# Laplace transform of a Taylor series expansion

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I'm reading a paper on tissue cell rheology ("Viscoelasticity of the human red blood cell") that models the creep compliance of the cell (in the s-domain) as

$$J(s) = \frac{1}{As+Bs^{a+1}}$$

where $0\leq a\leq 1$. Since there's no closed-form inverse Laplace transform for this expression, they explore early-time ($t\rightarrow 0$) and late-time ($t\rightarrow \infty$) behavior by using a Taylor series expansion around $s\rightarrow \infty$ and $s\rightarrow 0$, respectively. This is said to yield

$$J(t)\approx \frac{t^a}{B\Gamma(a+1)}-\frac{At^{2a}}{B^2\Gamma(2a+1)}+\frac{A^2t^{3a}}{B^3\Gamma(3a+1)}$$

for the early-time behavior and

$$J(t)\approx \frac{1}{A}-\frac{Bt^{-a}}{A^2\Gamma(1-a)}$$

for the late-time behavior. However, I just can't see how these expressions arise. I know that the Laplace transform of $t^a$ is

$$L[t^a]=\frac{\Gamma(a+1)}{s^{a+1}}$$

and so presumably

$$L\left[\frac{t^a}{\Gamma(a+1)}\right]=\frac{1}{s^{a+1}}\mathrm{,}\quad L\left[\frac{t^{-a}}{\Gamma(1-a)}\right]=\frac{1}{s^{-a+1}}$$

but I can't figure out where these terms would appear in a Taylor series expansion. When I try to expand $J(s)$ in the manner of

$$f(x+\Delta x)\approx f(x) + f^\prime(x)\Delta x +\frac{1}{2}f^{\prime\prime}(x)(\Delta x)^2$$

I get zero or infinity for each term. Unfortunately, Mathematica is no help in investigating an expansion around $s\rightarrow\infty$ or $s\rightarrow 0$; it just returns the original expression. Perhaps I'm making a silly error, or perhaps the paper skipped an important enabling or simplifying step. Any thoughts?

Last edited:

benorin
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Both series expansions below are geometric series: $\frac{1}{1-x}=\sum_{k=0}^{\infty}x^k\mbox{ for }|x|<1$.

For $\left| {\scriptstyle \frac{B}{A}}s^{a} \right| < 1,$ we have

$$J(s) = \frac{1}{As+Bs^{a+1}} = \frac{1}{As}\cdot\frac{1}{1+{\scriptstyle \frac{B}{A}}s^{a}} = \frac{1}{As}\sum_{k=0}^{\infty}\left(-1\right)^k \left(\frac{B}{A}}\right)^k s^{ak}=\frac{1}{A}\sum_{k=0}^{\infty}\left(-1\right)^k \left(\frac{B}{A}}\right)^k s^{ak-1}$$

$$J(s) = \frac{1}{As}-\frac{B}{A^2s^{1-a}}}+\frac{B^2}{A^3s^{1-2a}}}-\cdots$$

hence

$$J(t) = \frac{1}{A}u(t)-\frac{Bt^{-a}}{A^2\Gamma (1-a)}}+\frac{B^2t^{-2a}}{A^3\Gamma (1-2a)}}-\cdots$$​

where $u(t)$ is the unit step function...

And for $\left| {\scriptstyle \frac{A}{B}} s^{-a} \right| < 1,$ we have

$$J(s) = \frac{1}{As+Bs^{a+1}} = \frac{1}{Bs^{a+1}}\cdot\frac{1}{ {\scriptstyle \frac{A}{B}}s^{-a}}+1} =\frac{1}{Bs^{a+1}}\sum_{k=0}^{\infty}\left(-1\right)^k \left(\frac{A}{B}}\right)^k s^{-ak}=\frac{1}{B}\sum_{k=0}^{\infty}\left(-1\right)^k \left(\frac{A}{B}}\right)^k s^{-ak-a-1}$$

$$J(s) = \frac{1}{Bs^{a+1}}-\frac{A}{B^2s^{2a+1}}+\frac{A^2}{B^3s^{3a+1}}-\cdots$$​

hence

$$J(t) = \frac{t^{a}}{B\Gamma (a+1)}-\frac{At^{2a}}{B^2\Gamma (2a+1)}+\frac{A^2t^{3a}}{B^3\Gamma (3a+1)}-\cdots$$​