Inverse Laplace transform from a power series with finite radius of convergence

Question

Suppose we are given some function $f(t)$ define on $t\in [0,\infty)$. Then, the Laplace of this function is given by \begin{align} F(s)= \int_0^\infty e^{-st} f(t) dt \end{align} where $s=\sigma+i\omega$, and the inverst Laplace transform is given by \begin{align} L^{-1}[F(s)](t)=\frac{1}{ 2 \pi i} \lim_{T \to \infty} \oint_{c-iT}^{c+iT} e^{ts} F(s) ds, \end{align} where $c$ is some real constant in the region of convergnce of $F(s)$.

My question

Suppose that we are given a Laplace transform in the form of a power series only: \begin{align} F(s)=\sum_{n=0}^\infty a_n s^n \end{align} where the radius of convergence is given by $|s|<r$.

How can we invert $F(s)$ from this the power series representation? The issue I have here is that contour integration needs to be from $c-iT$ to $c+iT$ but the radius of convergence is finite.

Can the inversion be done here in some way?

Edit: Here is a concrete example.
Consider \begin{align} F(s)=\sum_{n=0}^\infty a_n s^n \end{align} where $a_n=2^{-n-2} (i(1+i)^{n+1}+(i-1)^n) $. This power series has a radis of convergence $r=\sqrt{2}$.

In fact, the above power series corepsonds to a function \begin{align} F(s)= \frac{1}{1+(1+s)^2}, \end{align} which has an inverse trasform given by \begin{align} f(t)= e^{-t} \sin(t) u(t) \end{align} where $u(t)$ is the step function.

However, if we don't know what the actual function is, how would we use the power series representation to find the inverse.

Answer 1

The question can be rephrased as

When the Taylor series of $F(s)$ is known in a neighbourhood of $s=\alpha$, how can the inverse Laplace transform of $F(s)$ be found?

For $\displaystyle{F(s)= \int_0^\infty e^{-st} f(t) dt}$, the region of convergence of $F(s)$ must be of the form $\text{Re } s>\sigma$ (call this set $H_\sigma$).

Theoretically, with knowledge of Taylor series of $F$ in a neighbourhood of $\alpha$, we can analytically continue $F$ to $H_\sigma$. However, Taylor series, which was born to have circular region of convergence, is hard to be analytically continued to half-planes.

Therefore, my idea is:

1. Conformally map $H_\sigma$ to the unit circle, as well as $\alpha$ to the origin.
2. Expand the 'conformally-mapped $F$' as a Maclaurin series.
3. Due to the absence of singularities in the unit circle, the Maclaurin series has radius of convergence $1$.
4. Inverse the conformal map. Now, the functional form of $F$ converges on $H_\sigma$.
5. Perform inverse Laplace transform.

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Theorem If $F(s)$, the Laplace transform of $f(t)$, has abscissa $\sigma$ and can be expanded as a $\sum^\infty_{j=0}a_j(s-\alpha)^j$ (where $\text{Re }\alpha>\sigma$), then $$f(t)=u(t)\cdot\lambda e^{(\alpha-\lambda)t}\sum^\infty_{n=1}b_n\cdot \ell_n(\lambda t)$$ where $u(t)$ is the Heaviside step function and $$\lambda=2(\text{Re }\alpha-\sigma)$$ $$b_n=\sum^n_{k=1}\binom{n-1}{k-1}a_k\lambda^k$$ $$\ell_n(x)=\frac{dL_n(x)}{dx}$$ (NB: $L_n$ are Laguerre polynomials.)

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Conformal mapping

All confomral maps from upper-half plane $\mathbb H$ to the unit disk $\mathbb D$ have the form $$M(z)=e^{i\theta}\frac{z-b}{z-\bar b}\qquad \theta\in\mathbb R, b\in \mathbb H$$

First, map $H_\sigma$ to $\mathbb H$ by $g(z)=i(z-\sigma)$ (equivalent to a leftward translation and 90-degree anticlockwise rotation). Then, map $\mathbb H$ to $\mathbb D$ by $m(z)=\frac{z-a}{z-\bar a}$, where $a=i(\alpha-\sigma)$.

You may check that these two maps together map $\alpha$ to $0$.

Hence, $$F:H_\sigma\mapsto\mathbb C\implies F\circ g^{-1}\circ m^{-1}:\mathbb D\mapsto \mathbb C$$ (This could be a little counter-intuitive, but I will leave it to you to figure out.)

Here, $$m^{-1}(z)=\frac{\bar a z-a}{z-1}\qquad g^{-1}(z)=-iz+\sigma$$

Therefore, the 'conformally-mapped $F$' is $$F\circ g^{-1}\circ m^{-1}=F\left(\frac{(2\sigma-\bar\alpha)z-\alpha}{z-1}\right)$$ (by substituting in $a=i(\alpha-\sigma)$ and doing some algebra.)

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Maclaurin series expansion

Let $\lambda=\alpha+\bar\alpha-2\sigma=2(\text{Re }\alpha-\sigma)$.

Given that $\displaystyle{F(s)=\sum_{n=0}^\infty a_n (s-\alpha)^n}$, after some algebra, we have $$F\left(\frac{(2\sigma-\bar\alpha)z-\alpha}{z-1}\right) =\sum_{n=0}^\infty a_n 2^n(\text{Re }\alpha-\sigma)^n\left(\frac{z}{1-z}\right)^n \equiv \sum_{n=0}^\infty b_nz^n $$

where

$$b_n=\sum^n_{k=1}\binom{n-1}{k-1}a_k\lambda^k$$

This Maclaurin series necessarily has radius of convergence $1$.

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Inverse conformal mapping

Since $F\circ g^{-1}\circ m^{-1}=\sum_{n=0}^\infty b_nz^n$, we have $$F(s)=\sum_{n=0}^\infty b_n(m\circ g(s))^n=\sum_{n=0}^\infty b_n\left(\frac{s-\alpha}{s+\bar\alpha-2\sigma}\right)^n$$

which is valid on the entire $H_\sigma$.

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Inverse Laplace transform

(It is assumed $t>0$. The proof for $t<0$ is trivial.)

Note that $\lim_{s\to+\infty}F(s)=0$, therefore $$\sum^\infty_{n=0}b_n=0$$

Thus, $F(s)$ can be rewritten as $$F(s)=\sum_{n=0}^\infty b_n\left[\left(\frac{s-\alpha}{s+\bar\alpha-2\sigma}\right)^n-1\right]$$

By residue theorem and Jordan's lemma, $$\frac1{2\pi i}\lim_{T\to\infty}\int^{\sigma+iT}_{\sigma-iT}\left[\left(\frac{s-\alpha}{s+\bar\alpha-2\sigma}\right)^n-1\right]e^{st}ds=\text{Res}_n$$

where

$$\text{Res}_{n}=\frac1{n!}\lim_{z\to 2\sigma-\bar\alpha}\frac{d^{n}}{dz^{n}}(z-\alpha)^{n+1}e^{zt}=\lambda e^{(\alpha-\lambda)t}\sum^n_{r=1}\binom{n}{r}\frac{(-1)^r(\lambda t)^{r-1}}{(r-1)!}$$

To this end, we notice the resemblance with Laguerre polynomials $$L_n(x)=\sum^n_{r=0}\binom{n}{r}\frac{(-1)^rx^r}{r!}$$

In essence, $$\text{Res}_{n}=\lambda e^{(\alpha-\lambda)t}\frac{dL_n(x)}{dx}\bigg\vert_{x=\lambda t}:=\lambda e^{(\alpha-\lambda)t}\ell_n(\lambda t)$$

and hence the theorem.

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Example

Let us verify the theorem for $\mathcal L\{u(t)\}(s)=\frac1s$.

Suppose we know that $$F(s)=\sum^\infty_{j=0}(-1)^j(s-1)^j$$ Then what is $f(t)$?

Here, $\sigma=0$, $\alpha=1$, $\lambda=2$, $a_n=(-1)^n$.

Hence, $$\begin{align} b_n &=\sum^n_{k=1}\binom{n-1}{k-1}(-1)^k2^k \\ &=\sum^{n-1}_{k=0}\binom{n-1}{k}(-1)^{k+1}2^{k+1} \\ &=2\cdot(-1)^{n}\sum^{n-1}_{k=0}\binom{n-1}{k}(-1)^{n-1-k}2^{k} \\ &=2\cdot(-1)^{n}(2-1)^{n-1}=2\cdot(-1)^n \end{align} $$

According to the theorem, $$f(t)=u(t)\cdot 4e^{-t}\sum^\infty_{n=1}(-1)^n\ell_n(2t)$$

Recall the generating function of Laguerre polynomials $$\sum^\infty_{n=0}z^nL_n(x)=\frac1{1-z}e^{-zx/(1-z)}$$ Thus, $$\sum^\infty_{n=1}z^n\ell_n(x)=-\frac{z}{(1-z)^2}e^{-zx/(1-z)}$$

Consequently, $$f(t)=u(t)\cdot 4e^{-t}\cdot \frac1{4}e^{2t/2}=u(t)$$ as expected.

Answer 2

Concretely the only hope you have is to find a smooth function $g$ such that $G(s) = \int_{-\infty}^\infty g(t)e^{-st}dt$ is analytic at $s=0$ and such that $$f(t)=\lim_{K\to \infty} \sum_{k=0}^K (-1)^k b_k g^{(k)}(t)\qquad \text{converges in } L^1_{loc} \text{ or in the sense of distributions}$$ where $$\frac{F(s)}{G(s)} = \sum_{k=0}^\infty b_k s^k$$ In that case $f(t)$ is the inverse Laplace transform of $F(s)$