f\left(t\right)}{t}\right]={\int }_{0}^{\mathrm{\infty }}$
F(w)dw : where F(w) is the Laplace transformation of f(t)
Now , Laplace transformation of $f\left(t\right)={e}^{-t}-1$ is
$F\left(s\right)=L\left\{f\left(t\right)\right\}=L\left\{{e}^{-t}-1\right\}={\int }_{0}^{\mathrm{\infty }}\left({e}^{-t}-1\right){e}^{-st}dt$
$={\int }_{0}^{\mathrm{\infty }}\left({e}^{-\left(s+1\right)t}-{e}^{-st}\right)dt$
$={\left[\frac{{e}^{-\left(s+1\right)t}}{-\left(s+1\right)}+\frac{{e}^{-st}}{s}\right]}_{0}^{\mathrm{\infty }}$

Now , linear transformation of $\left\{\frac{f\left(t\right)}{t}\right\}$ is
$L\left\{\frac{f\left(t\right)}{t}\right\}=L\left\{\frac{{e}^{-t}-1}{t}\right\}={\int }_{s}^{\mathrm{\infty }}F\left(w\right)dw$
$={\int }_{s}^{\mathrm{\infty }}\left[\frac{1}{w+1}-\frac{1}{w}\right]dw$
$={\left[\mathrm{ln}\left(w+1\right)-\mathrm{ln}\left(w\right)\right]}_{s}^{\mathrm{\infty }}$
$={\left[\mathrm{ln}\left(\frac{w+1}{w}\right)\right]}_{s}^{\mathrm{\infty }}$
$=-\mathrm{ln}\left(\frac{s+1}{s}\right).⟨\underset{w\to \mathrm{\infty }}{lim}\mathrm{ln}\left(\frac{w+1}{w}\right)=0⟩$
$=\mathrm{ln}\left(\frac{s}{s+1}\right)$
But for $g\left(t\right)={e}^{-t}-2$
$G\left(s\right)=L\left\{g\left(t\right)\right\}=\frac{1}{s+1}-\frac{2}{s}$ . Simply as previous

Now , $L\left\{\frac{g\left(t\right)}{t}\right\}=L\left\{\frac{{e}^{-t}-2}{t}\right\}={\int }_{s}^{\mathrm{\infty }}G\left(w\right)dw$

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