Mixing
Integral of Laplace transform
$begingroup$
Consider the following theorem:
Let $u: mathbb{R}rightarrow mathbb{C}$, $uin T$, the set of Laplace-transformable functions, and $v in T$ as well, $v(t)=u(t)/t$. Then $int^{infty}_{s}mathcal{L}{u}(sigma)dsigma=mathcal{L}{v}(s)$ 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 $u: mathbb{R}rightarrow mathbb{C}$, $u, vin T$, $v(t)=u(t)/t$. 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 $sigmain mathbb{R}$, right?
Thanks in advance!
complex-analysis proof-verification laplace-transform
asked Jan 11 at 12:35
LeonardoLeonardo
3449
$endgroup$
add a comment |
$begingroup$
Consider the following theorem:
Let $u: mathbb{R}rightarrow mathbb{C}$, $uin T$, the set of Laplace-transformable functions, and $v in T$ as well, $v(t)=u(t)/t$. Then $int^{infty}_{s}mathcal{L}{u}(sigma)dsigma=mathcal{L}{v}(s)$ 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 $u: mathbb{R}rightarrow mathbb{C}$, $u, vin T$, $v(t)=u(t)/t$. 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 $sigmain mathbb{R}$, right?
Thanks in advance!
complex-analysis proof-verification laplace-transform
asked Jan 11 at 12:35
LeonardoLeonardo
3449
$endgroup$
$begingroup$
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). $U(s) =lim_{T toinfty} U_T(s),V(s) = lim_{T toinfty} V_T(s)$. If $lim_{T toinfty}U_T(s_0),lim_{T toinfty}V_T(s_0)$ 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
add a comment |
$begingroup$
Consider the following theorem:
Let $u: mathbb{R}rightarrow mathbb{C}$, $uin T$, the set of Laplace-transformable functions, and $v in T$ as well, $v(t)=u(t)/t$. Then $int^{infty}_{s}mathcal{L}{u}(sigma)dsigma=mathcal{L}{v}(s)$ 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 $u: mathbb{R}rightarrow mathbb{C}$, $u, vin T$, $v(t)=u(t)/t$. 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 $sigmain mathbb{R}$, right?
Thanks in advance!
complex-analysis proof-verification laplace-transform
asked Jan 11 at 12:35
LeonardoLeonardo
3449
$endgroup$
Consider the following theorem:
Let $u: mathbb{R}rightarrow mathbb{C}$, $uin T$, the set of Laplace-transformable functions, and $v in T$ as well, $v(t)=u(t)/t$. Then $int^{infty}_{s}mathcal{L}{u}(sigma)dsigma=mathcal{L}{v}(s)$ 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 $u: mathbb{R}rightarrow mathbb{C}$, $u, vin T$, $v(t)=u(t)/t$. 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 $sigmain mathbb{R}$, right?
Thanks in advance!
complex-analysis proof-verification laplace-transform
complex-analysis proof-verification laplace-transform
asked Jan 11 at 12:35
LeonardoLeonardo
3449
asked Jan 11 at 12:35
LeonardoLeonardo
3449
asked Jan 11 at 12:35
LeonardoLeonardo
3449
asked Jan 11 at 12:35
LeonardoLeonardo
3449
asked Jan 11 at 12:35
LeonardoLeonardo
3449
3449
$begingroup$
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). $U(s) =lim_{T toinfty} U_T(s),V(s) = lim_{T toinfty} V_T(s)$. If $lim_{T toinfty}U_T(s_0),lim_{T toinfty}V_T(s_0)$ 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
add a comment |
$begingroup$
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). $U(s) =lim_{T toinfty} U_T(s),V(s) = lim_{T toinfty} V_T(s)$. If $lim_{T toinfty}U_T(s_0),lim_{T toinfty}V_T(s_0)$ 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
$begingroup$
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). $U(s) =lim_{T toinfty} U_T(s),V(s) = lim_{T toinfty} V_T(s)$. If $lim_{T toinfty}U_T(s_0),lim_{T toinfty}V_T(s_0)$ 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
$begingroup$
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). $U(s) =lim_{T toinfty} U_T(s),V(s) = lim_{T toinfty} V_T(s)$. If $lim_{T toinfty}U_T(s_0),lim_{T toinfty}V_T(s_0)$ 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
add a comment |
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$begingroup$
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). $U(s) =lim_{T toinfty} U_T(s),V(s) = lim_{T toinfty} V_T(s)$. If $lim_{T toinfty}U_T(s_0),lim_{T toinfty}V_T(s_0)$ 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