Integration with Laplace transform

Integration is related to laplace transform in two ways.

Integration in the Time Domain

If $F(s) = \mathcal{L}\{f(t)\}$ , the Laplace transform of the running integral of $f(t)$ from $0$ to $t$ is:

\mathcal{L}\left\{ \int_{0}^{t} f(\tau) \, d\tau \right\} = \frac{F(s)}{s}

Define an auxiliary function $g(t) = \int_{0}^{t} f(\tau) \, d\tau$ .

By the Fundamental Theorem of Calculus:

Now apply the standard first-derivative rule to $g(t)$ :

\mathcal{L}\{g'(t)\} = s G(s) - g(0)

Substitute $g'(t) = f(t)$ and $g(0) = 0$ :

F(s) = s G(s) - 0 \implies G(s) = \frac{F(s)}{s}

Dividing a time-domain function $f(t)$ by $t$ corresponds to integrating its Laplace transform $F(s)$ from $s$ to $\infty$ :

\mathcal{L}\left\{ \frac{f(t)}{t} \right\} = \int_{s}^{\infty} F(\sigma) \, d\sigma

(This property requires that $\lim_{t \to 0^+} \frac{f(t)}{t}$ exists and is finite.)

Start by expressing $F(\sigma)$ using the unilateral Laplace transform definition:

\int_{s}^{\infty} F(\sigma) \, d\sigma = \int_{s}^{\infty} \left( \int_{0}^{\infty} e^{-\sigma t} f(t) \, dt \right) d\sigma

Assuming Fubini's theorem holds, switch the order of integration:

\int_{s}^{\infty} F(\sigma) \, d\sigma = \int_{0}^{\infty} f(t) \left( \int_{s}^{\infty} e^{-\sigma t} \, d\sigma \right) dt

Evaluate the inner integral with respect to $\sigma$ :

\int_{s}^{\infty} e^{-\sigma t} \, d\sigma = \left[ \frac{e^{-\sigma t}}{-t} \right]_{s}^{\infty} = 0 - \left( \frac{e^{-st}}{-t} \right) = \frac{e^{-st}}{t}

Substitute this result back into the outer integral:

\int_{s}^{\infty} F(\sigma) \, d\sigma = \int_{0}^{\infty} e^{-st} \left( \frac{f(t)}{t} \right) dt = \mathcal{L}\left\{ \frac{f(t)}{t} \right\}