# Laplace Transform of Integrals

So I know the laplace transform of an integral is the product of the laplace transorm of the functions inside the integral, such as Laplace{integral from 0 to t [tau*e^(t - tau) d tau = Laplace[t] * Laplace[e^t], ignoring the tau

what if I have Laplace{integral from 0 to t [t*e^(-tau) d tau], how do I break it down into 2 laplace functions? The answer is (1/s) * (1/(s + 1)^2), I don't get it. :/

Sorry, I don't know how to type an equation but hopefully you know what I mean. Any help is appreciated.

Orodruin
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Do you mean
$$\mathcal L\left(\int_0^t \tau e^{-\tau} d\tau\right)?$$
In that case you would write this as
$$\mathcal L \left( \int_0^t g(t-\tau) f(\tau) d\tau\right),$$
where ##g(t-\tau) = 1## is the constant function 1 and ##f(\tau)## is the original integrand. By the convolution result you mentioned in the beginning, it follows that
$$\mathcal L\left(\int_0^t \tau e^{-\tau} d\tau\right) = \mathcal L(g) \mathcal L(f) = \mathcal L(1) \mathcal L(t e^{-t})= \frac{1}{s} \frac{1}{(s+1)^2},$$
which is the quoted result.

Note that this can also be shown in a different way for any integrand ##f(\tau)##, namely:
$$\mathcal L\left( \int_0^t f(\tau) d\tau\right) = \int_{t = 0}^\infty \int_{\tau = 0}^t e^{-st} f(\tau) d\tau \, dt.$$
Changing the order of the integrals results in
$$\int_{\tau = 0}^\infty f(\tau) \int_{t = \tau}^\infty e^{-st} dt\, d\tau = \frac{1}{s} \int_{\tau = 0}^\infty f(\tau)\left[-e^{-st}\right]_\tau^\infty d\tau = \frac{1}{s} \int_{\tau = 0}^\infty f(\tau) e^{-s\tau} d\tau,$$
which is equal to ##\mathcal L(f)/s## by definition.