The second shift theorem states that if \(\mathcal{L}\\{f(t)\\}=F(s)\) then \(\mathcal{L}\\{u(t-d) f(t-d)\\}=\mathrm{e}^{-s d} F(s), \quad d>0\) where \(u(t)\) is the unit step function. (a) Prove this theorem. (b) Find the Laplace transform of \(u(t-3)(t-3)^{5}\) (c) Find the inverse Laplace transform of $$ \frac{\mathrm{e}^{-4 s} 4 !}{s^{5}} $$

In order to prove the second shift theorem, we have to show the relationship \(\mathcal{L}\\{u(t-d) f(t-d)\\}=\mathrm{e}^{-s d} F(s)\). Starting with the definition of the Laplace transform for the function \(u(t-d)f(t-d)\): $$ \mathcal{L}\\{u(t-d) f(t-d)\\} = \int_{0}^{\infty} u(t-d)f(t-d) e^{-st} dt $$ Now consider two cases for the integration. If \(t < d\), then \(u(t-d) = 0\) (since the unit step function is equal to zero for negative arguments). If \(t ≥ d\), then \(u(t-d) = 1\). So, we can rewrite the integral: $$ \mathcal{L}\\{u(t-d) f(t-d)\\} = \int_{d}^{\infty} f(t-d) e^{-st} dt $$ To simplify the integral, we can substitute \(t-d = x\). This gives \(t = x + d\), and \(dt = dx\). The new limits of integration become \(0\) and \(\infty\): $$ \mathcal{L}\\{u(t-d) f(t-d)\\} = \int_{0}^{\infty} f(x) e^{-s(x + d)} dx $$ Now, we can separate the exponential function into two factors: $$ \mathcal{L}\\{u(t-d) f(t-d)\\} = \int_{0}^{\infty} f(x) \mathrm{e}^{-sx} \mathrm{e}^{-sd} dx $$ Since \(\mathrm{e}^{-sd}\) does not depend on \(x\), we can move it outside the integral: $$ \mathcal{L}\\{u(t-d) f(t-d)\\} = \mathrm{e}^{-sd} \int_{0}^{\infty} f(x) \mathrm{e}^{-sx} dx $$ Recall that \(\mathcal{L}\\{f(t)\\}=F(s)\). By changing back to the variable \(t\), we get: $$ \mathcal{L}\\{u(t-d) f(t-d)\\} = \mathrm{e}^{-sd} F(s) $$ This proves the second shift theorem.

To find the Laplace transform of the given function, we can use the second shift theorem with \(f(t) = t^5\) and \(d = 3\). The Laplace transform of \(t^5\) is: $$ F(s) = \mathcal{L}\\{t^5\\} = \frac{5!}{s^6} $$ So, using the theorem, we get: $$ \mathcal{L}\\{u(t-3)(t-3)^{5}\\} = \mathrm{e}^{-3s} \frac{5!}{s^6} $$

To find the inverse Laplace transform of the given expression, we can again use the second shift theorem, but this time in reverse. First, recognize that the given expression has the form: $$ \mathrm{e}^{-4s}\frac{4!}{s^5} = \mathrm{e}^{-4s}F(s) $$ We need to find a function \(f(t)\) such that \(F(s) = \frac{4!}{s^5}\). Since we know the Laplace transform of \(t^4\) is \(\frac{4!}{s^5}\), we have: $$ f(t) = t^4 $$ Now, we can use the second shift theorem to find the inverse Laplace transform of the given expression: $$ \mathcal{L}^{-1}\\{\mathrm{e}^{-4s}\frac{4!}{s^5}\\} = u(t-4)(t-4)^4 $$ This is the inverse Laplace transform of the given expression.