Laplace Transform involving 1/x

Question

I'm encountering a problem in which the following identity is used

$\frac{1}{x} = \int_0^\infty \exp(-xt) \mathrm{d}t$, where $x$ is real and positive,

which is derived "with a Laplace transform". However, I cannot see how this works out. The Laplace transform of 1/x is

$F(s) = \int_0^\infty \frac{1}{t} \exp(-st) \mathrm{d}t$,

which is a divergent integral. Then, how do you obtain the equation above?

Answer 1

You can compute the integral directly... No need to consider the Laplace transform.

$$ \int_0^{\infty} e^{-tx} dt = \left[-\frac{1}{x} e^{-tx}\right]_{t=0}^{\infty} = \frac 1x, \quad x>0 $$