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?

Thanks in advance!

  • $\begingroup$ 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 $V(s)=\int_s^\infty U(z)dz$ $\endgroup$ – reuns Jan 11 at 15:04

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