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Find the inverse Laplace Transform: $$\mathcal L^{-1} \left\lbrace 1\over s^4\right\rbrace$$

My attempt: I used the equation: $$\mathcal L\left\lbrace t^n\right\rbrace={n!\over s^{n+1}}$$
and played with some numbers until I got an answer that worked when I used the above equation. This is what I did to solve the problem and I don't know if it is the proper way to solve it. If it's not, could someone help me solve it using the "right" method?
$${\frac 16}t^3= {\frac 16}\left(3!\over s^{3+1}\right)={\frac 16}\left(6\over s^4\right)={1\over s^4}$$

Is working backwards a safe way to solve these problems or could it sometimes lead me in the wrong direction?

R.W
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Jc E
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    It's fine to use that equation. You can find that equation on a whole list of Laplace transforms (google it). And it is a good exercise to find the Laplace transform again of your answer to see if you did it right. You'll get the hang of it when you do couple of those problems. (Change the exponents every time) – imranfat Dec 09 '13 at 21:29

2 Answers2

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Yes, it's fine doing what you did. Answering the questions that you stated systematically:

  1. Find the Laplace Transform

$$\mathcal L\left\lbrace t^n\right\rbrace={n!\over s^{n+1}} \implies \mathcal L \{t^3\} = \frac{6}{s^4} \implies t^3 = 6\mathcal L^{-1}\left\{\frac{1}{s^4}\right\} \implies $$ $$\implies \mathcal L ^{-1}\left\{\frac{1}{s^4}\right\} = \frac{t^3}{6}$$

The first relation can be proved using integration by parts and induction.

  1. Is working backwards a safe way to solve these problems or could it sometimes lead me in the wrong direction?

It is not a completely safe way. This is because the inverse Laplace transform is not a unique operation, i.e., we can have different functions having the same Laplace transform, so moving back in calculations can lead you to a function that is not the function you're looking for. But there are some rules that make the inverse unique such that do what you did above is completely wright leading to the one function desired. The function has to be real valued piecewise continuous and piecewise smooth, of exponential order and defined by $\epsilon > 0$, $\frac{1}{2}\left(f(x_0 + \epsilon)+f(x_0-\epsilon)\right)$ at each jump discontinuity $x_0$. The two such functions with the same Laplace transform will be identical.

Benedict W. J. Irwin
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R.W
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If you apply the Mellin's inverse formula you can use the residue theorem and then you have the right result.

LC7
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