Inverse Laplace Transform of $\frac{1}{z^4 + 1}$

Question

How would I find the Inverse Laplace Transform of $F(z) = \frac{1}{z^4 + 1}$?
The given formula for the inverse laplace transform is $$f(t) = \lim_{R\rightarrow\infty}\frac{1}{2\pi i}\int_{\lambda}F(z)e^{zt}dz$$
where $\lambda$ is the line segment from $\sigma-iR$ to $\sigma+iR$, where $\sigma\in[A,\infty)$, and $A\in\mathbb{R}$ is the constant in which
$$|f(t)| \leq Ce^{At}$$ for some constant $C$ and for all $t>0$.

I'm not sure if there's a nice way to compute this, so I would appreciate any help.

Answer 1

When we've done these problems, we used a table of known Laplace transforms to obtain the inverse. Like this one http://tutorial.math.lamar.edu/Classes/DE/Laplace_Table.aspx.

That being said, I'd try using partial fraction decomposition to break down the expression into expressions that appear on the table.

$\frac{1}{z^4+1}=\frac{1}{(z^2+i)(z^2-i)}=...$

That should break the original expression into a summation of functions of which you know what the inverse Laplace transform is. Moreover properties of the Laplace transform allow you to take the inverse of each term separately to obtain the result.

Answer 2

By using $$\mathcal{L}[\sinh(at)] = \frac{a}{s^2-a^2}$$

We can find the original function can be substituted by $s\to z^2, a\to i$

and then $\frac{1}{z^4+1}$ turns to $-i\sinh(it)$