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What is the inverse laplace transform of $\frac{s}{s+1}$?

My work was: $$ X(s)=\frac{s}{s+1}\\ X(s)=s\frac{1}{s+1}\\ x(t)=\frac{d}{dt}e^{-t}=-e^{-t} $$

My only issue is that when I check my answer with wolfram alpha, it says that the inverse laplace transform of $\frac{s}{s+1}$ is actually $-e^{-t}+\delta(t)$. What is the correct way to find the inverse transform?

user1539179
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2 Answers2

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HINT: $$\frac s{s+1}=1-\frac1{s+1}$$

We know, $\mathcal{L}(e^{at})=\frac1{s-a}$

and we can prove $\mathcal{L}(\delta(t-b))=e^{-bs}$, put $b=0$

Henry Lee
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lab bhattacharjee
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Simple way to solve this problem sorry about the formatting

\begin{align} \frac s{s+a}&=\frac{(s+a)-a}{s+a}\\ &=\frac{s+a}{s+a}-\frac a{s+a}\\ &=1-\frac a{s+a}\\ &=\mathcal{L}[δ(t)-ae^{-at}] \end{align}

Gonçalo
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bob
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    Edit this into your answer to fix formatting: \begin{align} &=\frac s{s+a}\\ &=\frac{(s+a)-a}{s+a}\\ &=\frac{s+a}{s+a}-\frac a{s+a}\\ &=1-\frac a{s+a}\\ &=1-\mathcal{L}(ae^{-at}) \end{align} (You can put a line break before each & and before the \end{align) – Akiva Weinberger Feb 27 '15 at 05:33
  • The inverse Laplace transform of $1$ is $\delta(t)$ not $1$. – dustin Feb 27 '15 at 05:51
  • Your delta is up one too many lines. – dustin Feb 27 '15 at 06:10