The problem asks to find the Laplace transform of the given piecewise function $f(t)$, where:

• $f(t) = t$ for $0 < t < \tau$
• $f(t) = 1$ for $t > \tau$

Let us compute the Laplace transform $\mathcal{L}\{f(t)\}$, defined as:

$\mathcal{L}\{f(t)\} = \int_0^\infty f(t) e^{-st} dt$

1. Split the Laplace Transform Based on $f(t)$

We divide the Laplace transform into two parts based on the piecewise definition of $f(t)$:

$\mathcal{L}\{f(t)\} = \int_0^\tau t e^{-st} dt + \int_\tau^\infty 1 \cdot e^{-st} dt$

2. Compute Each Part

(a) First Integral: $\int_0^\tau t e^{-st} dt$

Using integration by parts:

• Let $u = t$, so $du = dt$
• Let $dv = e^{-st} dt$, so $v = \frac{e^{-st}}{-s}$

$\int t e^{-st} dt = \left[ \frac{-t e^{-st}}{s} \right]_0^\tau + \int \frac{e^{-st}}{s} dt$

Now compute: $\int \frac{e^{-st}}{s} dt = \frac{-e^{-st}}{s^2}$

So: $\int_0^\tau t e^{-st} dt = \left[ \frac{-t e^{-st}}{s} - \frac{e^{-st}}{s^2} \right]_0^\tau$ Evaluate the limits:

• At $t = \tau$: $\frac{-\tau e^{-s\tau}}{s} - \frac{e^{-s\tau}}{s^2}$
• At $t = 0$: $0 - \frac{1}{s^2}$

Thus: $\int_0^\tau t e^{-st} dt = \frac{-\tau e^{-s\tau}}{s} - \frac{e^{-s\tau}}{s^2} + \frac{1}{s^2}$

(b) Second Integral: $\int_\tau^\infty e^{-st} dt$

This is a simple exponential integral: $\int_\tau^\infty e^{-st} dt = \left[ \frac{-e^{-st}}{s} \right]_\tau^\infty$ Evaluate the limits:

• At $t = \infty$: $0$
• At $t = \tau$: $\frac{-e^{-s\tau}}{s}$

Thus: $\int_\tau^\infty e^{-st} dt = \frac{e^{-s\tau}}{s}$

3. Combine Results

Now combine the results from both parts: $\mathcal{L}\{f(t)\} = \left( \frac{-\tau e^{-s\tau}}{s} - \frac{e^{-s\tau}}{s^2} + \frac{1}{s^2} \right) + \frac{e^{-s\tau}}{s}$

Simplify: $\mathcal{L}\{f(t)\} = \frac{-\tau e^{-s\tau}}{s} + \frac{1}{s^2}$

Thus, the Laplace transform of $f(t)$ is: $\boxed{\mathcal{L}\{f(t)\} = \frac{-\tau e^{-s\tau}}{s} + \frac{1}{s^2}}$

Let me know if you'd like further steps or clarifications! Here are some related questions:

1. What is the physical interpretation of the Laplace transform in this context?
2. How does the Laplace transform simplify differential equation solving?
3. What happens to the Laplace transform if $\tau \to \infty$?
4. How would $f(t)$ be represented in the frequency domain?
5. Can this result be used for convolution integrals?

Tip: Use properties of the Laplace transform (linearity, shifts, etc.) to speed up solving similar problems.