# Find the inverse Laplace transformation and limit

The question is related to this post but can be solved independtly. I am trying to find a general expression in the time domain for the asymptotic behavior when $t \to \infty$ of $\bar{f}(s)$ defined in the Laplace domain.

$$\bar{f}(s) = \cfrac{\sigma}{\Gamma\left(\frac{n}{2} \right)}\cfrac{K_{\nu-1}\left(\sigma \right)}{K_\nu\left(\sigma\right)}$$

and

• $\bar{u}(s) = \cfrac{s\omega (\omega -1)+\lambda }{s(\omega -1)+\lambda }$ with $\omega < 1$ and $\lambda \in \mathbb{R^{+}}$
• $\sigma = \sqrt{s\bar{u}(s)}$
• $\nu = 1- \cfrac{n}{2}$ with $n \in ]2;3]$

Because $\nu < 0$ we can not use the approximated values available for small arguments of the modified Bessel function.

I tried to tackle the problem with approximations of the modified Bessel function, to check the properties of the incomplete gamma function but I am not familiar with this. I ended to plot the values of $f(t)$ for great values of $t$ as a function of n. The graph shows that for small values of $n$ (i.e. $n<2$) the limit of $f(t)$ tends to $0$, this can be found analyticaly with the approximations available for the modified Bessel function. It takes a specific solution when $n = 2$ (red circle on the figure), still the approx. of the modified Bessel function. Once $n>3$, $n$ influences the value of the limit of $f(t)$.

See the plot:

Thank you for your suggestions!

Viv

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The behavior at large times might depend if $\omega <1$ or $\omega>1$. Are you interested in one of the two regimes only? – Fabian Feb 1 '13 at 18:15
@Fabian, you are right, I am considering the case where $\omega < 1$, I edited my post to specify this particularity, thank you. – Vivh Feb 1 '13 at 18:29