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Detailedly compare the difficulties of different entire functions $F(s)$ where $F(0)\neq0$ when evaluating $\mathcal{L}^{-1}_{s\to x}\bigl\{\frac{F(s)}{s}\bigr\}$ by considering contour integration, e.g. $\mathcal{L}^{-1}_{s\to x}\bigl\{\frac{1}{s}\bigr\}$ , $\mathcal{L}^{-1}_{s\to x}\bigl\{\frac{e^{as}}{s}\bigr\}$ , $\mathcal{L}^{-1}_{s\to x}\bigl\{\frac{e^{as^2+bs}}{s}\bigr\}$ and $\mathcal{L}^{-1}_{s\to x}\left\{\frac{\cos as}{s}\right\}$ , where $a$ and $b$ are real numbers and $a\neq0$ .

dustin
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doraemonpaul
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  • A related problem. – Mhenni Benghorbal Apr 09 '13 at 23:00
  • @Mhenni Benghorbal: Really as simple as http://math.stackexchange.com/questions/248330??? But they are both satisfy $\lim\limits_{s\to0}F(s)e^{xs}=1$ . If the residue theorem is really as omnipotent as you imagined, why there are in fact still have many difficult inverse laplace transform problem? – doraemonpaul Apr 10 '13 at 03:32
  • Can you elaborate more on your problem? For instance the problem you consider $\mathcal{L}^{-1}_{s\to x}\left{\dfrac{1}{s}\right}$. – Mhenni Benghorbal Apr 18 '13 at 05:05
  • The residues of $\dfrac{1}{s}$ , $\dfrac{e^{as}}{s}$ , $\dfrac{e^{as^2+bs}}{s}$ and $\dfrac{\cos as}{s}$ are all equal to $1$ , but $1$ is only suitable for $\mathcal{L}^{-1}{s\to x}\left{\dfrac{1}{s}\right}$ but not $\mathcal{L}^{-1}{s\to x}\left{\dfrac{e^{as}}{s}\right}$ , $\mathcal{L}^{-1}{s\to x}\left{\dfrac{e^{as^2+bs}}{s}\right}$ and $\mathcal{L}^{-1}{s\to x}\left{\dfrac{\cos as}{s}\right}$ . – doraemonpaul Apr 18 '13 at 07:42
  • Do you know what's the inverse Laplace transform of $\mathcal{L}^{-1}{s\to x}\left{\dfrac{e^{as}}{s}\right}$ or $\mathcal{L}^{-1}{s\to x}\left{\dfrac{e^{as^2+bs}}{s}\right}$? – Mhenni Benghorbal Apr 18 '13 at 13:43
  • @Mhenni Benghorbal: The result of $\mathcal{L}^{-1}{s\to x}\left{\dfrac{e^{as}}{s}\right}$ is known as $u(x+a)$ when $a<0$ but not known when $a>0$ , and the result of $\mathcal{L}^{-1}{s\to x}\left{\dfrac{e^{as^2+bs}}{s}\right}$ is not known when both $a<0$ and $a>0$ . – doraemonpaul Apr 19 '13 at 06:34

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