Laplace Transform using t-shift (second shift)

Question

$$f(t) = tu(t-π)$$

I know I have to get t in terms of $$(t-π)$$ and to do that I have done

$$ t = a(t-π) + b$$ $$ t = at-aπ + b$$ $$ t = (a-π)t + b$$ $$ (a-π) = 1$$ and $$b = 0$$

Then I think I have done the write thing and being to t-shift with

$$ f(t) = (a-π)(t-π)u(t-π)$$ but when I apply the t-shift theorem I get lost. Is this the write method? have I done it right?

Answer 1

Assuming that $*$ doesn't mean convolution in your original statement, note that $$tu(t-\pi) = (t-\pi + \pi)u(t-\pi) = (t-\pi)u(t-\pi) + \pi u(t-\pi)$$ Thus,

$$\mathcal{L}\{tu(t-\pi)\} = \mathcal{L}\{(t-\pi)u(t-\pi)\} + \pi\mathcal{L}\{u(t-\pi)\} = \ldots$$