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This question is about inverse Laplace transform $\mathscr{L}^{-1}:s\rightarrow t$. Although I was not able to find appropriate contour to invert $K_0 \left(r s\right)$, I somehow know that $$\mathscr{L}^{-1}\{K_0 \left(r s\right)\}=\frac{\theta(t-r)}{\sqrt{t^2-r^2}}.$$

How do one show that statement rigorously? The main part of my question is: Is it possible, similarly, to express a similar inverse fourier transform $$\mathscr{L}^{-1}\left\{K_0 \left(r \sqrt{s^2-1}\right)\right\}$$

in terms of elementary functions? Thank you for suggestions. ($K_0$ is the modified Bessel function of second kind, $\theta$ is just Heaviside theta, $r>0$ is a positive real parameter).


Important note: The function $K_0$ in the second laplace transform is ill-defined for $s\in (0,1)$ and is ment to represent only its real part, equivalently, using common identities for Bessel functions,

$$K_0 \left(r \sqrt{s^2-1}\right) = -\frac{\pi}{2}Y_0\left(r\sqrt{1 - s^2}\right)$$

which extends the domain of the original function to $s\in (0,1)$.

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  • $\begingroup$ I am not getting a closed form, but something like $$\frac{\pi}{2} \int_{-1}^1 dx \, J_0{\left ( r \sqrt{1-x^2} \right )} \, e^{-t x} $$ $\endgroup$
    – Ron Gordon
    Jul 25 '18 at 19:15
  • $\begingroup$ This looks very nice. However, it cannot be true since if you do the Laplace transform on it and compare with the original function, the graphs looks differently for $s>1$ - and even for $s<1$ they do not coincide. $\endgroup$
    – Machinato
    Jul 25 '18 at 19:39
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The result given by Mariusz can be verified as follows. The substitution $t = r \cosh \tau$ gives $$F(s) = \int_0^\infty \frac {\cosh \sqrt {t^2 - r^2}} {\sqrt {t^2 - r^2}} \theta(t - r) e^{-s t} dt = \int_0^\infty e^{-r s \cosh \tau} \cosh(r \sinh \tau) d\tau, \\ \operatorname{Re} s > 1.$$ Converting $\cosh(r \sinh \tau)$ to $e^{\pm r \sinh \tau}$ and writing $a \sinh \tau + b \cosh \tau$ as $A \cosh(\tau + \tau_0)$ gives $$F(s) = \frac 1 2 \int_0^\infty e^{A \cosh(\tau - \tau_0)} d\tau + \frac 1 2 \int_0^\infty e^{A \cosh(\tau + \tau_0)}, \\ A = -r \sqrt {s^2 - 1}, \; \tau_0 = \operatorname{arcsinh} \frac 1 {\sqrt {s^2 - 1}}.$$ Since $\cosh$ is even, $$F(s) = \frac 1 2 \int_{-\tau_0}^\infty e^{A \cosh \tau} d\tau + \frac 1 2 \int_{\tau_0}^\infty e^{A \cosh \tau} d\tau = \int_0^\infty e^{A \cosh \tau} d\tau = \\ K_0(-A).$$

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With CAS help:

$$\mathcal{L}_s^{-1}\left[K_0\left(r \sqrt{s^2-1}\right)\right](t)=\frac{\theta (t-r) \cosh \left(\sqrt{t^2-r^2}\right)}{\sqrt{t^2-r^2}}$$

for $r>0$

Maple code:

`assuming`([inttrans:-invlaplace(BesselK(0, r*sqrt(s^2-1)), s, t)], [r > 0])

#Heaviside(t-r)*cosh(sqrt(-r^2+t^2))/sqrt(-r^2+t^2)
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  • $\begingroup$ The square root inside $\cosh$ is missing in the MathJax formula. $\endgroup$
    – Maxim
    Jul 26 '18 at 2:34
  • $\begingroup$ @Maxim. Thanks,it was a typo. $\endgroup$ Jul 26 '18 at 7:24
  • $\begingroup$ This is miraculous! Thank you, this is what I was looking for. $\endgroup$
    – Machinato
    Jul 26 '18 at 8:01

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