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I have always accepted that the bilateral Laplace Transform of a constant function $f(t) = c$ does not exist. How could the following integral possibly converge,

$$\mathcal{L}[f(t)]=\int\limits_\mathbb{R}ce^{-st}\,\mathrm{d}t\;?$$

Then I learned about distributions and how they are perfect candidates to find the Fourier Transform of "problematic" functions for which it is hard or even impossible to evaluate the usual Fourier integral. Here, a constant function can be transformed and yields the Dirac impulse $\delta(f)$ and by duality, this holds also in the other direction.

So the Laplace Transform of a Dirac impulse is easily found by employing the sifting property and the definition of the Dirac impulse:

$$\mathcal{L}[\delta(t)]=\int\limits_\mathbb{R}\delta(t)e^{-st}\,\mathrm{d}t=\int\limits_\mathbb{R}\delta(t)\underbrace{e^{-s\cdot0}}_{=1}\,\mathrm{d}t=1.$$

Now I was wondering, why the following does not hold,

$$\mathcal{L}[1]=\delta(s).$$

I looked up a few papers and lectures on the Laplace Transform of distributions but nowhere did I find a reason for why this is not true (I may have overlooked it, though). I then tried to find out, whether $\delta(s)$ is defined but all sources that I found defined the domain of both distributions and test functions (let us consider Schwartz functions) as the real line or subsets thereof.

I suspect that there is a reaon that prevents distributions to be defined on the complex plane. Maybe it has to do with complex integation, but I am not sure.

Another reason I was thinking of is the region of convergence. When viewing the Laplace Transform of $f(t)$ as the Fourier Transform of $f(t)e^{-\alpha t}$, where $\alpha=\mathrm{Re}(s)$, I think this can only dealt with in the context of distibutions when $\alpha=0$. Otherwise we could find a test function $\phi(t)$ which decreases expontially and thus the pairing $\langle 1\cdot e^{-\alpha t}, \phi(t)\rangle \; \forall \alpha \neq 0$ gives the integral over a constant function which will not converge. But if the region of convergence is just the imaginary axis, we cannot evaluate the integral in the inverse Laplace Transform (but I cannot really say, why. It is rather a gut feeling).

I am looking forward to enlightening answers why we cannot find the bilateral Laplace Transform of a constant function.

Edit: In the notes of my Signals & Systems class it was argued that it would be the same as the sum of the transforms of a usual and a reflected step function. The resulting region of convergence is the union of both regions but they do not overlap as these are the left half plane and the right half plane, respectively. Hence, the bilateral transform of a constant function cannot exist. But why does this rule out the use of distributions?

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The inverse Laplace transform of $1$ is $\delta(x)$.


Assuming the Fourier transform of $f(x)$ is defined by $\mathcal{F}_x[f(x)](t)=\int\limits_{-\infty }^{\infty } f(x)\ e^{-2 \pi i t x} \, dx$, then the Fourier transform of $f(x)=1$ is $\mathcal{F}_x[1](t)=\int\limits_{-\infty }^{\infty } 1\ e^{-2 \pi i t x} \, dx=\delta(t)$.


This is equivalent to the bilateral Laplace transform $\mathcal{L}_x[1](s)=\int\limits_{-\infty }^{\infty } 1\ e^{-s x} \, dx$ evaluated at $s=2 \pi i t$.


Note $\int\limits_{-1/(2 \pi \epsilon)}^{1/(2 \pi \epsilon)} 1\ e^{-2 i \pi t x} \, dx=\frac{\sin \left(\frac{t}{\epsilon }\right)}{\pi t}$ which is associated with the limit representation $\delta (t)=\underset{\varepsilon \to 0}{\text{lim}}\frac{\sin \left(\frac{t}{\varepsilon }\right)}{\pi t}$.


Assuming the Fourier transform $F(t)=\mathcal{F}_x[f(x)](t)$ defined above, the inverse Fourier transform is given by $f(x)=\mathcal{F}_t^{-1}[F(t)](x)=\int\limits_{-\infty }^{\infty } F(t)\ e^{2 \pi i x t} \, dt$.


For $F(t)=\delta(t)$ this becomes $f(x)=\mathcal{F}_t^{-1}[\delta(t)](x)=\int\limits_{-\infty }^{\infty } \delta(t)\ e^{2 \pi i x t} \, dt=1$ which is consistent with $F(t)=\mathcal{F}_x[1](t)=\delta(t)$.


The Dirac delta function $\delta(t)$ is only defined for $t\in \mathbb{R}$ and is undefined for $\Im(t)\ne 0$ (i.e. is only applicable in integrals over $t\in \mathbb{R}$). This is why it doesn't make sense to talk about the bilateral Laplace transform $\mathcal{L}_x[1](s)$, whereas it does make sense to talk about the bilateral Laplace transform $\mathcal{L}_x[1](2 \pi i t)$.

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  • $\begingroup$ Thanks for your answer but I think you misunderstood my question. I was not asking for the inverse Laplace Transform of 1 but the Laplace Transform of 1, the other direction. $\endgroup$ Jul 13 '20 at 19:45
  • $\begingroup$ Unfortunately, one cannot always obtain the Laplace from the Fourier Transform just by changing the variables from frequency to complex frequency and vice versa. This holds in some cases but it depends on the region of convergence (and some other things as well, I think). I searched many tables, lecture notes and papers but the bilateral Laplace Transform of a constant function was either not mentioned at all or claimed to be non-existing. I had written another comment but I appended it to the original question. $\endgroup$ Jul 13 '20 at 20:57
  • $\begingroup$ I think you're confused as $\mathcal{L}_x[f(x)](s)\ne \mathcal{L}_x[f(x)](2 \pi i t)$, i.e. the bilateral Laplace transform of $f(x)$ evaluated at $s$ is not the same thing as the bilateral Laplace transform of $f(x)$ evaluated at $2 \pi i t$. Also, the relationship between the Fourier and bilateral Laplace transforms of $f(x)$ holds for all definitions of $f(x)$ for which the Fourier transform is defined which is easily proven with a simple substitution of variables as I illustrated above. $\endgroup$ Jul 13 '20 at 22:30

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