Does $e^{1/t}$ have a Laplace Transform?

Question

I'm having a little trouble understanding why some functions have a Laplace transformation and others don't.

The definition I was given in class last week was "Given a suitable function $F(t)$ the Laplace transform, written $f(s)$ is defined by

$$f(s)=\int_0^{\infty} F(t)\,e^{-st} dt$$

We were told some conditions for "suitability", like "All polynomial functions and exponential functions (of the type $e^{kt}$, $k$=constant) are suitable, as well as all bounded functions". And that, "Excluded functions are those that have singularities such as $\ln(t)$ and $1/(t-1)$, and functions with a growth rate more rapid than the standard exponential such as $e^{t^2}$."

But these definitions seem a bit loose to me, and now I've encountered the following problem: "Does $e^{t^{-1}}$ have a Laplace Transform, if so then what is it, and if not then why not?"

I don't think it does, because it's discontinuous at $t=0$ and the growth-rate isn't like a "standard" exponential growth-rate... But I'm honestly not sure now.

(P.S. Apologies for the equations not being all pretty and such, I haven't worked out how to do that yet)

Answer 1

For some reason, your instructor doesn't seem to be giving the obvious definition, which is that a function has a Laplace transform iff the corresponding integral which defines the transform exists (and is not infinity, since some definitions of integral accept infinity as a value). Perhaps that is because there are different definitions of integrals (e.g., Riemann, Lebesgue, "improper", "principal value").

What you want to know is whether $$f(s)=\int_0^{\infty} e^{1/t}\,e^{-st} dt$$ exists and for what $s$. For domain of integration $[1,\infty)$ it's no sweat: for $t\ge 1$, $e^{1/t}\le e$, so the integral exists for $s>0$ because $\int_1^\infty e^{-st}\,dt$ does. For $[0,1]$, however, $e^{-st}\ge e^{-s}$, and so $f(s)\ge e^{-s}\int_a^1 e^{1/t}\,dt$ for all $0<a<1$. The latter quantity becomes infinitely large as we let $a$ go to zero, since $e^{1/t}\ge 1/t$ and $\int_0^1 dt/t$ diverges. Because of this, $e^{1/t}$, I'd say, lacks a Laplace transform.

Caveat: There may be some way I'm unaware of to give the Laplace transform of $e^{1/t}$ a meaning.

Answer 2

Short answer, no. There is a nonintegrable singularity at $t=0$.

Longer, much more subtle answer...we can define a LT of this function, just not in the usual way. Rather, it is an analytical continuation of another, seemingly unrelated result. Let me explain a bit.

We begin with the following known result which I will not derive here:

$$\int_0^{\infty} dt \, e^{-1/t} e^{-s t} = \frac{2 K_1 \left ( 2 \sqrt{s} \right )}{\sqrt{s}} $$

where $K_1$ the modified Bessel function of the second kind of first order. We may express this as an inverse LT:

$$\frac1{i 2 \pi} \int_{c-i \infty}^{c+i \infty} ds \, \frac{2 K_1 \left ( 2 \sqrt{s} \right )}{\sqrt{s}} e^{s t} = e^{-1/t} $$

Now, throwing caution to the wind, we replace $t \mapsto -t$:

$$e^{1/t} = \frac1{i 2 \pi} \int_{c-i \infty}^{c+i \infty} ds \, \frac{2 K_1 \left ( 2 \sqrt{s} \right )}{\sqrt{s}} e^{-s t}$$

To get this to be a proper-looking ILT, replace $s \mapsto -s$:

$$\begin{align}e^{1/t} &= \frac1{i 2 \pi} \int_{-c-i \infty}^{-c+i \infty} ds \, \left [i \frac{2 K_1 \left ( i 2 \sqrt{s} \right )}{\sqrt{s}}\right ] e^{s t} \\ &= \frac1{i 2 \pi} \int_{-c-i \infty}^{-c+i \infty} ds \, \left [-\frac{\pi Y_1 \left ( 2 \sqrt{s} \right )}{\sqrt{s}} - i \frac{\pi J_1 \left ( 2 \sqrt{s} \right )}{\sqrt{s}}\right ] e^{s t}\end{align}$$

where $Y_1$ and $J_1$ are, respectively, the Bessel functions of the second and first kind of first order. One more simplification: the $J_1$ term is analytic in the complex plane to the left of the line of integration and as such contributes nothing to this integral. Thus, we can say that

$$e^{1/t} = \frac1{i 2 \pi} \int_{-c-i \infty}^{-c+i \infty} ds \, \left [-\frac{\pi Y_1 \left ( 2 \sqrt{s} \right )}{\sqrt{s}} \right ] e^{s t} $$

Now, we haven't really done anything radical yet. We actually still need to close the contour to the right rather than the left in order to capture the singularity at $s=0$. (We assumed that $c \gt 0$.) However, we can for the sake of a definition of a true inverse map $c \mapsto -c$ so that we define the ILT of $e^{1/t}$ for $t \gt 0$ (even though, again, it really doesn't make sense from the original definition).

In this sense, we can define the LT of $e^{1/t}$ as

$$\mathcal{L}\left [e^{1/t} \right ] = -\frac{\pi Y_1 \left ( 2 \sqrt{s} \right )}{\sqrt{s}}$$

even though the integral is not defined.

ADDENDUM

Just for laughs, if you have Mathematica, type in LaplaceTransform[Exp[1/t], t, s] and it returns the above result. Trying to evaluate the original integral of course results in a "Divergent integral" message.