Inverse Laplace Transform properities

Question

If you have the inverse Laplace transform of a function of the form $F(\omega)/\omega$ , how to find the inverse Laplace transform of the function $F(\omega)/w$? I have attached the formula for the inverse laplace transform of the $((1/\omega)\cdot F(\omega))$ and mu function $F(\omega)$.

general formula

My function

Answer 1

Hello I hope I am not too late. In this answer I'll give you an example how to calculate $\mathcal{L}^{-1} \{F(s)\}(t)$ as you requested in the comments.

Definition

An integral formula for the inverse Laplace transform, is given by the line-integral$$f(t) = \mathcal{L}^{-1} \{F(s)\}(t) = \frac{1}{2\pi i}\lim_{T\to\infty}\int\limits_{\gamma-iT}^{\gamma+iT}e^{st}F(s)\,\mathrm{d}s,$$ where the integration is done along the vertical line $Re(s) = γ$ in the complex plane such that γ is greater than the real part of all singularities of $F(s)$ and $F(s)$ is bounded on the line, for example if contour path is in the region of convergence.

Example Let $F(\omega)=O(\lvert\omega\rvert^{-1})$ be entire. Evaluate $\mathcal{L}^{-1} \left\{\frac{F(\omega)}{\omega}\right\}(t).$

Proof. We will evaluate the integral by expressing it as a limit of contour integrals along the contour $\mathcal{C}$ that goes along the line from $\gamma−iT$ to $\gamma+iT$ and then counterclockwise along a semicircle $\Gamma$ centered at the origin from $\gamma+iT$ to $\gamma−iT$. Take $T$ to be greater than $\gamma$, so that $0$ is enclosed within the curve. The contour integral is $$\oint\limits_{\mathcal{C}}\frac{F(\omega)}{\omega}e^{\omega t}\,\mathrm{d}\omega$$ Since $F(\omega)e^{\omega t}$ is an entire function (having no singularities at any point in the complex plane), the integrand has singularities only where the denominator $\omega$ is zero. According to the residue theorem, then, we have $$\oint\limits_{\mathcal{C}}\frac{F(\omega)}{\omega}e^{\omega t}\,\mathrm{d}\omega=2\pi i\operatorname{Res}_{\omega=0}\left[\frac{F(\omega)}{\omega}e^{\omega t}\right]=2\pi i\lim_{\omega\to0}\left[\omega\cdot\frac{F(\omega)}{\omega}e^{\omega t}\right]=2\pi i F(0).$$ The contour $\mathcal{C}$ may be split into a "straight" part and a curved arc, so that $$\int\limits_{\gamma-iT}^{\gamma+iT}\frac{F(\omega)}{\omega}e^{\omega t}\,\mathrm{d}s+\int\limits_{\Gamma}\frac{F(\omega)}{\omega}e^{\omega t}\,\mathrm{d}\omega=2\pi iF(0) \Rightarrow \int\limits_{\gamma-iT}^{\gamma+iT}\frac{F(\omega)}{\omega}e^{\omega t}\,\mathrm{d}s=2\pi iF(0)-\int\limits_{\Gamma}\frac{F(\omega)}{\omega}e^{\omega t}\,\mathrm{d}\omega.$$ Now we can show that the integral over $\Gamma$ equals zero: $$\left\lvert\int\limits_{\Gamma}\frac{F(\omega)}{\omega}e^{\omega t}\,\mathrm{d}\omega\right\rvert = \int\limits_{\pi/2}^{3\pi/2}\frac{\lvert F(Te^{i\varphi})\rvert}{T}e^{T\cos(\varphi)t}\,\mathrm{d}\varphi\to0\qquad (T\to\infty).$$ This is because $\cos\varphi<0, \,\varphi\in(\pi/2,3\pi/2)$. In conclusion we have $$\lim_{T\to\infty}\frac{1}{2\pi i}\int\limits_{\gamma-iT}^{\gamma+iT}\frac{F(\omega)}{\omega}e^{\omega t}\,\mathrm{d}s=\mathcal{L}^{-1} \left\{\frac{F(\omega)}{\omega}\right\}(t)=F(0).$$