# Laplace transform of convolution with no function of t

Instructions: Evaluate the given Laplace transform. Do not evaluate the integral before transforming.
Problem Given: $\mathscr{L}\{\int_0^t e^{-\tau} cos\tau d\tau \}$
My Problem: To treat this as a convolution, I think I need to rewrite one of the two functions of $\tau$ as a function of $t-\tau$. My intuition is to use the following trig identity, but this could make things nasty come test time (i.e. tomorrow):

$cos{t-\tau} = {costcos\tau - sintsin\tau}$

$cos\tau = \frac{cos{(t-\tau)} - sintsin\tau}{cost}$

Do I just need to tough it out going this route, or am I approaching this poorly?

## 1 Answer

Use the fact that $$\mathscr{L} \left( \int_{0}^{t} e^{-\tau} \cos \tau d \tau \right) = \frac{ \mathscr{L} (e^{-\tau} \cos \tau ) }{s} = \frac{\frac{s+1}{(s+1)^2+1}}{s} = \frac{1}{s}\left (\frac{s+1}{(s+1)^2+1} \right).$$

In general, we have $$\mathscr{L} \left( \int_{0}^{t} f(\tau) d\tau \right) = \frac{\mathscr{L} ( f(\tau))}{s} .$$

• Ah! We just treat the product of the functions as one function, since they're in terms of only one variable. Thus, the identity in your first step didn't occur to me. Thanks for the quick reply! EDIT: Will up-vote when I have sufficient rep. – eenblam Apr 23 '14 at 21:02
• Let me edit for the general formula. – IAmNoOne Apr 23 '14 at 21:02
• Ha, I was just doing the same. I'll remove to allow for your answer to be self-contained. Thanks again! – eenblam Apr 23 '14 at 21:06