Consider the Laplace transform $\tilde{f}(s)$ of a function $f(t)$ defined as $$\tilde{f}(s)=\int_0^\infty f(t)e^{-st} \, dt.$$ Can this relation be inverted to obtain $f(t)$ in terms of $\tilde{f}(s)$?
The reason I ask this is as follows. The Fourier transform $$\tilde{f}(k) = \int_{-\infty}^\infty f(x)e^{-ikx} \, dx$$ is defined simulataneously with the inversion formula $$f(x)=\int_{-\infty}^\infty\tilde{f}(k)e^{ikx} \, dk.$$ But I haven't seen the same for the Laplace's transform.
On
If $\widetilde{f}$ converges on $\gamma+i\mathbb{R}$ with $\gamma\in\mathbb{R}$, the inverse can be obtained as $$f(t)=\frac{1}{2\pi i}\lim_{T\to\infty}\int_{\gamma-iT}^{\gamma+iT}\widetilde{f}(s)e^{st}ds.$$An alternative expression for the inverse Laplace transform is given here.
Yes, the Laplace transform can be inverted. Based on your earlier question I suspect you're actually curious whether the inverse is given by $$f(t)=\int_0^\infty\tilde f(s)e^{-st}\,ds.$$No, that is absolutely not how you invert the Laplace transform.