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What I've tried so far (for the first one):

$$\int^\infty_0e^{-sa} \int_0^t f(u)duda $$

$$= \int^\infty_0 \int_0^t e^{-sa} f(u)duda$$ $$= \int^t_0 \int_0^\infty e^{-sa} f(u)dadu$$

And now the inner integral looks like Laplace transform of $f(u)$ subject to the outer integral. I'm not sure if this is at all the right track, and if it is, I'm not sure how to proceed. Thanks in advance for any help.

  • $\begingroup$ The $t$ in the integral above should be replaced by $a$. $\endgroup$ – copper.hat Sep 11 '16 at 3:44
  • $\begingroup$ The second integral is straightforward to evaluate using a change of variables $\tau = at$. $\endgroup$ – copper.hat Sep 11 '16 at 4:12

Here is one way of evaluating the first transform:

let $g(t) = \int_0^t f(u) du$, then we have $\dot{g}(t) = f(t)$, with $g(0) = 0$, and taking the transform of both sides gives $s \hat{g}(s) - g(0) = \hat{f}(s)$, or, more simply, $\hat{g}(s) = {1 \over s} \hat{f}(s)$.

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  • $\begingroup$ That works! Thank you! $\endgroup$ – Damon Williams Sep 11 '16 at 9:24

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