Inverse Laplace Transform for $ sK_0(\sqrt{as}) $?

Question

Are there any ways to manipulate the following in order to make it invertible from Laplace space analytically?

$$ sK_0(\sqrt{as})$$

where $a$ is a constant.

Answer 1

The following Laplace transform pair appears in Erdelyi's bible of integral transforms (eq. (30) on p.516) :

$$f(t)=\frac{1}{2t}\exp\left(-\frac{a}{4t}\right)\leftrightarrow F(s) = K_0(\sqrt{as})=\int_0^{\infty}e^{-s t}f(t)\,dt.$$

I write the explicit transform to recall that integration-by-parts implies

$$F(s)=-\left[\frac{1}{s}e^{-st}f(t)\right]_{0}^\infty+\frac{1}{s}\int_0^\infty e^{-st}f'(t)\,dt$$ The behavior as $t\to 0$ isn't as nice as one would like, but the exponential factor suppresses the $1/t$ divergence and so $f(0)=0$. Hence we have

$$s K_0(\sqrt{as})=s F(s)=\int_0^\infty e^{-s t}f'(t)\,dt$$

and therefore the requisite inverse Laplace transform is $$f'(t)=\left(-\frac{1}{2t^2}+\frac{a}{8t^3}\right)\exp\left(-\frac{a}{4t}\right)=\frac{a-4t}{8t^3}\exp\left(-\frac{a}{4t}\right).$$

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Here is a partial proof of the identity. (I regard it as a partial proof because I can't justify some of the formulas used.)

I begin with the integral representation

$$F(s)=K_0(\sqrt{as})=\int_0^\infty \frac{\cos x}{\sqrt{x^2+as}}\,dx.$$

According to Mathematica, one has $$\mathcal{L}^{-1}\left[\frac{1}{\sqrt{x^2+as}}\right]=\frac{e^{-tx^2/a}}{\sqrt{a \pi t}}$$ and therefore the inverse transform of $F(s)$ may be represented as $$f(t)=\int_0^\infty \frac{e^{-t x^2/a}}{\sqrt{a \pi t}}\cos x\,dx.$$

Leaving this up to Mathematica, we obtain $f(t) = (2t)^{-1} e^{-a/4t}$ as claimed.

Notes:

• (1) is not the standard form of this representation (compare equation 10.32.E6, but I've rescaled the integration variable to suit my purpose. For the derivation, the verification shown in the answer to this earlier question may provide a lead.

• The functions in (2) are simple enough that one may perhaps verify the Laplace transform by hand and the inverse Laplace transform via the Bromwich integral.

• If I was being rigorous, I'd need to justify writing $$\int_0^\infty f(x)\mathcal{L}[g(x,t)](s)\,dx=\mathcal{L}\left[\int_0^\infty f(x) g(x,t)\,dt\right](s)$$ ...but you'd better believe I can't be bothered.

• If we consider $\cos x$ as the real part of $e^{i x}$, then we may view (3) as the real part of a computable Gaussian integral. So this is tractable by hand. (We may also regard it as a Fourier cosine transform, and as such is almost certainly in Erdelyi's table as well.)

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I couldn't resist one more try, and the result is worth it. We start from the integral representation

$$K_0(\sqrt{as})=\frac12 \int_0^\infty\exp\left(-x-\frac{as}{4x}\right)\frac{dx}{x}$$

(see eqn. 10.32.10 of the DLMF). We then make the substitution $t=a/(4x)$, obtaining

$$K_0(\sqrt{as})=\frac12 \int_0^\infty\exp\left(-\frac{a}{4t}-st \right)\frac{dt}{t}$$

which upon reflection is nothing other than the Laplace transform of $f(t)=\dfrac{1}{2t}\exp\left(-\dfrac{a}{4t}\right)$.