If a function for s-domain is defined as,
$$\frac{F(s)}{(s^2+\nu^2)}$$
how can I perform inverse laplace transform for above function?
I think it can be transformed into time domain function, but I can't transform.
My attemp was,
$$ \mathscr{L}^{-1}[\frac{F(s)}{(s^2+\nu^2)}]=\frac{1}{2i\nu}(\frac{1}{2\pi i} \lim_{Tt\to\infty} \int^{c+it}_{c-it} \frac{F(s)}{(s-i\nu)} e^{st} ds-\frac{1}{2\pi i} \lim_{Tt\to\infty} \int^{c+it}_{c-it} \frac{F(s)}{(s+i\nu)} e^{st} ds) $$
I tried to go a little further, but I couldn't make it. How do we deal with this?
There is a property of use here, which is the convolution property: $$L^{-1}[F(s)\cdot G(s)]=\int_0^t f(\tau)\,g(t-\tau)\,d\tau.$$ The operation on the RHS is called the convolution. Note here that \begin{align*} L^{-1}[F(s)]&=f(t)\\ L^{-1}[G(s)]&=g(t). \end{align*} So applying this to your scenario would yield the following (setting $G(s)=1/(s^2+\nu^2)$): \begin{align*} L^{-1}\!\left[\frac{F(s)}{s^2+\nu^2}\right] &=L^{-1}\!\left[F(s)\cdot \frac{1}{s^2+\nu^2}\right]. \end{align*} Note that $$L^{-1}\!\left[\frac{1}{s^2+\nu^2}\right]=\frac1\nu\,\sin(\nu t)\,u(t),$$ where $u(t)$ is the unit step function. So the best we can do, not knowing $F(s),$ is that \begin{align*} L^{-1}\!\left[\frac{F(s)}{s^2+\nu^2}\right] &=\frac1\nu\int_0^t f(\tau)\,\sin(\nu(t-\tau))\,u(t-\tau)\,d\tau. \end{align*}