Question

Using integration by parts, find the Laplace transform of the given function; \(n\) is a positive integer and \(a\) is a real constant. $$ t e^{a t} $$

Short answer

Question: Determine the Laplace transform of the function \(te^{at}\) when \(s > a\). Answer: The Laplace transform of the given function is \(\mathcal{L}\{t e^{at}\} = \frac{1}{(a-s)^2}\) when \(s > a\).

Step-by-step solution

Integration by parts requires us to choose \(u\) and \(dv\) functions. In this case, let \(u = t\) and \(dv = e^{(a-s)t}dt\). #Step 2: Find the derivatives and integrals of the chosen functions#

We need to compute \(du\) and \(v\). So, \(du = dt\) and to find \(v\), integrate \(dv = e^{(a-s)t}dt\). We get \(v = \frac{1}{a-s}e^{(a-s)t}\). #Step 3: Apply the integration by parts formula#

The integration by parts formula tells us that: $$ \int u dv = uv - \int v du $$ Applying this to our case, we get: $$ \int_{0}^{\infty} t e^{(a-s)t} dt = \left. t\frac{1}{a-s}e^{(a-s)t} \right|_{0}^{\infty} - \int_{0}^{\infty} \frac{1}{a-s} e^{(a-s)t} dt $$ #Step 4: Evaluate the first term and solve the remaining integral#

Evaluate the limit in the first term: $$ \lim_{t \to \infty} t \frac{1}{a-s} e^{(a-s)t} = 0 $$ This is because \(lim_{t \to \infty} e^{(a-s)t}=0\) when \(s>a\) (as \(t\) goes to infinity, the exponential term goes to 0 if \(s>a\)). So, the first term becomes zero and we are left with: $$ \int_{0}^{\infty} \frac{1}{a-s} e^{(a-s)t} dt $$ Now, integrate the remaining integral: $$ F(s) = - \frac{1}{(a-s)^2}e^{(a-s)t} \Big|_0^{\infty} $$ #Step 5: Apply the limits and simplify the expression#

Applying the limits, we get: $$ F(s) = - \frac{1}{(a-s)^2} ( e^{(a-s)\infty} - e^{(a-s)0} ) $$ Now, \(e^{(a-s)\infty} = 0\) when \(s>a\) and \(e^{(a-s)0} = 1\). Thus, we obtain: $$ F(s) = \frac{1}{(a-s)^2} $$ #Final answer#

The Laplace transform of the given function is: $$ \mathcal{L}\{t e^{at}\} = \frac{1}{(a-s)^2} $$

Key concepts

Integration by Parts

Integration by parts is a helpful technique in calculus that simplifies the evaluation of integrals where the integrands are products of two functions. The formula for integration by parts stems from the product rule for differentiation and is written as:\[ \int u \, dv = uv - \int v \, du \] Here's how it works:

• Choose which parts of your integral function will become \( u \) and \( dv \). In our example, \( u = t \) and \( dv = e^{(a-s)t} dt \).
• Differentiating \( u \) gives \( du = dt \).
• Integrating \( dv \) to find \( v \), we obtain \( v = \frac{1}{a-s}e^{(a-s)t} \).
• Substitute into the integration by parts formula and simplify.

This method helps transform a complex integral into one or more simpler ones, making it easier to solve.

Exponential Functions

Exponential functions are mathematical expressions where a constant base is raised to a variable exponent. These functions are widely used in various scientific fields due to their growth properties.

The general form is \( f(t) = e^{at} \), where \( e \) is Euler's number, approximately 2.71828, and \( a \) is a constant that affects the rate of growth or decay.

• In the given problem, the term \( e^{(a-s)t} \) plays a crucial role in transforming the integral because it dictates how quickly the function approaches zero as \( t \) increases.
• When \( a-s < 0 \), this results in exponential decay, meaning the function will approach zero, simplifying the limits evaluation.

Exponential functions have the unique property that their derivative is proportional to the function itself, which helps in solving differential equations and, as seen here, in integration problems.

Laplace Transform Properties

The Laplace Transform is a powerful integral transform used to switch a function from the time domain to the frequency domain. This is especially useful in simplifying complex linear differential equations.

Several properties make the Laplace transform particularly useful:

• Linearity: The Laplace transform is linear, meaning that \( \mathcal{L}\{af(t) + bg(t)\} = a\mathcal{L}\{f(t)\} + b\mathcal{L}\{g(t)\} \) for any two functions \( f \) and \( g \), and constants \( a \) and \( b \).
• Shifting: Shifts in the time domain correspond to multiplication by an exponential in the Laplace domain. For example, \( \mathcal{L}\{e^{at}f(t)\} = F(s-a) \).
• Convolution: Used for the solution of linear differential equations, the Laplace transform of a convolution is the product of the Laplace transforms.

In our problem, after using integration by parts, we apply the property of limits by taking the transforming function \( e^{(a-s)t} \) where it neatly eliminates terms due to its decay, leaving a simple rational Laplace function as the result:\[ \mathcal{L}\{t e^{at}\} = \frac{1}{(a-s)^2} \] Understanding these properties allows us to effectively manipulate and solve related equations.