# 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 $$f''(x)$$?

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Whether or not relevant, it is interesting to note the following (not too well-known) result: the Laplace transform of the product $f_1 (t) f_2 (t)$ is equal to the convolution of $F_1 (s)$ and $F_2 (s)$. See atp.ruhr-uni-bochum.de/rt1/syscontrol/… – Shai Covo Apr 13 '11 at 20:24

Hint: use integration by parts.

See here for a full solution.

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already tried that – user9551 Apr 13 '11 at 19:10
@user9551 It is all you need, really. I'll put the solution in the answer then. – Glen Wheeler Apr 13 '11 at 19:11
i do not find solution of this particular form there ,n is for power & not for order of diff. function. – user9551 Apr 13 '11 at 19:19
@user9551 Really? Apply the formula there to the second derivative, use the boundary condition you have, and then take the $n$-th power of both sides. – Glen Wheeler Apr 13 '11 at 19:30
really sorry but i don't get you ,can you please explain ,i'll be really thankful – user9551 Apr 13 '11 at 19:57

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.