Please provide steps to solve this integral:
$$3\int^t_0{\sin{u}(t-u)e^{-(t-u)}du}.$$
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Mhenni Benghorbal
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user1211030
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1What progress have you made? – Peter Woolfitt Jul 01 '14 at 21:42
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None. I fail at the integral having 3 factors and none of u' or v' for partial integration disappear – user1211030 Jul 01 '14 at 21:44
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This is the convolution of the two function $\sin t$ and $t e^t$. – Mhenni Benghorbal Jul 01 '14 at 21:46
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@MhenniBenghorbal correct. But how do we solve it? – user1211030 Jul 01 '14 at 21:48
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@user1211030: You can use integration techniques. – Mhenni Benghorbal Jul 01 '14 at 21:52
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@MhenniBenghorbal That's a great suggestion, thank you – user1211030 Jul 01 '14 at 21:53
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@user1211030: I post an answer. – Mhenni Benghorbal Jul 01 '14 at 21:59
1 Answers
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Here is a start.
$$ f(t) = 3\int^t_0{\sin{u}(t-u)e^{-(t-u)}du} = 3e^{-t}\int^t_0{\sin{u}(t-u)e^{u}du} $$
$$ = 3te^{-t} \int^t_0{\sin{u}\,e^{u}du} - 3e^{-t} \int^t_0{u\sin{u}\,e^{u}du}. $$
To make it easier to evaluate the last two integrals use the identity
$$ \sin u = \frac{e^{iu}-e^{-iu}}{2i} $$
Another approach:
We can use the fact
$$ \mathcal{L} (g*h) = \mathcal{L}(g) \mathcal{L}(h).$$
Taking the Laplace of
$$ f(t) = 3\int^t_0{\sin{u}(t-u)e^{-(t-u)}du} $$
gives
$$ F(s) = \frac{1}{(s^2+1)(s-1)^2}. $$
Using partial fraction we can have the form
$$ - \frac{1}{2(s-1)}+\frac{1}{2(s-1)^2}+\frac{1}{4(s-i)}+\frac{1}{4(s+i)}.$$
Now you can use the tables of Laplace transform. I think you can finish the problem.
Mhenni Benghorbal
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