Inverse Laplace transform of F(s)=exp(-as) as delta(t-a)

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Main Question or Discussion Point

This is mostly a procedural question regarding how to evaluate a Bromwich integral in a case that should be simple.

I'm looking at determining the inverse Laplace transform of a simple exponential F(s)=exp(-as), a>0. It is known that in this case f(t) = delta(t-a). Using the Bromwich formula, with s = x+iw:
$$2\pi f(t) = \int_{-\infty} ^\infty F(x+iw)exp(xt)exp(iwt)dw$$
For this specific F(s):
$$2\pi f(t) = \int_{-\infty} ^\infty exp((t-a)x)exp(iw(t-a))dw$$
$$2\pi f(t) = exp((t-a)x) \int_{-\infty} ^\infty exp(iw(t-a))dw$$
At t = a, this becomes:
$$2\pi f(a) = \int_{-\infty} ^\infty dw$$
Thus $$f(a) -> \infty$$
This is so far consistent with a delta(t-a) function.
For all t <> a:
$$\pi f(t) = lim_{T->\infty} exp((t-a)x) sin(T(t-a))/(t-a)$$
which clearly does not converge rather than being zero as expected for f(t) = delta(t-a).

Can anyone explain what I am doing wrong?

jasonRF
Gold Member
The real problem here is that you are dealing with generalized functions (also called distributions) like the delta function. The standard techniques that work for nice functions don't really work for generalized functions the same way. I can think of three ways to do this, but don't know which one would be the most useful for you. Here are the methods

method 1: Do this the way I would want to do this using hte theory of distributions. This would be a good way only if you have been introduced to the subject.

method 2: Use a leap of faith, by claiming that,
\begin{eqnarray}
\frac{1}{2\pi i}\int_{\sigma_0-i \infty}^{\sigma_0+i \infty}ds \, e^{s(t-a)} & = & \lim_{\epsilon\rightarrow 0}\frac{1}{2\pi i}\int_{\sigma_0-i \infty}^{\sigma_0+i \infty}ds \, e^{s(t-a)} \, e^{-\epsilon|s|}
\end{eqnarray}
The reason this is true follows from the topic of method 1. Anyway, if you use this approach, then you also need to know that
\begin{eqnarray}
\lim_{\epsilon\rightarrow 0}\frac{\epsilon}{\pi\left(\epsilon^2 + (t-a)^2\right)} & = & \delta(t-a).
\end{eqnarray}
This is provable with things that you learned in calculus (but basic real analysis would be better).

method 3: realize that the nice integral form of the inversion integral is "easy" to use for continuous ##f(t)##. So instead of directly finding the laplace transform of your ##f(t)## that we know is essentially a delta function, we can instead find the inverse transform of ##G(s) = F(s)/s^2##, then realize that ##f(t) = \frac{d^2}{dt^2}g(t)##. We know that ##g(t)## will be continuous because we really know the answer and the second integral of the delta function will be ##(t-a)\, u(t-a)## which is continuous. The only issue with this approach for your problem is that you need to know countour integration (specifically the residue theorem) to do the resulting integral of ##e^{s(t-a)}/s^2##.

Note that for a general ##F(s)## you won't know the answer. It turns out that if it is bounded above by a polynomial in ##|s|## of order ##m## as ##s\rightarrow \infty##, then we will want to find the inverse transform of ##G(s) = F(s)/s^{m+2}## and then take ##m+2## derivatives of the resulting ##g(t)##. In your case, ##m=0##.

I am happy to help more, especially if you let me know what option makes the most sense to you.

Jason

EDIT: I think I am implicitly assuming you are using hte unilateral Laplace transform, not the bilateral, for method 3. Perhaps I should learn more about it.

Also, for method 2, the limit
\begin{eqnarray}
\lim_{\epsilon\rightarrow 0}\frac{\epsilon}{\pi\left(\epsilon^2 + (t-a)^2\right)} & = & \delta(t-a).
\end{eqnarray}
really means that for a nice function ##\phi(t)##,
\begin{eqnarray}
\lim_{\epsilon\rightarrow 0}\int dt \, \phi(t) \frac{\epsilon}{\pi\left(\epsilon^2 + (t-a)^2\right)} & = & \phi(t-a).
\end{eqnarray}

Last edited:
Hi Jason, and thanks for the very detailed and helpful answer. You are correct I do not have much exposure to the theory of distributions, and to answer your question regarding what I'm trying to do - this is connected to another thread I have posted here at https://www.physicsforums.com/threa...sform-of-a-piecewise-defined-function.904387/

I eventually want to express a piecewise defined function as a discrete series of exponentials of the type below as a single Laplace transform, i.e.
$$\int_0 ^\infty f(t)\exp (-st) ds = F(s) =\begin{cases} \sum_{n=1,2..} A_n \exp(-a_n s) &\text{ if }0\le s\le s_c \text{ and}\\ \sum_{n=1,2..} B_n \exp(-b_n s) &\text{ if } s>s_c \end{cases}$$

Note that on a per-piece basis, the inverse Laplace transform of the RHS is easy to derive, namely it is a sum of delta functions centered on a_1, a_2, etc., b_1,b_2, etc., but I'd like to look at the whole function and on the LHS t is of course on the continuum.

This is the solution of a differential equation for a system involving diffusion I am working on - it is made of two pieces in contact with each other and which allow mass transfer across an interface. In one of the phases the solution is given by the RHS, in the other by the LHS, and so I'm trying to find the relationship between them.

This also answers your last question which is that I am indeed using unilateral transforms.

jasonRF
Gold Member
This is a very unusual problem. What is the value of ##F(s)## for ##s<0##? Are any (or all) of the ##(A_n,a_n)## pairs duplicated in the ##(B_n,b_n)##?

One thing we know is that the unilateral Laplace transform of a function is an analytic function of ##s## for a right-half of the complex plane, usually denoted ##\Re(s) > \sigma_0##. Also, if ##f## is zero outside of a finite interval, then the Laplace transform is analytic over the entire complex plane. It turns out that both of these statements also hold if ##f## is a generalized function. Thus, a sum of delta functions will have a Laplace transform that is analytic over the entire complex plane; indeed, it will be a sum of exponentials valid over the entire complex plane. Your Laplace transform does not satisfy that criterion. Perhaps it is possible to find an ##f## that yields your transform, but I feel like there must be some missing information that we need to properly understand how to interpret and invert this transform. I am wondering how you arrived at this functional form of your transform. Perhaps you should post the actual problem you are trying to solve. There may be a better way to solve it.

EDIT: Is there anything that you know about the general properties of ##f(t)##? For example, is it non-zero for all ##t>0##, or is it non-zero only for a finite interval of time? Any extra info would be helpful here.

Jason

Last edited:
Hi again Jason,

I'll create a thread with the original problem as you advise, but that is a bit involved and relatively specific, so we'll have to see if people will have the patience to go through it. But to answer your other questions for now - F(s) is zero for s < 0. The equality in question, for s>0, is:

$$\int_0^{\infty{}}\exp{\left(-\tau{}\xi{}\right)}\sum_n\frac{T_n(0)}{2\sqrt{\tau{}}}\left\{A_n^*(\tau{})\sin{\left(\sqrt{\frac{\tau{}}{D_p}}z\right)}+B_n^*(\tau{})\cos{\left(\sqrt{\frac{\tau{}}{D_p}}z\right)}\right\}d\tau{}=\left\{\begin{array}{l}\frac{1}{2}B_0-\alpha{}C_S+\sum_{n=0,1,2…}^{\infty{}}{\epsilon{}}_n\left\{\frac{\sqrt{k}}{{\lambda{}}_n}\sin{\left(\frac{{\lambda{}}_n}{\sqrt{k}}z\right)}+\alpha{}\frac{l_zD_p}{D_g}\cos{\left(\frac{{\lambda{}}_n}{\sqrt{k}}z\right)}\right\}\exp{\left(-{{\lambda{}}_n}^2\xi{}\right)}if\ 0<\xi{}\leq{}L_x \\ -\sum_{m=1,2,..}^{\infty{}}B_m\cos{\left(\frac{m\pi{}}{L_{zp}}z\right)}\exp{\left(kL_x{\left(\frac{m\pi{}}{L_{zp}}\right)}^2\right)}\exp{\left(-k{\left(\frac{m\pi{}}{L_{zp}}\right)}^2\xi{}\right)}if\ \xi{}>L_x\end{array}\right.$$

The f(t) is the sum in the LHS, s is referred to here as csi, B0/2-alpha*Cs are constants such that the RHS is continuous and converges to zero as csi --> Inf, which is a requirement of any Laplace transform. Alpha, k, Lz, Dp, Dg are constants.

The b_n are proportional to n^2 as seen in the second part of RHS above, and the lambda above which is what the a_n is related to is the solution to:
$${\lambda{}}_n=\frac{D_g}{l_zD_p}\frac{\sqrt{k}}{\alpha{}}\cot{\left(\frac{{\lambda{}}_n}{\sqrt{k}}L_{zp}\right)},\ \frac{{\lambda{}}_n}{\sqrt{k}}L_{zp}\in{}\left(n\pi{},\left(n+\frac{1}{2}\right)\pi{}\right),n=0,1,2…$$
That means that all b_n are different from all a_n to answer another question you had.

They are both solutions by Fourier (and Fourier-like for the a_n) series of the convection-diffusion equation for my specific system at steady-state.

Finally the amplitudes epsilon_n and B_m are given by:

$${\epsilon{}}_n=\frac{2\alpha{}C_S}{{\left(\alpha{}\frac{l_zD_p}{D_g}\right)}^2\frac{{{\lambda{}}_n}^2}{k}L_{zp}+\alpha{}\frac{l_zD_p}{D_g}+L_{zp}}$$
$$B_0=2\alpha{}C_S-\frac{2k}{L_{zp}}\sum_{n=0,1,2…}^{\infty{}}\frac{{\epsilon{}}_n}{{{\lambda{}}_n}^2}\exp{\left(-{{\lambda{}}_n}^2L_x\right)}$$
$$B_m=2\sum_{n=0,1,2…}^{\infty{}}{\epsilon{}}_n\exp{\left(-{{\lambda{}}_n}^2L_x\right)}\frac{L_{zp}}{{\pi{}}^2m^2-\frac{{{\lambda{}}_n}^2}{k}{L_{zp}}^2}$$

jasonRF