Inverse Laplace transform of an exponential function

Question

What is the inverse Laplace transform of $$\frac{e^{\frac{-2}{s}}}{s}$$ I have seen an answer using Maclaurin series expansion of this function. This function is not analytic at $0$, so, is such expansion legal?

Answer 1

To answer your question: it is not a Maclurin expansion of this function, or the integrand of the ILT. Such an expansion does not exist because of the essential singularity at $s=0$. Rather, as I will demonstrate below, it is a Laurent expansion of the integrand. A Laurent expansion of a function about an essential singularity has nonzero coefficients at all negative powers. However, to compute the ILT, we need the residue, which is just the coefficient of $s^{-1}$.

Write out the ILT as

$$\frac1{i 2 \pi}\int_{c-i \infty}^{c+i \infty} \frac{ds}{s} e^{-2/s} e^{s t} $$

Consider the contour integral over the Bromwich contour $C$:

$$\frac1{i 2 \pi}\oint_C \frac{dz}{z} e^{-2/z} e^{t z} $$

which is equal to the above ILT as well as the residue of the singularity at $z=0$. In this case, the integrand has an essential singularity at $z=0$. Thus, we may determine the residue from a Laurent expansion of the integrand at $z=0$:

$$\frac1{z} e^{2/z} e^{t z} = \frac1{z} \sum_{n=0}^{\infty} \frac1{n!} \left (-\frac{2}{z}+t z \right )^n$$

Recall that the residue is the coefficient of $\frac1{z}$ in the Laurent expansion. Thus, because we have a factor of $\frac1{z}$ outside the sum, we just need to find the coefficient of the constant term in the sum. In this case, we find that we get $z^0$ terms in the expansion of the sum for even $n$. For even $n$, the $z^0$ term is the center term. Thus, to compute this residue, we simply sum over these center terms for each even term in the sum:

$$\frac1{i 2 \pi}\int_{c-i \infty}^{c+i \infty} \frac{ds}{s} e^{-2/s} e^{s t} = \sum_{n=0}^{\infty} \frac1{(2 n)!} \binom{2 n}{n} (-2 t)^n = \sum_{n=0}^{\infty} (-1)^n \frac{\left (\sqrt{2 t}\right )^{2 n}}{(n!)^2}$$

The sum on the RHS is a Bessel function. Thus, we conclude that

$$\frac1{i 2 \pi}\int_{c-i \infty}^{c+i \infty} \frac{ds}{s} e^{-2/s} e^{s t} = J_0 \left (2 \sqrt{2 t} \right ) $$

ADDENDUM

To prove that the Laplace transform of that Bessel function produces the result we started with, one may use a Maclurin series of the Bessel about $t=0$. Note that it is indeed a Maclurin series despite that square root because all powers of the Bessel series are even.

Answer 2

$\mathcal{L}^{-1}\left\{\dfrac{e^{\frac{-2}{s}}}{s}\right\}$

$=\mathcal{L}^{-1}\left\{\sum\limits_{n=0}^\infty\dfrac{(-1)^n2^n}{n!s^{n+1}}\right\}$

$=\sum\limits_{n=0}^\infty\dfrac{(-1)^n2^nt^n}{(n!)^2}$

$=J_0(2\sqrt{2t})$