Question

As in Example 2, use the definition to find the Laplace transform for \(f(t)\), if it exists. In each exercise, the given function \(f(t)\) is defined on the interval \(0 \leq t<\infty\). If the Laplace transform exists, give the domain of \(F(s)\). In Exercises 9-12, also sketch the graph of \(f(t)\). $$ f(t)=t e^{-t} $$

Short answer

Question: Determine the Laplace transform of the function \(f(t) = t e^{-t}\) for \(0 \leq t < \infty\). Answer: The Laplace transform of the given function is \(F(s) = \frac{1}{(s+1)^2}\) with the domain \(s > -1\).

Step-by-step solution

Write down the definition of the Laplace Transform

The Laplace transform of a given function \(f(t)\) is defined as: $$ F(s) = \int_{0}^{\infty} f(t)e^{-st} dt $$

Substitute \(f(t)\) into the definition

Now, we will substitute our given function \(f(t) = te^{-t}\) into this definition: $$ F(s) = \int_{0}^{\infty} (te^{-t})e^{-st} dt $$

Merge the exponential terms

To simplify the integral, merge the exponential terms \(e^{-t}\) and \(e^{-st}\): $$ F(s) = \int_{0}^{\infty} te^{-(s+1)t} dt $$

Apply integration by parts

Now, we will apply integration by parts to solve this integral. We can match the terms as follows: $$ u = t, \quad dv = e^{-(s+1)t} dt \\ du = dt,\quad v = \frac{1}{-(s+1)}e^{-(s+1)t} $$ Now, we plug these into the integration by parts formula: \(\int u dv=uv-\int v du\). $$ F(s) = \left. -\frac{t}{s+1} e^{-(s+1)t}\right|_{0}^{\infty} + \int_{0}^{\infty} \frac{1}{s+1} e^{-(s+1)t} dt $$

Evaluate the first term and solve the remaining integral

To determine the first term, first consider the limit as \(t \to \infty\). As \(e^{-(s+1)t}\) tends to 0 when \(s > -1\) and the first term is \(t=\infty \times 0\), which is an indeterminate form, we use L'Hôpital's Rule as follows: $$ \lim_{t \to \infty} \frac{t}{e^{(s+1)t}} = \lim_{t \to \infty} \frac{1}{(s+1)e^{(s+1)t}} = 0, \quad if \; s > -1 $$ The second part can be easily integrated as we've already solved a similar integral: $$ \int_{0}^{\infty} \frac{1}{s+1} e^{-(s+1)t} dt = \left. -\frac{1}{(s+1)^2}e^{-(s+1)t}\right|_{0}^{\infty} $$ Now, we add both terms together and substitute the limits of integration: $$ F(s) = 0 - \left(-\frac{1}{(s+1)^2}e^{-(s+1)t}\right)\Big|_{0}^{\infty} = \frac{1}{(s+1)^2} $$

Find the domain of \(F(s)\)

As \(s > -1\), we have convergence, and thus, the domain of \(F(s)\) is \(s > -1\). And our final answer is: $$ F(s) = \frac{1}{(s+1)^2}, \quad s > -1 $$

Key concepts

Integration by Parts

Integration by parts is a technique derived from the product rule of differentiation. It's particularly useful when dealing with integrals of products of functions like the one in the given Laplace transform problem. The formula for integration by parts is:\[ \int u \ dv = uv - \int v \ du \]This method works by selecting parts of the integral for \( u \) and the differential \( dv \) based on which components can be differentiated and integrated easily:

• For \( u \), choose a function that simplifies upon differentiation.
• For \( dv \), select a part that is easy to integrate.

In our example, we chose \( u = t \) and \( dv = e^{-(s+1)t} dt \). As soon as these are chosen, differentiate \( u \) to get \( du = dt \) and integrate \( dv \) to get \( v = \frac{1}{-(s+1)}e^{-(s+1)t} \).

Applying this strategy simplifies otherwise complicated integrals into more manageable ones. Remember to evaluate \( uv \) at the limits of integration to complete the step.

Domain of Transforms

The domain of a Laplace transform tells us for what values of \( s \) the transformed function \( F(s) \) is valid and convergent. Ensuring convergence is key to solving Laplace transforms correctly. In our example, \( f(t) = t e^{-t} \), the parameter \( s \) must satisfy a condition derived from the behavior of exponential terms.
The given function results in the formation of \( te^{-(s+1)t} \), which means we are primarily concerned about the term \( e^{-(s+1)t} \). To ensure convergence of the integral from \( 0 \) to \( \infty \), the exponential expression must tend to zero as \( t \rightarrow \infty \).

• For this, \( s+1 \) must be positive, leading to the condition \( s > -1 \).

Therefore, the domain of \( F(s) = \frac{1}{(s+1)^2} \) is \( s > -1 \), ensuring that our transformed function is both convergent and valid within this range.

Exponential Functions

Exponential functions, especially in the context of Laplace transforms, often indicate growth or decay dynamics based on their power (i.e., \( e^{at} \)). They are crucial due to their integral and differential properties which make them easy to handle mathematically.

In our Laplace example, the function \( f(t) = t e^{-t} \) involves an exponential decay term \( e^{-t} \). When combined with another exponential term \( e^{-st} \), it forms a more complex term \( e^{-(s+1)t} \). This compound exponent influences both the behavior and the domain of the Laplace transform.

• Exponential growth occurs when the exponent is positive, i.e., \( e^{at} \) where \( a > 0 \).
• Exponential decay, like \( e^{-t} \), indicates a rapid reduction as a variable (like time \( t \)) increases.

The fact that exponential functions tend towards infinity or zero so sharply and predictably makes them highly valuable tools in mathematical calculations, such as integration or transformation, where behaviors at infinity play a critical role.