Laplace transform of $[f''[x]]^n$ Can anyone help me get this Laplace transform,
$$
L[(f''(x))^n]
$$
where $f'(0)=0$ and $f''(0)=0$ and $n$ is power of $$f''(x)$$?
 A: Hint: use integration by parts.
See here for a full solution.
A: I don't think there is a simple enough closed form (i.e. the case $f''(x)^{10}$ would be a sucession of integrals involving products and sums of the powers of all derivatives from $10$ to $0$ (the original function). 
Integrating by parts gives:
$$\mathcal{L}\left\{ {f{{\left( t \right)}^n}} \right\} = \frac{{f{{\left( 0 \right)}^n}}}{s} + \frac{n}{s}\int\limits_0^\infty  {{e^{ - st}}f{{\left( t \right)}^{n - 1}}f'\left( t \right)dt} $$
But then you'd need a new formula for
$$\mathcal{L}\left\{ {f{{\left( t \right)}^{n - 1}}f'\left( t \right)} \right\}\left( s \right)$$
Which would be, if I'm not mistaken
$$L\left\{ {f{{\left( t \right)}^{n - 1}}f'\left( t \right)} \right\}\left( s \right) = \frac{1}{s}f{\left( 0 \right)^{n - 1}}f'\left( 0 \right) + \frac{{n - 1}}{s}L\left\{ {f{{\left( t \right)}^{n - 2}}f'{{\left( t \right)}^2}} \right\}\left( s \right) + \frac{1}{s}L\left\{ {f{{\left( t \right)}^{n - 1}}f''\left( t \right)} \right\}\left( s \right)$$
And now you'd need one for the two new arguments. The recursion would be rather chaotic.
EDITED: Didn't read question properly.
