Question

Show that $$\mathscr{L}\left\\{e^{-a t} f(t)\right\\}=F(s+a)$$

Short answer

The Laplace transform of \( e^{-at}f(t) \) is \( F(s+a) \).

Step-by-step solution

Understand the Laplace Transform

The Laplace transform of a function \( f(t) \) is given by \( \, \mathscr{L}\{f(t)\} = \int_{0}^{\infty} e^{-st} f(t) \, dt \, \). Our goal is to show that for the function \( g(t) = e^{-at}f(t) \), the Laplace transform would be \( F(s+a) \), where \( F(s) = \mathscr{L}\{f(t)\} \).

Apply the Laplace Transform to g(t)

Substitute \( g(t) = e^{-at}f(t) \) in the Laplace transform definition: \[ \mathscr{L}\{e^{-at}f(t)\} = \int_{0}^{\infty} e^{-st} e^{-at} f(t) \, dt = \int_{0}^{\infty} e^{-(s+a)t} f(t) \, dt \]

Simplify the Integrand

Notice that the exponential terms \( e^{-st} \) and \( e^{-at} \) combine into a single exponential term: \[ e^{-(s+a)t} \]. This simplifies our integral to:\[ \mathscr{L}\{e^{-at}f(t)\} = \int_{0}^{\infty} e^{-(s+a)t} f(t) \, dt \] which is identical to the Laplace transform of \( f(t) \) but evaluated at \( s+a \) instead of \( s \).

Recognize the Transformed Function

The integrand \( e^{-(s+a)t} f(t) \) implies that the transform is simply \( F(s+a) \). Therefore, the Laplace transform of \( g(t) = e^{-at}f(t) \) is:\[ \mathscr{L}\{e^{-at}f(t)\} = F(s+a) \] where \( F(s) = \mathscr{L}\{f(t)\} \).

Key concepts

Exponential functions

Exponential functions are mathematical expressions in which a constant base is raised to a variable exponent.They take the form of \( e^{x} \), where \( e \) is the base.This base, approximately equal to 2.718, is known as Euler's number and is a crucial constant in mathematics.
Exponential functions are profoundly important in various fields of science and engineering for modeling growth and decay processes. They can represent phenomena that increase rapidly or decrease rapidly over time.
These functions have a specific property where the rate of growth or decay is proportional to the current amount. For instance, in the equation \( e^{-at} \), \( a \) is a constant that determines how fast the function decreases. When used within the context of Laplace transforms, they help shift functions in the frequency domain, allowing simplification in solving differential equations.

Integral calculus

Integral calculus deals with the accumulation of quantities, such as areas under a curve and volumes. At its core, it involves determining the integral, which mathematically represents the area under a function on a graph.
The definite integral is often represented as \( \int_{a}^{b} f(t) \, dt \), which computes the accumulation between \( a \) and \( b \). This plays a significant role in the Laplace transform calculation, as seen in the formula: \( \mathscr{L}\{f(t)\} = \int_{0}^{\infty} e^{-st} f(t) \, dt \).
Here, the integral from 0 to infinity captures the entire effect of the function \( f(t) \) over time, weighted by the exponential term \( e^{-st} \). The idea is to transform a time-based function into a frequency-based representation. This helps simplify complex differential equations by converting them into algebraic equations.

Laplace transform shifting property

The Laplace transform shifting property is a powerful tool that helps in solving differential equations involving exponential functions. The property is formally expressed as \( \mathscr{L}\{e^{-at} f(t)\} = F(s+a) \). In simpler terms, it states that multiplying the time function by \( e^{-at} \) in the time domain shifts the function by \( a \) units in the frequency domain.
This property allows engineers and scientists to adjust a function's position in the 's' domain easily. By adjusting the parameter \( a \), different frequency components of the function can be analyzed or solved.
Understanding this property is crucial as it helps simplify integral expressions into forms that are easier to handle. This is especially useful in control systems and signal processing, where managing and shifting signals is a foundational task. Applying this property also facilitates solving linear differential equations by changing complex time functions into simpler algebraic forms.