Prove the inverse transform of unilateral Laplace transform












0












$begingroup$


I'm reading this article and having a question.



Consider a function $f$ and its Laplace transform



$hspace{3.0cm} F(s) = int_0^infty f(t) e^{-st} dt$, with ${s|text{Re}(s) = 0} in text{ROC}[F(s)]$



The inverse Laplace transform would be $f(t) = lim_{omega to infty} frac{1}{2pi i}int_{sigma - i omega}^{sigma + i omega} F(s) e^{st} ds$.



Now, consider the case where $text{Re}(s) = 0$, we can see that



$hspace{3.0cm} f(t) = lim_{omega to infty} frac{1}{2pi i}int_{- i omega}^{i omega} F(s) e^{st} ds = frac{1}{2pi}int_{-infty}^infty F(i omega) e^{i omega t} domega$



is an inverse Fourier transform.



According to Fourier transform, we know that



$hspace{3.0cm} F(i omega) = cal{F}{f(t)} = int_{-infty}^infty f(t) e^{-iomega t} dt$



But the Laplace transform of $f$, when $text{Re}(s) = 0$, suggests that



$hspace{3.0cm} F(i omega) = int_0^infty f(t) e^{-i omega t} dt$



It turns out that $int_{-infty}^infty f(t) e^{-i omega t} dt = int_0^infty f(t) e^{-i omega t} dt$ ? This seems wrong to me ? Am I getting wrong somewhere ?






fourier-analysis laplace-transform fourier-transform inverselaplace







share|cite|improve this question











$endgroup$












  • $begingroup$
    The answer lies in the fact that your function $f$, if you want to define it on $(-infty,0]$, would be $0$ there. Then everything works. By the way, I edited your question to include factors of $2pi$ that you were missing. However, you are stilling missing some in the second part of the answer.
    $endgroup$
    – DisintegratingByParts
    Jan 17 at 16:03












  • $begingroup$
    If you define $f$ in that way, $f(t) = f(t) u(t)$ where $u(t)$ is the heavy-side function. But this assumption isn't stated in the definition of unilateral Laplace transform.
    $endgroup$
    – HOANG GIANG
    Jan 18 at 7:18












  • $begingroup$
    As written in some text, if $f(t) = f(t) u(t)$ then unilateral transform is the same as bilateral transform. But in the other case, unilateral transform is said to be different from bilateral transform
    $endgroup$
    – HOANG GIANG
    Jan 18 at 7:21










  • $begingroup$
    In your case, you started with $f(t)=f(t)u(t)$ and, in that case, the Laplace transform is a holomorphic function on a right half-plane. Then your inverse Laplace transform is performed on that holomorphic function, with the result being a function $f$ such that $f(t)=f(t)u(t)$. If you want to start with a different $f$, then the issues are different.
    $endgroup$
    – DisintegratingByParts
    Jan 19 at 21:49
















0












$begingroup$


I'm reading this article and having a question.



Consider a function $f$ and its Laplace transform



$hspace{3.0cm} F(s) = int_0^infty f(t) e^{-st} dt$, with ${s|text{Re}(s) = 0} in text{ROC}[F(s)]$



The inverse Laplace transform would be $f(t) = lim_{omega to infty} frac{1}{2pi i}int_{sigma - i omega}^{sigma + i omega} F(s) e^{st} ds$.



Now, consider the case where $text{Re}(s) = 0$, we can see that



$hspace{3.0cm} f(t) = lim_{omega to infty} frac{1}{2pi i}int_{- i omega}^{i omega} F(s) e^{st} ds = frac{1}{2pi}int_{-infty}^infty F(i omega) e^{i omega t} domega$



is an inverse Fourier transform.



According to Fourier transform, we know that



$hspace{3.0cm} F(i omega) = cal{F}{f(t)} = int_{-infty}^infty f(t) e^{-iomega t} dt$



But the Laplace transform of $f$, when $text{Re}(s) = 0$, suggests that



$hspace{3.0cm} F(i omega) = int_0^infty f(t) e^{-i omega t} dt$



It turns out that $int_{-infty}^infty f(t) e^{-i omega t} dt = int_0^infty f(t) e^{-i omega t} dt$ ? This seems wrong to me ? Am I getting wrong somewhere ?






fourier-analysis laplace-transform fourier-transform inverselaplace







share|cite|improve this question











$endgroup$












  • $begingroup$
    The answer lies in the fact that your function $f$, if you want to define it on $(-infty,0]$, would be $0$ there. Then everything works. By the way, I edited your question to include factors of $2pi$ that you were missing. However, you are stilling missing some in the second part of the answer.
    $endgroup$
    – DisintegratingByParts
    Jan 17 at 16:03












  • $begingroup$
    If you define $f$ in that way, $f(t) = f(t) u(t)$ where $u(t)$ is the heavy-side function. But this assumption isn't stated in the definition of unilateral Laplace transform.
    $endgroup$
    – HOANG GIANG
    Jan 18 at 7:18












  • $begingroup$
    As written in some text, if $f(t) = f(t) u(t)$ then unilateral transform is the same as bilateral transform. But in the other case, unilateral transform is said to be different from bilateral transform
    $endgroup$
    – HOANG GIANG
    Jan 18 at 7:21










  • $begingroup$
    In your case, you started with $f(t)=f(t)u(t)$ and, in that case, the Laplace transform is a holomorphic function on a right half-plane. Then your inverse Laplace transform is performed on that holomorphic function, with the result being a function $f$ such that $f(t)=f(t)u(t)$. If you want to start with a different $f$, then the issues are different.
    $endgroup$
    – DisintegratingByParts
    Jan 19 at 21:49














0












0








0





$begingroup$


I'm reading this article and having a question.



Consider a function $f$ and its Laplace transform



$hspace{3.0cm} F(s) = int_0^infty f(t) e^{-st} dt$, with ${s|text{Re}(s) = 0} in text{ROC}[F(s)]$



The inverse Laplace transform would be $f(t) = lim_{omega to infty} frac{1}{2pi i}int_{sigma - i omega}^{sigma + i omega} F(s) e^{st} ds$.



Now, consider the case where $text{Re}(s) = 0$, we can see that



$hspace{3.0cm} f(t) = lim_{omega to infty} frac{1}{2pi i}int_{- i omega}^{i omega} F(s) e^{st} ds = frac{1}{2pi}int_{-infty}^infty F(i omega) e^{i omega t} domega$



is an inverse Fourier transform.



According to Fourier transform, we know that



$hspace{3.0cm} F(i omega) = cal{F}{f(t)} = int_{-infty}^infty f(t) e^{-iomega t} dt$



But the Laplace transform of $f$, when $text{Re}(s) = 0$, suggests that



$hspace{3.0cm} F(i omega) = int_0^infty f(t) e^{-i omega t} dt$



It turns out that $int_{-infty}^infty f(t) e^{-i omega t} dt = int_0^infty f(t) e^{-i omega t} dt$ ? This seems wrong to me ? Am I getting wrong somewhere ?






fourier-analysis laplace-transform fourier-transform inverselaplace







share|cite|improve this question











$endgroup$




I'm reading this article and having a question.



Consider a function $f$ and its Laplace transform



$hspace{3.0cm} F(s) = int_0^infty f(t) e^{-st} dt$, with ${s|text{Re}(s) = 0} in text{ROC}[F(s)]$



The inverse Laplace transform would be $f(t) = lim_{omega to infty} frac{1}{2pi i}int_{sigma - i omega}^{sigma + i omega} F(s) e^{st} ds$.



Now, consider the case where $text{Re}(s) = 0$, we can see that



$hspace{3.0cm} f(t) = lim_{omega to infty} frac{1}{2pi i}int_{- i omega}^{i omega} F(s) e^{st} ds = frac{1}{2pi}int_{-infty}^infty F(i omega) e^{i omega t} domega$



is an inverse Fourier transform.



According to Fourier transform, we know that



$hspace{3.0cm} F(i omega) = cal{F}{f(t)} = int_{-infty}^infty f(t) e^{-iomega t} dt$



But the Laplace transform of $f$, when $text{Re}(s) = 0$, suggests that



$hspace{3.0cm} F(i omega) = int_0^infty f(t) e^{-i omega t} dt$



It turns out that $int_{-infty}^infty f(t) e^{-i omega t} dt = int_0^infty f(t) e^{-i omega t} dt$ ? This seems wrong to me ? Am I getting wrong somewhere ?






fourier-analysis laplace-transform fourier-transform inverselaplace




fourier-analysis laplace-transform fourier-transform inverselaplace






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 17 at 16:04









DisintegratingByParts

60.4k42681




60.4k42681










asked Jan 17 at 6:57









HOANG GIANGHOANG GIANG

175




175












  • $begingroup$
    The answer lies in the fact that your function $f$, if you want to define it on $(-infty,0]$, would be $0$ there. Then everything works. By the way, I edited your question to include factors of $2pi$ that you were missing. However, you are stilling missing some in the second part of the answer.
    $endgroup$
    – DisintegratingByParts
    Jan 17 at 16:03












  • $begingroup$
    If you define $f$ in that way, $f(t) = f(t) u(t)$ where $u(t)$ is the heavy-side function. But this assumption isn't stated in the definition of unilateral Laplace transform.
    $endgroup$
    – HOANG GIANG
    Jan 18 at 7:18












  • $begingroup$
    As written in some text, if $f(t) = f(t) u(t)$ then unilateral transform is the same as bilateral transform. But in the other case, unilateral transform is said to be different from bilateral transform
    $endgroup$
    – HOANG GIANG
    Jan 18 at 7:21










  • $begingroup$
    In your case, you started with $f(t)=f(t)u(t)$ and, in that case, the Laplace transform is a holomorphic function on a right half-plane. Then your inverse Laplace transform is performed on that holomorphic function, with the result being a function $f$ such that $f(t)=f(t)u(t)$. If you want to start with a different $f$, then the issues are different.
    $endgroup$
    – DisintegratingByParts
    Jan 19 at 21:49


















  • $begingroup$
    The answer lies in the fact that your function $f$, if you want to define it on $(-infty,0]$, would be $0$ there. Then everything works. By the way, I edited your question to include factors of $2pi$ that you were missing. However, you are stilling missing some in the second part of the answer.
    $endgroup$
    – DisintegratingByParts
    Jan 17 at 16:03












  • $begingroup$
    If you define $f$ in that way, $f(t) = f(t) u(t)$ where $u(t)$ is the heavy-side function. But this assumption isn't stated in the definition of unilateral Laplace transform.
    $endgroup$
    – HOANG GIANG
    Jan 18 at 7:18












  • $begingroup$
    As written in some text, if $f(t) = f(t) u(t)$ then unilateral transform is the same as bilateral transform. But in the other case, unilateral transform is said to be different from bilateral transform
    $endgroup$
    – HOANG GIANG
    Jan 18 at 7:21










  • $begingroup$
    In your case, you started with $f(t)=f(t)u(t)$ and, in that case, the Laplace transform is a holomorphic function on a right half-plane. Then your inverse Laplace transform is performed on that holomorphic function, with the result being a function $f$ such that $f(t)=f(t)u(t)$. If you want to start with a different $f$, then the issues are different.
    $endgroup$
    – DisintegratingByParts
    Jan 19 at 21:49
















$begingroup$
The answer lies in the fact that your function $f$, if you want to define it on $(-infty,0]$, would be $0$ there. Then everything works. By the way, I edited your question to include factors of $2pi$ that you were missing. However, you are stilling missing some in the second part of the answer.
$endgroup$
– DisintegratingByParts
Jan 17 at 16:03






$begingroup$
The answer lies in the fact that your function $f$, if you want to define it on $(-infty,0]$, would be $0$ there. Then everything works. By the way, I edited your question to include factors of $2pi$ that you were missing. However, you are stilling missing some in the second part of the answer.
$endgroup$
– DisintegratingByParts
Jan 17 at 16:03














$begingroup$
If you define $f$ in that way, $f(t) = f(t) u(t)$ where $u(t)$ is the heavy-side function. But this assumption isn't stated in the definition of unilateral Laplace transform.
$endgroup$
– HOANG GIANG
Jan 18 at 7:18






$begingroup$
If you define $f$ in that way, $f(t) = f(t) u(t)$ where $u(t)$ is the heavy-side function. But this assumption isn't stated in the definition of unilateral Laplace transform.
$endgroup$
– HOANG GIANG
Jan 18 at 7:18














$begingroup$
As written in some text, if $f(t) = f(t) u(t)$ then unilateral transform is the same as bilateral transform. But in the other case, unilateral transform is said to be different from bilateral transform
$endgroup$
– HOANG GIANG
Jan 18 at 7:21




$begingroup$
As written in some text, if $f(t) = f(t) u(t)$ then unilateral transform is the same as bilateral transform. But in the other case, unilateral transform is said to be different from bilateral transform
$endgroup$
– HOANG GIANG
Jan 18 at 7:21












$begingroup$
In your case, you started with $f(t)=f(t)u(t)$ and, in that case, the Laplace transform is a holomorphic function on a right half-plane. Then your inverse Laplace transform is performed on that holomorphic function, with the result being a function $f$ such that $f(t)=f(t)u(t)$. If you want to start with a different $f$, then the issues are different.
$endgroup$
– DisintegratingByParts
Jan 19 at 21:49




$begingroup$
In your case, you started with $f(t)=f(t)u(t)$ and, in that case, the Laplace transform is a holomorphic function on a right half-plane. Then your inverse Laplace transform is performed on that holomorphic function, with the result being a function $f$ such that $f(t)=f(t)u(t)$. If you want to start with a different $f$, then the issues are different.
$endgroup$
– DisintegratingByParts
Jan 19 at 21:49










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