Laplace transform, bromwich contour - basic question
$begingroup$
I am confused about the following two observations which seem
contradictory:
It is stated that the region of convergence of the Laplace transform
is a half space. That is $L(s)$ is defined for all $s$ with $Re(s)>c$, while
undefined for all s with $Re(s)<c$.In the discussion of the inverse laplace transform, it is stated that the
integral
$$frac{1}{2pi i}int_{c-iinfty}^{c+iinfty}e^{st}L(s)mathrm{d}s$$
can be evaluated by closing the "bromwich contour" (the line of integration) via a
semicircle to the left of $c$ and then letting the radius tend to infinity.
(Afterwards using residue theorem)
Now here is my problem: In order for 2.) to be feasible it is necessary that the integrand $e^{st}L(s)$ is defined
on the whole complex plain (because if the radius of the semicircle becomes larger, then eventually every point - even
those points with $Re(s)<<c$ will be enclosed by the circle.). But the Laplace transform L(s) is only defined for $Re(s)>c$.
Hence, how is it possible to use 2.) when $L(s)$ is not defined in some parts of the complex plane?
Can anyone shed some light on this?
laplace-transform contour-integration
asked Dec 3 '18 at 15:04
P.JoP.Jo
61
$endgroup$
add a comment |
$begingroup$
I am confused about the following two observations which seem
contradictory:
It is stated that the region of convergence of the Laplace transform
is a half space. That is $L(s)$ is defined for all $s$ with $Re(s)>c$, while
undefined for all s with $Re(s)<c$.In the discussion of the inverse laplace transform, it is stated that the
integral
$$frac{1}{2pi i}int_{c-iinfty}^{c+iinfty}e^{st}L(s)mathrm{d}s$$
can be evaluated by closing the "bromwich contour" (the line of integration) via a
semicircle to the left of $c$ and then letting the radius tend to infinity.
(Afterwards using residue theorem)
Now here is my problem: In order for 2.) to be feasible it is necessary that the integrand $e^{st}L(s)$ is defined
on the whole complex plain (because if the radius of the semicircle becomes larger, then eventually every point - even
those points with $Re(s)<<c$ will be enclosed by the circle.). But the Laplace transform L(s) is only defined for $Re(s)>c$.
Hence, how is it possible to use 2.) when $L(s)$ is not defined in some parts of the complex plane?
Can anyone shed some light on this?
laplace-transform contour-integration
asked Dec 3 '18 at 15:04
P.JoP.Jo
61
$endgroup$
$begingroup$
See this answer.
$endgroup$
– Maxim
Dec 6 '18 at 15:59
add a comment |
$begingroup$
I am confused about the following two observations which seem
contradictory:
It is stated that the region of convergence of the Laplace transform
is a half space. That is $L(s)$ is defined for all $s$ with $Re(s)>c$, while
undefined for all s with $Re(s)<c$.In the discussion of the inverse laplace transform, it is stated that the
integral
$$frac{1}{2pi i}int_{c-iinfty}^{c+iinfty}e^{st}L(s)mathrm{d}s$$
can be evaluated by closing the "bromwich contour" (the line of integration) via a
semicircle to the left of $c$ and then letting the radius tend to infinity.
(Afterwards using residue theorem)
Now here is my problem: In order for 2.) to be feasible it is necessary that the integrand $e^{st}L(s)$ is defined
on the whole complex plain (because if the radius of the semicircle becomes larger, then eventually every point - even
those points with $Re(s)<<c$ will be enclosed by the circle.). But the Laplace transform L(s) is only defined for $Re(s)>c$.
Hence, how is it possible to use 2.) when $L(s)$ is not defined in some parts of the complex plane?
Can anyone shed some light on this?
laplace-transform contour-integration
asked Dec 3 '18 at 15:04
P.JoP.Jo
61
$endgroup$
I am confused about the following two observations which seem
contradictory:
It is stated that the region of convergence of the Laplace transform
is a half space. That is $L(s)$ is defined for all $s$ with $Re(s)>c$, while
undefined for all s with $Re(s)<c$.In the discussion of the inverse laplace transform, it is stated that the
integral
$$frac{1}{2pi i}int_{c-iinfty}^{c+iinfty}e^{st}L(s)mathrm{d}s$$
can be evaluated by closing the "bromwich contour" (the line of integration) via a
semicircle to the left of $c$ and then letting the radius tend to infinity.
(Afterwards using residue theorem)
Now here is my problem: In order for 2.) to be feasible it is necessary that the integrand $e^{st}L(s)$ is defined
on the whole complex plain (because if the radius of the semicircle becomes larger, then eventually every point - even
those points with $Re(s)<<c$ will be enclosed by the circle.). But the Laplace transform L(s) is only defined for $Re(s)>c$.
Hence, how is it possible to use 2.) when $L(s)$ is not defined in some parts of the complex plane?
Can anyone shed some light on this?
laplace-transform contour-integration
laplace-transform contour-integration
asked Dec 3 '18 at 15:04
P.JoP.Jo
61
asked Dec 3 '18 at 15:04
P.JoP.Jo
61
asked Dec 3 '18 at 15:04
P.JoP.Jo
61
asked Dec 3 '18 at 15:04
P.JoP.Jo
61
asked Dec 3 '18 at 15:04
P.JoP.Jo
61
61
$begingroup$
See this answer.
$endgroup$
– Maxim
Dec 6 '18 at 15:59
add a comment |
$begingroup$
See this answer.
$endgroup$
– Maxim
Dec 6 '18 at 15:59
$begingroup$
See this answer.
$endgroup$
– Maxim
Dec 6 '18 at 15:59
$begingroup$
See this answer.
$endgroup$
– Maxim
Dec 6 '18 at 15:59
add a comment |
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$begingroup$
See this answer.
$endgroup$
– Maxim
Dec 6 '18 at 15:59