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First let N(t) be a function such that $\int_0^a N(t)dt=0\ \forall a>0$, this is what I call a null function.

I must prove that, for every null function N(t), $\mathcal{L}\{N(t)\}=0$ (supposing that every null function can have a Laplace transform). I tried to prove it using the definition of the Laplace transform and partwise integration, and just want to check if something is wrong.

My attempt:

$\mathcal{L}\{N(t)\}=\lim_{b\to\infty}\int_0^b e^{-st}N(t)\,dt$

If we make $u=e^{-st}, dv=N(t)\,dt$, then $du=-se^{-st}dt, v=\int_0^b N(t)\,dt=0$. So we have:

$\mathcal{L}\{N(t)\}=\lim_{b\to\infty}[(0\cdot u)\mid_0^b - \int_0^b 0\cdot du]=\lim_{b\to\infty}[0]=0$

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  • $\begingroup$ Your proof looks fine, but it is more straightforward to show that $N(t) = 0$ ae. $\endgroup$
    – copper.hat
    Commented Oct 17, 2019 at 23:14

2 Answers 2

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You need more conditions on $N$, such as $N$ being integrable.

If $N$ is integrable, then $\phi(a) = \int_0^a N(t)dt$ is absolutely continuous, and $\phi'(a) = N(a)$ ae. In particular, $\phi'(a) = 0$ for ae. $a$. Hence $N(t)=0$ ae. and so ${\cal L} N = 0$.

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  • $\begingroup$ Why the downvote? $\endgroup$
    – copper.hat
    Commented Oct 17, 2019 at 23:09
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The Laplace transform, either unilateral or bilateral, of $f(t)=0$ is $F(s)=0$, simply because of linearity, by multiplying any known Laplace pair by the scalar $0$.

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