The problem asks to find the Laplace transform of the given function $f(t)$. Based on the visible content, the function $f(t)$ is defined piecewise as follows:

1. $f(t) = t/\tau$ when $0 < t < \tau$,
2. $f(t) = 1$ when $t > \tau$.

The Laplace transform is given by: $\mathcal{L}\{f(t)\} = \int_0^\infty f(t) e^{-st} \, dt$

Step 1: Split the integral based on the definition of $f(t)$

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

Step 2: Solve the first integral $\int_0^\tau \frac{t}{\tau} e^{-st} \, dt$

Let $I_1 = \int_0^\tau \frac{t}{\tau} e^{-st} \, dt$. Perform integration by parts, where:

• Let $u = \frac{t}{\tau}$ and $dv = e^{-st} \, dt$,
• Then, $du = \frac{1}{\tau} \, dt$ and $v = \frac{-1}{s} e^{-st}$.

Using the integration by parts formula $\int u \, dv = uv - \int v \, du$, we get: $I_1 = \left[ \frac{t}{\tau} \cdot \frac{-1}{s} e^{-st} \right]_0^\tau + \int_0^\tau \frac{1}{\tau} \cdot \frac{1}{s} e^{-st} \, dt$

Simplify the two terms:

1. Evaluate $\left[ \frac{t}{\tau} \cdot \frac{-1}{s} e^{-st} \right]_0^\tau$: $= \frac{-\tau}{\tau s} e^{-s\tau} - 0 = \frac{-1}{s} e^{-s\tau}.$

2. Solve $\int_0^\tau \frac{1}{\tau s} e^{-st} \, dt$: $\int_0^\tau e^{-st} \, dt = \left[ \frac{-1}{s} e^{-st} \right]_0^\tau = \frac{-1}{s} e^{-s\tau} + \frac{1}{s}.$

Substitute these results back into $I_1$: $I_1 = \frac{-1}{s} e^{-s\tau} + \frac{1}{s^2 \tau} \left( 1 - e^{-s\tau} \right).$

Step 3: Solve the second integral $\int_\tau^\infty e^{-st} \, dt$

Let $I_2 = \int_\tau^\infty e^{-st} \, dt$. This is a simple exponential integral: $I_2 = \left[ \frac{-1}{s} e^{-st} \right]_\tau^\infty = 0 - \frac{-1}{s} e^{-s\tau} = \frac{1}{s} e^{-s\tau}.$

Step 4: Combine $I_1$ and $I_2$

The total Laplace transform is: $\mathcal{L}\{f(t)\} = I_1 + I_2 = \left( \frac{-1}{s} e^{-s\tau} + \frac{1}{s^2 \tau} \left( 1 - e^{-s\tau} \right) \right) + \frac{1}{s} e^{-s\tau}.$

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

This is the Laplace transform of $f(t)$.

Let me know if you need a deeper explanation or additional steps clarified.

Related Questions:

1. What is the Laplace transform of a piecewise-defined function in general?
2. How do you solve improper integrals involving exponential functions?
3. Can you explain the properties of the Laplace transform for functions with a time shift?
4. How is integration by parts used in solving Laplace transforms?
5. What are practical applications of Laplace transforms in engineering?

Tip:

Always check the continuity of piecewise functions when finding their Laplace transform to ensure the integral limits are correctly applied.