Laplace transform of $\int_t^{+\infty}\,\psi(\tau)\,d\tau$

Question

Why this equality?

$$\mathcal{L}\left(\int_t^{+\infty}\,\psi(\tau)\,d\tau\right)=\frac{1-\psi(s)}{s}$$

Can you help me with the steps?

Answer 1

Denoting $\Psi(t) = \int_t^\infty \psi(\tau) \, d\tau$, we have by definition of $\mathcal L$, that \begin{align*} \mathcal L(\Psi)(s) &= \int_0^\infty \exp(-st)\Psi(t)\, dt\\ &= \int_0^\infty \exp(-st)\int_t^\infty \psi(\tau)\, d\tau\, dt\\ &= \int_0^\infty \int_t^\infty \exp(-st)\psi(\tau)\, d\tau\, dt\\ &= \int_0^\infty \int_0^\tau \exp(-st)\psi(\tau)\, dt\, d\tau\\ &= \int_0^\infty \int_0^\tau\exp(-st)\, dt\cdot \psi(\tau)\, d\tau\\ &= \int_0^\infty \left(\frac 1s -\frac{1}s\exp(-s\tau)\right)\psi(\tau)\, d\tau\\ &=\frac{\int_0^\infty \psi(\tau)\, d\tau - \mathcal L(\psi)(s)}{s} \end{align*} So your equation is true, if $\int_0^\infty \psi(\tau)\, d\tau = 1$ and $\mathcal L(\psi) = \psi$.