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

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|.

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.

• This is an interesting question seeing as a Laplace transform is basically the continuous version of a power series. Commented Oct 29, 2019 at 20:32
• @Math1000 any idea on how to approach something like this?
– Boby
Commented Oct 29, 2019 at 20:38
• Note that the $\lim_{T\to\infty}$ applies to the imaginary part of $c-iT$ and $c+iT$. So as long as we choose $c<r$, there is no problem with convergence. Commented Oct 29, 2019 at 21:14
• @Math1000 Is that correct though? The region of convergence of power series is a ball in a complex plane of radius $r$. Is in it the case that once $|c-iT|>r$ the power series no longer converges?
– Boby
Commented Oct 29, 2019 at 21:40
• Oh, I was thinking of the power series as as real function. In that case I am no longer sure. Commented Oct 29, 2019 at 21:41

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.

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.)

## 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.)

## 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 $$1$$.

## 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 $$H_\sigma$$.

## 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.

## 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.

• Thanks. Very cool answer. I will try to go over this soon and ask some questions. Meanwhile, it would be cool if you could finish the last step
– Boby
Commented Nov 4, 2019 at 21:25
• @Boby please see my latest edit. Commented Nov 4, 2019 at 23:54
• I have a power series that converges everywhere. Can I still use your theorem? Analogously to the power series expansion of the exponential function (r=Infinity and neighbourhood alpha=0) As far as I can tell sigma =-Infinity and alpha=0 which makes your formula hard to use. I'm a physicist and no expert at all so I would really appreciate your advice. Thank you. Commented May 28, 2020 at 9:31

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 $$F(s)$$

• Thanks. I don't follow every think but will think about this approach. Could maybe outline how this would work with a power series that I provided?
– Boby
Commented Oct 30, 2019 at 21:58
• No. It works with $f(t)=e^{-t^2}, g(t)= e^{-2t^2}$ because the quotient of their bilateral Laplace transform is $F(s)/G(s)= \sqrt{1/2} e^{s^2/4}/e^{s^2/8}$ which is entire and rapidly decreasing on vertical lines, from the power series at $0$ we have as its inverse Laplace transform the series of dirac deltas derivatives $\sum_kb_k(-1)^n\delta^{(k)}$ which converges only in the sense of analytic functionals but when taking the convolution with $g(t)$ we obtain $\sum_k b_k (-1)^n g^{(k)}$ which does converge in the sense of distributions, to $f$. Finding $g$ is not much easier than finding $f$. Commented Oct 30, 2019 at 22:14