Inverse Laplace transform of a rational function

• I
• room_
In summary, the conversation discusses the struggle to find an appropriate inverse Laplace transform for a given equation. WolframAlpha provides a solution using the confluent hypergeometric function, but it only works for small values of n. The speaker also attempts to use a Laplace transform handbook for a solution, but it may not be legitimate to use. The conversation concludes with a suggestion to use the exponential shift theorem to further explore the solution.
room_
TL;DR Summary
I struggle to find an appropriate inverse Laplace transform of the following
$$2^n a^n \frac{p^{n-1}}{(p+a)^{2n}}, \quad a>0$$
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!

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_
benorin said:
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
I had and idea to write it out using exponential shift theorem, so here is my reasoning
$$\frac{d^{n-1}}{dt^{n-1}} (e^{-\omega_c t}t^{2n-1}) = D^{n-1}(e^{-\omega_c t}t^{2n-1})= e^{-\omega_c t}(D-\omega_c)^{n-1} t^{2n-1} = \\ = e^{-\omega_c t}\sum_{k=0}^{n-1}(-1)^{n-1-k}\omega_c^{n-1-k}C_{n-1}^{k}D^{k}t^{2n-1} = \\ = e^{-\omega_c t}\sum_{k=0}^{n-1}(-1)^{n-1-k}\omega_c^{n-1-k}C_{n-1}^{k} (2n-1)\ldots (2n-k)t^{2n-1-k},$$
where ##C_{n-1}^{k}## are binomial coefficients. Doesn't seem equivalent to ##_1F_1##.

1. What is an inverse Laplace transform?

An inverse Laplace transform is a mathematical operation that converts a function from the Laplace domain to the time domain. It essentially "undoes" the Laplace transform, allowing us to find the original function.

2. What is a rational function?

A rational function is a function that can be expressed as the ratio of two polynomials. In other words, it is a fraction where the numerator and denominator are both polynomials.

3. How do you find the inverse Laplace transform of a rational function?

To find the inverse Laplace transform of a rational function, we use partial fraction decomposition to break it down into simpler fractions. Then, we use a table of Laplace transforms to find the inverse of each individual term. Finally, we combine these inverse transforms to get the overall inverse Laplace transform of the rational function.

4. What is the importance of the inverse Laplace transform of a rational function?

The inverse Laplace transform of a rational function is important in many areas of science and engineering, particularly in the study of systems and signals. It allows us to understand the behavior of a system or signal in the time domain, which is often more intuitive and easier to interpret than the Laplace domain.

5. Are there any limitations to using the inverse Laplace transform of a rational function?

Yes, there are some limitations to using the inverse Laplace transform of a rational function. It can only be used for functions that have a Laplace transform, and it may not work for functions with complicated or undefined behavior. Additionally, the inverse Laplace transform may not always be unique, meaning that different functions can have the same Laplace transform.

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