Laplace Transform under the integral

Question

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:-)

Answer 1

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 $g(t)=\sin 2t$.