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I want to ask the inverse Laplace transform of the following function:

$$F(s) = \frac{1}{s \cdot (1 + a \cdot s)^{m} \cdot (1 + b \cdot s)^{m-k}} \cdot \Bigl[\exp{(\frac{- c \cdot s}{ 1 + b \cdot s } )}\Bigr]^{m-k}$$

where $a,b,c,m,k$ are all positive constants. $m$ and $k$ are integers with $m\ge k$.

It is possible to get an approximate solution using the Euler summation-based technique. But I want to have an exact solution for the inverse transform.

Some similar questions but in a much less simple form are also found on Stackexchane, i.e., Inverse Laplace Transform of e$^{-c \sqrt{s}}/(\sqrt{s}(a - s))$ or find the inverse Laplace transform of complex function . Especially I am wondering how to deal with the point $s = \frac{1}{b}$ since it is a pole of order $(m-k)$ but also appears in the exponential function and also I am not sure about the contour integral considering that $s$ exists in both the nominator and denominator of the argument of the exponential function.

Thanks a lot.

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Vic
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1 Answers1

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It might be easier to solve by series approach:

Hint:

$\mathcal{L}^{-1}\left\{\dfrac{\left(e^\frac{-cs}{1+bs}\right)^{m-k}}{s(1+as)^m(1+bs)^{m-k}}\right\}$

$=\mathcal{L}^{-1}\left\{\dfrac{e^\frac{c(k-m)s}{1+bs}}{s(1+as)^m(1+bs)^{m-k}}\right\}$

$=\mathcal{L}^{-1}\left\{\sum\limits_{n=0}^\infty\dfrac{c^n(k-m)^ns^{n-1}}{n!(1+as)^m(1+bs)^{m+n-k}}\right\}$

doraemonpaul
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  • Hi, @doraemonpaul . Thanks a lot and could you give any hints on the next step to the final answer. For me, what I will do after the step 3 in your derivation is to work out the inverse Laplace using the contour integral and residue theorem (except that when $n=0$, there are three poles $(s = 0, s = -\frac{1}{a}, s = -\frac{1}{b})$ and for other values of $n$, there are only two poles $(s = -\frac{1}{a}, s = -\frac{1}{b})$ and no branch point). Is my way to work it out correct, Or do you have any better and simpler way to solve the inverse Laplace transform after the step 3 in your derivation? – Vic Oct 19 '15 at 17:16
  • @Vic Perhaps you could handle the $n=0$ term separately. For $n=0$, Mathematica gives the conditional result $\mathcal{L}_s^{-1}\left\frac{(a s+1)^{-m} (b s+1)^{k-m}}{s}\right=\theta (x)$ which assumes $\log(a)\neq 0\land \log(b)\neq 0\land (m+1) \log(a b)\neq k \log(b)$. – Steven Clark May 28 '22 at 16:08
  • @Vic But $\mathcal{L}_x\theta (x)=\frac{1}{s}$, so Mathematica's conditional result seems to imply $\frac{(a s+1)^{-m} (b s+1)^{k-m}}{s}$ is somehow equivalent to $\frac{1}{s}$ which I'm not sure is correct. – Steven Clark May 28 '22 at 16:43