Integral of Laplace transform

Consider the following theorem:

Let $$u: \mathbb{R}\rightarrow \mathbb{C}$$, $$u\in T$$, the set of Laplace-transformable functions, and $$v \in T$$ as well, $$v(t)=u(t)/t$$. Then $$\int^{\infty}_{s}\mathcal{L}\{u\}(\sigma)d\sigma=\mathcal{L}\{v\}(s)$$ for $$s \in \mathbb{R}$$, $$s>\lambda_u$$, the abscissa of convergence of $$u$$.

One has, given $$s \in \mathbb{C}$$ with $$\text{Re}(s)>\lambda_u$$:

$$\mathcal{L}\{u\}(s)=\mathcal{L}\{t$$ $$u(t)/t\}(s)=$${derivative of Laplace transform}$$=-\frac{d}{ds}\mathcal{L}\{u(t)/t\}(s)=-\frac{d}{ds}\mathcal{L}\{v\}(s)$$.

This shows that $$-\mathcal{L}\{v\}$$ is a primitive for $$\mathcal{L}\{u\}$$ and therefore, by the complex version of the fundamental theorem of calculus, and for a sufficiently smooth curve $$\Gamma: [a,b] \rightarrow \mathbb{C}$$:

$$\int_{\Gamma}\mathcal{L}\{u\}(z)dz= \mathcal{L}\{v\}(\Gamma(a)) - \mathcal{L}\{v\}(\Gamma(b))$$ where $$\Gamma(a), \Gamma(b)$$ must of course fall in the domain of $$\mathcal{L}\{v\}$$. Choosing $$\Gamma(a)=s$$ (the same $$s$$ as above), and letting $$\text{Re}({\Gamma(b)})\rightarrow + \infty$$ one concludes (with a small change in notation for the integral) the following complex version of the theorem above:

Let $$u: \mathbb{R}\rightarrow \mathbb{C}$$, $$u, v\in T$$, $$v(t)=u(t)/t$$. Then $$\int^{\infty}_{s}\mathcal{L}\{u\}(z)dz=\mathcal{L}\{v\}(s)$$ for $$s \in \mathbb{C}$$, $$\text{Re}(s)>\lambda_u$$.

The argument still needs a proof of the existence of the integral $$\int^{\infty}_{s}\mathcal{L}\{u\}(z)dz$$, doesn't it?

However, is my reasoning correct or should I settle with the original version of the theorem?

Furthermore, in the original version $$\sigma\in \mathbb{R}$$, right?

• Let $U_T(s)=\int_0^Tu(t)e^{-st}dt,V_T(s)=\int_0^T \frac{u(t)}{t} e^{-st}dt$ assumed to converge absolutely (condition A). $U(s) =\lim_{T \to\infty} U_T(s),V(s) = \lim_{T \to\infty} V_T(s)$. If $\lim_{T \to\infty}U_T(s_0),\lim_{T \to\infty}V_T(s_0)$ converge (B) then integrating by parts $U(s_0+s) = \int_0^\infty U_t(s_0)s e^{-s t}dt,V(s_0+s) = \int_0^\infty U_t(s_0)(\frac{se^{-st}}{t}+\frac{e^{-st}}{t^2}) dt$ (no problem in $t=0$ by (A)) both converge absolutely for $\Re(s) > 0$ and $V'(s_0+s)=-U(s_0+s)$. Since $\lim_{\Re(s)\to \infty}V(s)= 0$ then $V(s)=\int_s^\infty U(z)dz$ – reuns Jan 11 at 15:04