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The Fourier transform for tempered distributions is well-known. It's defined by $$\langle \mathcal{F}T , \phi\rangle = \langle T,\mathcal{F}\phi\rangle$$For any Schwartz function $\phi$. For ordinary functions, it's defined by $$\mathcal{F}f(s) = \int_{-\infty}^{+\infty}e^{-2\pi ist}f(t)dt$$On the other hand, the unilateral Laplace transform for the ordinary functions is $$\mathcal{L}f(s) = \int_{0^{-}}^{+\infty}e^{-st}f(t)dt$$Where $s \in \mathbb{C}$. Is it possible to take Laplace transform of distributions? How is it defined, then? It's known that $\mathcal{L}\delta(t) = 1$ but I don't know if it's rigorous since $\delta(t)$ is not an ordinary function.

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    $\begingroup$ @downvoter Why is my question inappropriate? $\endgroup$
    – S.H.W
    Sep 30 '20 at 20:28
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    $\begingroup$ Question seems right to me, I don't know why it has downvotes without an comment for the downvote-reason. Uptoving $\endgroup$ Sep 30 '20 at 21:16
  • $\begingroup$ My friend, downvoters never waste their time with anything helpfull like leaving a comment, they are selfish and arrogant, they never will leave a comment $\endgroup$ Sep 30 '20 at 21:17
  • $\begingroup$ @LuisFelipe Unfortunately they don't leave a comment. I appreciate your support. $\endgroup$
    – S.H.W
    Oct 1 '20 at 9:41
  • $\begingroup$ The eight chapter of the book "Théorie des distributions" of Laurent Schwartz is entirely devoted to the Laplace transform. It's in French, though. I don't think it has ever been translated, curiously. $\endgroup$ Oct 1 '20 at 11:24
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If $T$ is a distribution with compact support then $\langle T(t), e^{-st} \rangle$ is well-defined. One can take $\rho \in C_c^\infty$ such that $\rho \equiv 1$ on a neighborhood of the support of $T$ and define $\langle T(t), e^{-st} \rangle = \langle T(t), \rho(t) e^{-st} \rangle$. The result doesn't depend on the choice of $\rho$.

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  • $\begingroup$ Thanks. Would you elaborate more, please? What do you mean by $\langle T(t), e^{-st} \rangle$? $\endgroup$
    – S.H.W
    Oct 1 '20 at 9:44
  • $\begingroup$ $\langle T(t), e^{-st} \rangle$ denotes $T$ applied to the function $t \mapsto e^{-st}$. For example, $\langle \delta(t), e^{-st} \rangle = e^{-s\cdot 0} = 1.$ $\endgroup$
    – md2perpe
    Oct 1 '20 at 10:32
  • $\begingroup$ So you are saying that if $T$ is a distribution with compact support the pairing is well-defined, otherwise we define $\langle T(t), \rho(t) e^{-st} \rangle$, right? Is this the way the Laplace transform is defined in the books? Also how have we taken into account the interval of integration which is from $0$ to $+\infty$ instead of $-\infty$ to $+\infty$? $\endgroup$
    – S.H.W
    Oct 1 '20 at 13:11
  • $\begingroup$ Not exactly. An ordinary distribution has a defined action only on compactly supported smooth functions. But if the distribution has compact support, we can extend the action to non-compactly supported smooth function by multiplying the test function with such a $\rho$. Since two such choices of $\rho$ only differ outside of the support of the distribution, this extension is well-defined. $\endgroup$
    – md2perpe
    Oct 1 '20 at 13:25

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