# The Laplace transform of any null function is 0?

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:

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

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

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 $${\cal L} N = 0$$.
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 $$0$$.