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As a result of a former question, I am still confused about the following problem:

Deduce(!) (without using direct integration) the value of the following integral:

$$\int_{0}^{t} \tau \sin2(t-\tau) d\tau$$

by using the Laplace transform of the convolution integral. (Hint from Mark Viola).

So starting of: $$\mathcal{L}\bigg[\int_{0}^{t} \tau \sin2(t-\tau) d\tau\bigg]$$

$$\mathcal{L}[\sin(2t)]\mathcal{L}(f(t))$$

But I'am not sure what I am writing is correct. Perhaps somebody can help me further. (I know the answer by using integration by parts, but that is not allowed:-)

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  • $\begingroup$ What would $f(t)$ be? $\endgroup$ – Martin Argerami Mar 10 '18 at 19:08
  • $\begingroup$ I thought that $f(t)$ is the remainder of the convolution. But like I said, I don't know whether what I write is correct. I was a first thought. $\endgroup$ – Joe Goldiamond Mar 10 '18 at 19:17
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What you did was correct, but you introduced a "mysterious" $f(t)$ out of the blue. In general, you have $$ \mathcal L\left[\int_0^t f(\tau)\,g(t-\tau)\,d\tau\right]=\mathcal L[f]\,\mathcal L[g]. $$ Here, $f(t)=t$, and $g(t)=\sin 2t$.

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  • $\begingroup$ So I can write: $\mathcal{L}[t] = \frac{1}{s^2}$ and $\mathcal{L}[sin 2t]=\frac{2}{s^2 +4}$? $\endgroup$ – Joe Goldiamond Mar 10 '18 at 21:23
  • $\begingroup$ Yes. $\ \ \ \ \ $ $\endgroup$ – Martin Argerami Mar 10 '18 at 21:34

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