What is the laplace transform of the following function

Question

$f(t)= t^{10}\cosh(t)$

Is there another method that doesn't include differentiating the Laplace transform of $\cosh t$ ten times?

Answer 1

Yes. Remember that $\cosh(t) = \dfrac{e^t + e^{-t}}2$. Use this to rewrite $f(t)$ and use a relatively simple Laplace transform for functions of the form $t^n e^{at}$.

Answer 2

Note that

$$\cosh(t) = \frac{e^t + e^{-t}}2$$

Thus, applying linearity and this identity,

$$\mathcal L \{ t^{10}\cosh(t) \} = \frac 1 2 \mathcal L \{ t^{10} e^t \} + \frac 1 2 \mathcal L \{ t^{10}e^{-t} \}$$

Taking the derivatives here should be much easier. Other identities could also prove helpful, such as these, for even easier alternatives. (Note: $n\in \Bbb Z^+$ and $F = \mathcal L \{ f \}$.)

• $\mathcal L \{ t^n e^{at} \} = n!/(s-a)^{n+1}$
• $\mathcal L \{ e^{ct} f(t) \} = F(s-c)$

These are identities $(23),(27)$ on this page I have bookmarked.

Answer 3

$$\begin{align}f(t)&=t^{10}\cosh t\\&=t^{10}\left(\dfrac{e^t+e^{-t}}2\right)\\&=\dfrac12t^{10}e^t+\dfrac12t^{10}e^{-t}\\\mathcal{L}(f(t))&=\dfrac12\cdot\dfrac{10!}{(s-1)^{11}}+\dfrac12\cdot\dfrac{10!}{(s+1)^{11}}\end{align}$$