Question

Find the Laplace transforms of the following functions by evaluating the integral \(F(s)=\int_{0}^{\infty} e^{-s t} f(t) d t\). (a) \(t\) (b) \(t e^{-t}\) (c) \(\sinh b t\) (d) \(e^{2 t}-3 e^{t}\) (e) \(t^{2}\)

Short answer

#tag_title#Function (f) - cos(kt)#tag_content#For this function, \(f(t) = \cos(kt)\). Therefore, the integral becomes: \(F(s) = \int_0^\infty e^{-st}\cos(kt) dt\) To solve this integral, we can use integration by parts. First, let \(u = \cos(kt)\) and \(dv = e^{-st} dt\). Then, \(du = -k\sin(kt) dt\) and \(v = -\frac{1}{s}e^{-st}\). Now, apply integration by parts: \(F(s) = \left[-\frac{1}{s}e^{-st}\cos(kt)\right]_0^\infty + \frac{k}{s}\int_0^\infty e^{-st}\sin(kt) dt\) As \(t \rightarrow \infty\), \(-\frac{e^{-st}\cos(kt)}{s}\) approaches \(\lim_{t \to \infty}[-\frac{1}{s}e^{-st}\cos(kt)] = 0\). Now, we need to deal with the second integral. Let \(u = \sin(kt)\) and \(dv = e^{-st} dt\), so that \(du = k\cos(kt) dt\) and \(v = -\frac{1}{s}e^{-st}\). Then, apply integration by parts again: \(\frac{k}{s}\int_0^\infty e^{-st}\sin(kt) dt = \frac{k}{s}\left[-\frac{1}{s}e^{-st}\sin(kt)\right]_0^\infty - \frac{k^2}{s}\int_0^\infty e^{-st}\cos(kt) dt\) As \(t \rightarrow \infty\), the limit of the first term approaches 0: \(\lim_{t \to \infty}[-\frac{k}{s^2}e^{-st}\sin(kt)] = 0\). Now, we have the following equation for \(F(s)\): \(F(s) = 0 + \frac{k^2}{s}\int_0^\infty e^{-st}\cos(kt) dt\) This equation represents an integral equation for \(F(s)\). To obtain the Laplace transform, we can multiply both sides of the equation by \(s\) and then solve for \(F(s)\): \(sF(s) = k^2 \int_0^\infty e^{-st}\cos(kt) dt\) Rearranging, \(\int_0^\infty e^{-st}\cos(kt) dt = \frac{s}{k^2}F(s)\) Substituting the original expression for \(F(s)\) back into this equation, \(\frac{s}{k^2}\left[\left(-\frac{1}{s}e^{-st}\cos(kt)\right]_0^\infty - \left(-\frac{k^2}{s}\right)\int_0^\infty e^{-st}\cos(kt) dt\right] = -\frac{k^2}{s^2}\int_0^\infty e^{-st}\cos(kt) dt\) Notice that both sides of the equation are equal. Therefore, we have found a solution to the integral equation for \(F(s)\): \(F(s) = \frac{s}{s^2 + k^2}\) So, the Laplace Transform \(F(s) = \frac{s}{s^2 + k^2}\).

Step-by-step solution

Function (a) - t

For this function, \(f(t) = t\). Therefore, the integral becomes: \(F(s) = \int_0^\infty e^{-st}t dt\). To solve this integral, we can use integration by parts, with \(u=t\) and \(dv=e^{-st} dt\). First, we find \(du = dt\) and \(v = -\frac{1}{s}e^{-st}\). Now, apply integration by parts: \( F(s) = \left[-\frac{1}{s}e^{-st}t\right]_0^\infty + \frac{1}{s}\int_0^\infty e^{-st} dt \) As \(t \rightarrow \infty\), \(te^{-st}\) approaches \(0\). The first term, therefore, evaluates to \(\lim_{t \to \infty}[-\frac{1}{s}e^{-st}t] = 0\). The second term, we compute by integrating the exponent: \( \frac{1}{s}\int_0^\infty e^{-st} dt = \frac{1}{s}\left[\frac{-1}{s}e^{-st}\right]_0^\infty = \frac{1}{s^2}\left[1-e^{-\infty}\right] = \frac{1}{s^2} \) So, the Laplace Transform \(F(s) = \frac{1}{s^2}\).

Function (b) - t*e^(-t)

For this function, \(f(t) = te^{-t}\). Therefore, the integral becomes: \(F(s) = \int_0^\infty e^{-st}(te^{-t}) dt\). First, we can simplify the integral: \(F(s) = \int_0^\infty te^{-(s+1)t} dt\). Now, again use integration by parts, with \(u=t\) and \(dv=e^{-(s+1)t} dt\). First, we find \(du = dt\) and \(v = -\frac{1}{s+1}e^{-(s+1)t}\). Now, apply the integration by parts: \( F(s) = \left[-\frac{1}{s+1}e^{-(s+1)t}t\right]_0^\infty + \frac{1}{s+1}\int_0^\infty e^{-(s+1)t} dt \) As \(t \rightarrow \infty\), \(-t e^{-(s+1)t}\) approaches 0 for any \(s>-1\). Thus, \(\lim_{t \to \infty}\left[-\frac{1}{s+1}e^{-(s+1)t}t\right] = 0\). For the integral part, we compute by integrating the exponent: \(\frac{1}{s+1}\int_0^\infty e^{-(s+1)t} dt = \frac{1}{s+1}\left[\frac{-1}{s+1}e^{-(s+1)t}\right]_0^\infty = \frac{1}{(s+1)^2}\left[1-e^{-\infty}\right] = \frac{1}{(s+1)^2}\) Thus, the Laplace Transform \(F(s) = \frac{1}{(s+1)^2}\). +

Function (c) - sinh(bt)

For this function, \(f(t) = \sinh(bt)\). Recall that \(\sinh(bt) = \frac{1}{2}\left(e^{bt}-e^{-bt}\right)\). Therefore, the integral becomes: \(F(s)=\int_0^\infty e^{-st}\left(\frac{1}{2}\left(e^{bt}-e^{-bt}\right)\right) dt\) Now, split the integral into two parts: \(F(s) = \frac{1}{2}\left(\int_0^\infty e^{-st}e^{bt} dt - \int_0^\infty e^{-st}e^{-bt} dt\right)\) Simplify the integrals: \(F(s) = \frac{1}{2}\left(\int_0^\infty e^{(b-s)t} dt - \int_0^\infty e^{(-b-s)t} dt\right)\) Now, integrate the exponentials: \(F(s) = \frac{1}{2}\left[\frac{1}{b-s}\left(e^{(b-s)t}\right)_0^\infty - \frac{1}{-b-s}\left(e^{(-b-s)t}\right)_0^\infty\right]\) As \(t \rightarrow \infty\), both exponential terms in the brackets approach 0. Thus, the Laplace Transform \(F(s) = \frac{1}{2}\left(\frac{-1}{b-s} - \frac{1}{-b-s}\right)\), which simplifies to \(F(s) = \frac{1}{2}\left(\frac{s}{s^2-b^2}\right)\).

Function (d) - e^(2t) - 3*e^(t)

For this function, \(f(t) = e^{2t}-3e^t\). Now, split the integral into two parts: \(F(s) = \int_0^\infty e^{-st}\left(e^{2t}-3e^t\right) dt = \int_0^\infty e^{-st}e^{2t} dt - \int_0^\infty 3 e^{-st}e^t dt\) Simplify the integrals: \(F(s) = \int_0^\infty e^{(2-s)t} dt - 3\int_0^\infty e^{(1-s)t} dt\) Now, integrate the exponentials: \(F(s) = \frac{1}{2-s}\left(e^{(2-s)t}\right)_0^\infty - 3\frac{1}{1-s}\left(e^{(1-s)t}\right)_0^\infty\) Both exponential terms in the brackets approach 0 as \(t \rightarrow \infty\). Therefore: \(F(s) = \frac{1}{s-2} - \frac{3}{s-1}\) is the Laplace Transform of the function e^(2t) - 3*e^(t).

Function (e) - t^2

For this function, \(f(t) = t^2\). Therefore, the integral becomes: \(F(s) = \int_0^\infty e^{-st}t^2 dt\) To solve this integral, we can apply integration by parts two times. First, let \(u = t^2\) and \(dv = e^{-st}dt\). Then \(du = 2t dt\) and \(v = -\frac{1}{s}e^{-st}\). Now, apply integration by parts once: \(F(s)=\left[-\frac{1}{s}e^{-st}t^2\right]_0^\infty + \frac{1}{s}\int_0^\infty e^{-st}(2t) dt\) As \(t \rightarrow \infty\), \(-t^2e^{-st}\) approaches \(0\). For this second integral, again, let \(u = 2t\), \(dv = e^{-st}dt\), so that \(du = 2dt\) and \(v = -\frac{1}{s}e^{-st}\). Now, apply integration by parts again: \(F(s) = 0 + \frac{1}{s}\left[\frac{-2}{s}e^{-st}t\right]_0^\infty - \frac{2}{s}\int_0^\infty e^{-st}(1) dt\) As \(t \rightarrow \infty\), \(-2te^{-st}\) approaches \(0\) for any \(s>0\). Then, integrate the remaining exponential term: \(F(s) = -\frac{2}{s^2}\left[e^{-st}\right]_0^\infty = -\frac{2}{s^2}\left[1-e^{-\infty}\right] = \frac{2}{s^3}\) So, the Laplace Transform \(F(s) = \frac{2}{s^3}\).

Key concepts

Integration by Parts

Integration by parts is a powerful technique used to solve integrals that are products of functions. It is particularly useful for solving Laplace transforms, which often include such functions. The fundamental principle behind integration by parts is derived from the product rule of differentiation.
The formula is given by:

• \( \int u \, dv = uv - \int v \, du \)

Here, the choice of \( u \) and \( dv \) is crucial, and it is generally recommended to choose \( u \) as a polynomial or logarithmic part if present, and \( dv \) as the other part of the function, especially if it's an exponential.
When applying integration by parts, it's essential to correctly calculate the differentials \( du \) and \( v \). Depending on the complexity of the original function, you may need to use the method more than once, as demonstrated with \( t^2 \) in the original exercise. You start by setting \( u = t^2 \) and break it down further with repeated use of integration by parts until the function is fully integrated.

Exponential Functions

Exponential functions are functions of the form \( f(t) = a e^{kt} \), where \( e \) is the base of the natural logarithm, \( a \) is a constant, and \( k \) is a parameter that dictates the rate of growth or decay. They appear frequently in the context of Laplace transforms due to their behavior under integration.
When dealing with Laplace transforms, the properties of exponential functions can significantly simplify integration. For example, integrating \( e^{-(s+t)}t \) involves manipulating exponents, which leads to more manageable forms after integration by parts is performed. The existing constants and coefficients in these functions, such as \( s \) and \( t \), contribute to the ultimate computation of the Laplace transform.
It's worth noting the integral of an exponential function over \( [0, \infty) \) approaches 0 as \( t \to \infty \), providing crucial boundary conditions for many transforms. Consider, for instance, how the function \( e^{-st} \) naturally diminishes to zero, which aids in resolving improper integrals.

Improper Integrals

Improper integrals involve integrals of functions over an infinite interval or those involving a singularity. In the context of Laplace transforms, these are critical, as they often integrate over the range \( [0, \infty) \). To solve these integrals, especially with functions like exponential decay, one utilizes the fact that the integrals converge as \( t \rightarrow \infty \).
The Laplace transform itself is defined as:

• \( \int_0^\infty e^{-st}f(t) dt \)

In the original exercise, the convergence of the integral is assured due to the factor \( e^{-st} \), which tends to zero as \( t \to \infty \) for \( s > 0 \). This decay holds all the components in check, making it possible to evaluate integrals with finite values.
You must be aware that while evaluating, terms that ostensibly would contribute at infinity (like polynomials in exponentials) are managed, often shown via limits, to curb their impact. For example, \( te^{-st} \) and \( t^2e^{-st} \) approach 0 as \( t \rightarrow \infty \), ensuring the integral is not divergent.

Hyperbolic Functions

Hyperbolic functions, such as \( \sinh(x) \) and \( \cosh(x) \), are analogs to trigonometric functions but are based on exponential functions. They are defined using exponential equations:

• \( \sinh(x) = \frac{e^x - e^{-x}}{2} \)
• \( \cosh(x) = \frac{e^x + e^{-x}}{2} \)

These functions are crucial in solving Laplace transforms involving hyperbolic terms.
For example, consider calculating the Laplace transform of \( \sinh(bt) \). It requires breaking down the hyperbolic sine into its exponential components, \( \frac{1}{2}(e^{bt} - e^{-bt}) \), simplifying the integration process. The linearity of the Laplace transform allows for individual computation of each part, making it simpler to manage.
The properties of hyperbolic functions, such as symmetry, differentiate them from trigonometric functions. Unlike \( \sin \) and \( \cos \), \( \sinh \) and \( \cosh \) are unbounded, meaning they grow without bound as the argument increases, further compounding their utility in solving differential equations.