There was a similar question in the past, which was not resolved. The Laplace Transform pair is well known: $$ \mathcal{L}_{t \mapsto s}: \frac{e^{-\frac{x^2}{4t}}}{2t} \div K_0( x \sqrt{s}) $$ Is it possible to give meaning to the Inverse Laplace Transform of the other modified Bessel function $I_0(\sqrt{s})$? Since $s$ is complex a complex rotation $ I_0(\sqrt{s}) \mapsto J_0(\sqrt{s})$, which is oscillating, so may be integration along the Browmwitch contour is possible.

Explanations are welcome.

  • $\begingroup$ I just changed \frac{{{ e}^{-\frac{{{x}^{2}}}{4 t}}}}{2 \, t} to \frac{e^{-\frac{x^2}{4t}}}{2t}. All the purposeless complications in the former are absurd. $\endgroup$ Jul 9 '18 at 0:47

The integral of $I_0(\sqrt {\sigma + i \omega}) e^{t(\sigma + i \omega)}$ from $\omega=-A$ to $\omega=A$ will behave as a subexponentially growing factor times an oscillating factor, thus the Bromwich integral diverges. Roughly speaking, this is because the absolute value of $I_0(\sqrt {\sigma + i \omega})$ grows as $|\omega|^{-1/4} \exp(\sqrt {|\omega|/2})/\sqrt {2\pi}$, while the oscillation frequency does not grow fast enough to compensate for that.

If the Laplace transform is defined through the Fourier transform, the inverse transform will be given by $$\mathcal L_{s \to t}^{-1}[F] = e^{\sigma t} \mathcal F_{\omega \to t}^{-1}[F], \\ s = \sigma + i \omega,$$ showing that the inverse transform of $I_0(\sqrt s)$ doesn't exist in terms of tempered distributions either.

  • $\begingroup$ How did you get the last claim? Can you elaborate? $\endgroup$
    – user48672
    Jul 9 '18 at 15:09
  • $\begingroup$ The reasoning is that $F(\omega) = I_0(\sqrt {\sigma + i \omega})$ does not induce a tempered distribution, because of the subexponentially growing factor. $\endgroup$
    – Maxim
    Jul 9 '18 at 16:15

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