# Inverse Laplace transform of a rational function

• I

## Summary:

I struggle to find an appropriate inverse Laplace transform of the following
$$2^n a^n \frac{p^{n-1}}{(p+a)^{2n}}, \quad a>0$$

## Main Question or Discussion Point

I struggle to find an appropriate inverse Laplace transform of the following
$$F(p)= 2^n a^n \frac{p^{n-1}}{(p+a)^{2n}}, \quad a>0.$$
$$f(t)= 2^n a^n t^n \frac{_1F_1 (2n;n+1;-at)}{\Gamma(n+1)}, \quad (_1F_1 - \text{confluent hypergeometric function})$$
which seem to work for my original problem, but only for small ##n## (<50). Whenever I go close or above 50 (I use Matlab to plot things) the outcome gets bizarre (##n=50##):
although the actual simulation result (the process which is described by ##F(p)##) looks as it meant to be, smoothly getting to the shape on the following figure (##n=49##).
Therefore I conclude there is something wrong with the ##f(t)## given out by Wolfram. Using the Laplace transform handbook (russian one, idk if it's reasonable to link) the best I could do was$$\hat{F}(p)= 2^n a^n \frac{p^{n}}{(p+a)^{2n}} \div \frac{2^n a^n}{\Gamma(2n)}\frac{d^{n-1}}{dt^{n-1}}(e^{-at}t^{2n-1})=\hat{f}(t),$$
since (as the handbook suggests)
$$\left( \frac{p}{p+1} \right)^{n+1}\frac{1}{(p+1)^a}\div \frac{n!}{\Gamma(n+a+1)}e^{-t}t^a L_n^a(t), \ L_n^a(t)=\frac{e^tt^{-a}}{n!}\frac{d^n}{dt^n} (e^{-t}t^{n+a}),$$
which I'm not sure is legitimate to use (because of ##n+1## power), yet it yields somewhat fine solution for my original problem (which is strange, because obviously ##F(p)\ne \hat{F}(p)##). I tried to plot it, and for small ##n## it is fine, but Matlab refuses to calculate the derivative (it spits out 0) for ##n>30##, so I cannot check whether it really suits. So my question is whether the Wolfram answer is wrong. I'm not really familiar with the confluent hypergeometric function, so I don't know how to interpret the outcome. What would you suggest to calculate the ##F(p)##? Is it even possible to do in elementary functions? I'd appreciate any help in the solution. I am aware of possibility that my reasoning is wrong, so in case of that let me know. Thank you for help!

benorin
Homework Helper
Gold Member
Well, this will not answer all your queries: but it’ll give you somewhere to start.

From the definition of ##_1F_1## here (with p=q=1) $$f(t)= 2^n a^n t^n \frac{_1F_1 (2n;n+1;-at)}{\Gamma(n+1)} = \tfrac{2^n a^n t^n}{\Gamma(n+1)}\sum_{k=0}^{\infty}(-1)^k \tfrac{\left( 2n\right) _{k}}{\left( n+1\right) _{k}k!}a^k t^k$$

where ##(a)_{k}:= \tfrac{\Gamma (a+k)}{\Gamma (a)}## is a pochhammer symbol (there are some otherwise defined for different domains and a simplification on that page).

I encourage you to write out that ##\tfrac{d^{n-1}}{dt^{n-1}}(\cdots )## product rule as the binomial-like identity in terms of the sum of powers of differential operators (on this page) for the inverse Laplace transform you had and see if it’s not equivalent to- or a truncated form of- the wolfram confluent hypergeometric Function above. GL.

room_
$$\frac{d^{n-1}}{dt^{n-1}} (e^{-\omega_c t}t^{2n-1}) = D^{n-1}(e^{-\omega_c t}t^{2n-1})= e^{-\omega_c t}(D-\omega_c)^{n-1} t^{2n-1} = \\ = e^{-\omega_c t}\sum_{k=0}^{n-1}(-1)^{n-1-k}\omega_c^{n-1-k}C_{n-1}^{k}D^{k}t^{2n-1} = \\ = e^{-\omega_c t}\sum_{k=0}^{n-1}(-1)^{n-1-k}\omega_c^{n-1-k}C_{n-1}^{k} (2n-1)\ldots (2n-k)t^{2n-1-k},$$