In this previous question, it was shown that the Laplace transform of $f(t^2)$ can be expressed in terms of the Laplace transform $F$ of $f$, namely:
$$ H_2(s) = \int_0^\infty f(t^2) e^{-st} dt = \int_0^\infty\frac{F(t)}{2 \sqrt{\pi t}} e^{-s^2/(4t)} dt$$
Note that it is possible to rewrite this equation by using the modified Bessel function $K_\nu$:
$$ H_2(s) = \int_0^\infty f(t^2) e^{-st} dt = \frac{{s}}{2^{3/2}\pi} \int_{0}^{\infty} \frac{F(t)}{{t}} K_{1/2} \Bigg( \frac{s^{2}} {{4t}} \Bigg) dt$$
Using a similar argument than the one found in the previous question, one can find the following for the Laplace transform of $f(t^3)$:
$$H_3(s) = \int_{0}^{\infty} f(t^3) e^{-st} dt = \frac{\sqrt{s}}{3\pi} \int_{0}^{\infty} \frac{F(t)}{\sqrt{t}} K_{1/3} \Bigg( \frac{2}{3} \frac{s^{3/2}} {\sqrt{3t}} \Bigg) dt $$
Question: is it possible to find similar equations for the Laplace transform of $f(t^n)$ for $n>=4$ involving the modified Bessel function ?
I found the case for $n=3$ with the help of Mathematica by looking for the inverse Laplace transform of $\exp(-s*t^{1/n})/n/t^{(n-1)/n}$, which gave me an answer in terms of the Airy function, which can then be written in terms of $K_\nu$.
However, for case $n=4$, the result is much more complex (in terms of hypergeometric functions).
